Linear Algebra - IIT Bombaydey/linearalg04.pdf · Triangularization Closely related to the problem of diagonalization is the problem of triangularization.We shall use this concept

Linear Algebra

Santanu Dey

February 18, 2011

1/58

Spectral Theorem

Similarity does not necessarily preserve the distance.

In terms of matrices, thismay be noticed in the fact that an arbitrary conjugate C−1AC of a Hermitianmatrix may not be Hermitian. Thus the diagonalization problem for specialmatrices such as Hermitian matrices needs a special treatment viz., we need torestrict C to those matrices which preserve the inner product. In this section weshall first establish an important result that says that a Hermitian matrix can bediagonalized using unitary matrices. We shall then see some applications of this togeometry.

ARS (IITB) MA106-Linear Algebra February 14, 2011 68 / 99

Spectral Theorem

Similarity does not necessarily preserve the distance. In terms of matrices, thismay be noticed in the fact that an arbitrary conjugate C−1AC of a Hermitianmatrix may not be Hermitian.

Thus the diagonalization problem for specialmatrices such as Hermitian matrices needs a special treatment viz., we need torestrict C to those matrices which preserve the inner product. In this section weshall first establish an important result that says that a Hermitian matrix can bediagonalized using unitary matrices. We shall then see some applications of this togeometry.


Spectral Theorem

Similarity does not necessarily preserve the distance. In terms of matrices, thismay be noticed in the fact that an arbitrary conjugate C−1AC of a Hermitianmatrix may not be Hermitian. Thus the diagonalization problem for specialmatrices such as Hermitian matrices needs a special treatment viz., we need torestrict C to those matrices which preserve the inner product.

In this section weshall first establish an important result that says that a Hermitian matrix can bediagonalized using unitary matrices. We shall then see some applications of this togeometry.


Spectral Theorem

Similarity does not necessarily preserve the distance. In terms of matrices, thismay be noticed in the fact that an arbitrary conjugate C−1AC of a Hermitianmatrix may not be Hermitian. Thus the diagonalization problem for specialmatrices such as Hermitian matrices needs a special treatment viz., we need torestrict C to those matrices which preserve the inner product. In this section weshall first establish an important result that says that a Hermitian matrix can bediagonalized using unitary matrices.

We shall then see some applications of this togeometry.


Spectral Theorem

Similarity does not necessarily preserve the distance. In terms of matrices, thismay be noticed in the fact that an arbitrary conjugate C−1AC of a Hermitianmatrix may not be Hermitian. Thus the diagonalization problem for specialmatrices such as Hermitian matrices needs a special treatment viz., we need torestrict C to those matrices which preserve the inner product. In this section weshall first establish an important result that says that a Hermitian matrix can bediagonalized using unitary matrices. We shall then see some applications of this togeometry.


Triangularization

Closely related to the problem of diagonalization is the problem oftriangularization. We shall use this concept as a stepping stone toward thesolution of diagonalization problem.

DefinitionTwo n × n matrices A and B are said to be congruent if there exists a unitarymatrix C such that C∗AC = B.

DefinitionWe say A is triangularizable if there exists an invertible matrix C such thatC−1AC is upper triangular.

RemarkObviously all diagonalizable matrices are triangularizable. The following resultsays that triangularizability causes least problem:


Triangularization




RemarkObviously all diagonalizable matrices are triangularizable.

The following resultsays that triangularizability causes least problem:


Triangularization




RemarkObviously all diagonalizable matrices are triangularizable. The following resultsays that triangularizability causes least problem:


Triangularization

Proposition

Over the complex numbers every square matrix is congruent to an uppertriangular matrix.

Proof: Let A be an n × n matrix with complex entries. We have to find aunitary matrix C such that C∗AC is upper triangular. We shall prove this byinduction. For n = 1 there is nothing to prove. Assume the result for n − 1.Let µ be an eigenvalue of A and let v ∈ Cn \ {0} be such that Av1 = µv1.(Here we need to work with complex numbers; real numbers won’t do. why?)


Triangularization

Proposition


Proof: Let A be an n × n matrix with complex entries.

We have to find aunitary matrix C such that C∗AC is upper triangular. We shall prove this byinduction. For n = 1 there is nothing to prove. Assume the result for n − 1.Let µ be an eigenvalue of A and let v ∈ Cn \ {0} be such that Av1 = µv1.(Here we need to work with complex numbers; real numbers won’t do. why?)


Triangularization

Proposition


Proof: Let A be an n × n matrix with complex entries. We have to find aunitary matrix C such that C∗AC is upper triangular.

We shall prove this byinduction. For n = 1 there is nothing to prove. Assume the result for n − 1.Let µ be an eigenvalue of A and let v ∈ Cn \ {0} be such that Av1 = µv1.(Here we need to work with complex numbers; real numbers won’t do. why?)


Triangularization

Proposition


Proof: Let A be an n × n matrix with complex entries. We have to find aunitary matrix C such that C∗AC is upper triangular. We shall prove this byinduction.

For n = 1 there is nothing to prove. Assume the result for n − 1.Let µ be an eigenvalue of A and let v ∈ Cn \ {0} be such that Av1 = µv1.(Here we need to work with complex numbers; real numbers won’t do. why?)


Triangularization

Proposition


Proof: Let A be an n × n matrix with complex entries. We have to find aunitary matrix C such that C∗AC is upper triangular. We shall prove this byinduction. For n = 1 there is nothing to prove.

Assume the result for n − 1.Let µ be an eigenvalue of A and let v ∈ Cn \ {0} be such that Av1 = µv1.(Here we need to work with complex numbers; real numbers won’t do. why?)


Triangularization

Proposition


Proof: Let A be an n × n matrix with complex entries. We have to find aunitary matrix C such that C∗AC is upper triangular. We shall prove this byinduction. For n = 1 there is nothing to prove. Assume the result for n − 1.

Let µ be an eigenvalue of A and let v ∈ Cn \ {0} be such that Av1 = µv1.(Here we need to work with complex numbers; real numbers won’t do. why?)


Triangularization

Proposition


Proof: Let A be an n × n matrix with complex entries. We have to find aunitary matrix C such that C∗AC is upper triangular. We shall prove this byinduction. For n = 1 there is nothing to prove. Assume the result for n − 1.Let µ be an eigenvalue of A and let v ∈ Cn \ {0} be such that Av1 = µv1.

(Here we need to work with complex numbers; real numbers won’t do. why?)


Triangularization

Proposition


Proof: Let A be an n × n matrix with complex entries. We have to find aunitary matrix C such that C∗AC is upper triangular. We shall prove this byinduction. For n = 1 there is nothing to prove. Assume the result for n − 1.Let µ be an eigenvalue of A and let v ∈ Cn \ {0} be such that Av1 = µv1.(Here we need to work with complex numbers; real numbers won’t do. why?)


Triangularization

We can choose v1 to be of norm 1.

We can then complete it to an orthonormalbasis {v1, . . . , vn}. (Here we use Gram-Schmidt.) We then take C1 to be thematrix whose i th column is vi , 1 ≤ i ≤ n. Then as seen earlier, C1 is a unitarymatrix. Put A1 = C−11 AC1. ThenA1e1 = C−11 AC1e1 = C−11 Av1 = µC−11 (v1) = µe1. This shows that the firstcolumn of A1 has all entries zero except the first one which is equal to µ. Let Bbe the matrix obtained from A1 by cutting down the first row and the firstcolumn, so that A1 is a block matrix of the form

A1 =

[µ ?0n−1 B

]where 0n−1 is the column of size n − 1 consisting of zeros.


Triangularization

We can choose v1 to be of norm 1. We can then complete it to an orthonormalbasis {v1, . . . , vn}.

(Here we use Gram-Schmidt.) We then take C1 to be thematrix whose i th column is vi , 1 ≤ i ≤ n. Then as seen earlier, C1 is a unitarymatrix. Put A1 = C−11 AC1. ThenA1e1 = C−11 AC1e1 = C−11 Av1 = µC−11 (v1) = µe1. This shows that the firstcolumn of A1 has all entries zero except the first one which is equal to µ. Let Bbe the matrix obtained from A1 by cutting down the first row and the firstcolumn, so that A1 is a block matrix of the form

A1 =

[µ ?0n−1 B



Triangularization

We can choose v1 to be of norm 1. We can then complete it to an orthonormalbasis {v1, . . . , vn}. (Here we use Gram-Schmidt.)

We then take C1 to be thematrix whose i th column is vi , 1 ≤ i ≤ n. Then as seen earlier, C1 is a unitarymatrix. Put A1 = C−11 AC1. ThenA1e1 = C−11 AC1e1 = C−11 Av1 = µC−11 (v1) = µe1. This shows that the firstcolumn of A1 has all entries zero except the first one which is equal to µ. Let Bbe the matrix obtained from A1 by cutting down the first row and the firstcolumn, so that A1 is a block matrix of the form

A1 =

[µ ?0n−1 B



Triangularization

We can choose v1 to be of norm 1. We can then complete it to an orthonormalbasis {v1, . . . , vn}. (Here we use Gram-Schmidt.) We then take C1 to be thematrix whose i th column is vi , 1 ≤ i ≤ n.

Then as seen earlier, C1 is a unitarymatrix. Put A1 = C−11 AC1. ThenA1e1 = C−11 AC1e1 = C−11 Av1 = µC−11 (v1) = µe1. This shows that the firstcolumn of A1 has all entries zero except the first one which is equal to µ. Let Bbe the matrix obtained from A1 by cutting down the first row and the firstcolumn, so that A1 is a block matrix of the form

A1 =

[µ ?0n−1 B



Triangularization

We can choose v1 to be of norm 1. We can then complete it to an orthonormalbasis {v1, . . . , vn}. (Here we use Gram-Schmidt.) We then take C1 to be thematrix whose i th column is vi , 1 ≤ i ≤ n. Then as seen earlier, C1 is a unitarymatrix.

Put A1 = C−11 AC1. ThenA1e1 = C−11 AC1e1 = C−11 Av1 = µC−11 (v1) = µe1. This shows that the firstcolumn of A1 has all entries zero except the first one which is equal to µ. Let Bbe the matrix obtained from A1 by cutting down the first row and the firstcolumn, so that A1 is a block matrix of the form

A1 =

[µ ?0n−1 B



Triangularization

We can choose v1 to be of norm 1. We can then complete it to an orthonormalbasis {v1, . . . , vn}. (Here we use Gram-Schmidt.) We then take C1 to be thematrix whose i th column is vi , 1 ≤ i ≤ n. Then as seen earlier, C1 is a unitarymatrix. Put A1 = C−11 AC1. ThenA1e1 = C−11 AC1e1 = C−11 Av1 = µC−11 (v1) = µe1.

This shows that the firstcolumn of A1 has all entries zero except the first one which is equal to µ. Let Bbe the matrix obtained from A1 by cutting down the first row and the firstcolumn, so that A1 is a block matrix of the form

A1 =

[µ ?0n−1 B



Triangularization

We can choose v1 to be of norm 1. We can then complete it to an orthonormalbasis {v1, . . . , vn}. (Here we use Gram-Schmidt.) We then take C1 to be thematrix whose i th column is vi , 1 ≤ i ≤ n. Then as seen earlier, C1 is a unitarymatrix. Put A1 = C−11 AC1. ThenA1e1 = C−11 AC1e1 = C−11 Av1 = µC−11 (v1) = µe1. This shows that the firstcolumn of A1 has all entries zero except the first one which is equal to µ.

Let Bbe the matrix obtained from A1 by cutting down the first row and the firstcolumn, so that A1 is a block matrix of the form

A1 =

[µ ?0n−1 B



Triangularization

We can choose v1 to be of norm 1. We can then complete it to an orthonormalbasis {v1, . . . , vn}. (Here we use Gram-Schmidt.) We then take C1 to be thematrix whose i th column is vi , 1 ≤ i ≤ n. Then as seen earlier, C1 is a unitarymatrix. Put A1 = C−11 AC1. ThenA1e1 = C−11 AC1e1 = C−11 Av1 = µC−11 (v1) = µe1. This shows that the firstcolumn of A1 has all entries zero except the first one which is equal to µ. Let Bbe the matrix obtained from A1 by cutting down the first row and the firstcolumn, so that A1 is a block matrix of the form

A1 =

[µ ?0n−1 B



Triangularization

By induction there exists a (n− 1)× (n− 1) unitary matrix M such that M−1BM

is an upper triangular matrix.

Put M1 =

[1 0t

n−10n−1 M

].

Then M1 is unitary and hence C = C1M1 is also unitary. ClearlyC−1AC = M−11 C−11 AC1M1 = M−11 A1M1 which is of the form[

1 0tn−1

0n−1 M−1

] [µ ∗0n−1 B

] [1 0t

n−10n−1 M

]=

[µ ∗

0n−1 M−1BM

]and hence is upper triangular. ♠


Triangularization


is an upper triangular matrix. Put M1 =

[1 0t

n−10n−1 M

].


1 0tn−1

0n−1 M−1

] [µ ∗0n−1 B

] [1 0t

n−10n−1 M

]=

[µ ∗

0n−1 M−1BM



Triangularization



[1 0t

n−10n−1 M

].

Then M1 is unitary and hence C = C1M1 is also unitary.

ClearlyC−1AC = M−11 C−11 AC1M1 = M−11 A1M1 which is of the form[

1 0tn−1

0n−1 M−1

] [µ ∗0n−1 B

] [1 0t

n−10n−1 M

]=

[µ ∗

0n−1 M−1BM



Triangularization



[1 0t

n−10n−1 M

].


1 0tn−1

0n−1 M−1

] [µ ∗0n−1 B

] [1 0t

n−10n−1 M

]=

[µ ∗

0n−1 M−1BM



Triangularization

Remark

Assume now that A is a real matrix with all its eigenvalues real.

Then from lemma13, it follows that we can choose the eigenvector v1 to be a real vector and thencomplete this into a orthonormal basis for Rn

. Thus the matrix C1 correspondingto this basis will have real entries. By induction M will have real entries and hencethe product C = MC1 will also have real entries. Thus we have proved:

Proposition

For a real square matrix A with all its eigenvalues real, there exists an orthogonalmatrix C such that C tAC is upper triangular.


Triangularization

Remark

Assume now that A is a real matrix with all its eigenvalues real. Then from lemma13, it follows that we can choose the eigenvector v1 to be a real vector and thencomplete this into a orthonormal basis for Rn

.

Thus the matrix C1 correspondingto this basis will have real entries. By induction M will have real entries and hencethe product C = MC1 will also have real entries. Thus we have proved:

Proposition



Triangularization

Remark


. Thus the matrix C1 correspondingto this basis will have real entries. By induction M will have real entries and hencethe product C = MC1 will also have real entries.

Thus we have proved:

Proposition



Triangularization

Remark


. Thus the matrix C1 correspondingto this basis will have real entries. By induction M will have real entries and hencethe product C = MC1 will also have real entries. Thus we have proved:

Proposition



Normal Matrices

DefinitionA square matrix A is called normal if A∗A = AA∗.

Remark

(i) Normality is congruence invariant. This means that if C is unitary and A isnormal then C−1AC is also normal. This is easy to verify.(ii) Any diagonal matrix is normal. Therefore it follows that in order that a matrixA is congruent to a diagonal matrix, it is necessary that A is normal. Amazingly,this condition turns out to be sufficient as well. That is one simple reason todefine this concept.(iii) Observe that product of two normal matrices may not be normal. For

example take A =

[0 11 0

]; B =

[1 00 2

].

(iv) Certainly Hermitian matrices are normal. Of course, there are normal matrices

which are not Hermitian. For example, take A =

[ı 00 −ı

].


Normal Matrices


Remark

(i) Normality is congruence invariant. This means that if C is unitary and A isnormal then C−1AC is also normal. This is easy to verify.

(ii) Any diagonal matrix is normal. Therefore it follows that in order that a matrixA is congruent to a diagonal matrix, it is necessary that A is normal. Amazingly,this condition turns out to be sufficient as well. That is one simple reason todefine this concept.(iii) Observe that product of two normal matrices may not be normal. For

example take A =

[0 11 0

]; B =

[1 00 2

].



[ı 00 −ı

].


Normal Matrices


Remark

(i) Normality is congruence invariant. This means that if C is unitary and A isnormal then C−1AC is also normal. This is easy to verify.(ii) Any diagonal matrix is normal. Therefore it follows that in order that a matrixA is congruent to a diagonal matrix, it is necessary that A is normal.

Amazingly,this condition turns out to be sufficient as well. That is one simple reason todefine this concept.(iii) Observe that product of two normal matrices may not be normal. For

example take A =

[0 11 0

]; B =

[1 00 2

].



[ı 00 −ı

].


Normal Matrices


Remark

(i) Normality is congruence invariant. This means that if C is unitary and A isnormal then C−1AC is also normal. This is easy to verify.(ii) Any diagonal matrix is normal. Therefore it follows that in order that a matrixA is congruent to a diagonal matrix, it is necessary that A is normal. Amazingly,this condition turns out to be sufficient as well. That is one simple reason todefine this concept.

(iii) Observe that product of two normal matrices may not be normal. For

example take A =

[0 11 0

]; B =

[1 00 2

].



[ı 00 −ı

].


Normal Matrices


Remark

(i) Normality is congruence invariant. This means that if C is unitary and A isnormal then C−1AC is also normal. This is easy to verify.(ii) Any diagonal matrix is normal. Therefore it follows that in order that a matrixA is congruent to a diagonal matrix, it is necessary that A is normal. Amazingly,this condition turns out to be sufficient as well. That is one simple reason todefine this concept.(iii) Observe that product of two normal matrices may not be normal.

For

example take A =

[0 11 0

]; B =

[1 00 2

].



[ı 00 −ı

].


Normal Matrices


Remark


example take A =

[0 11 0

]; B =

[1 00 2

].



[ı 00 −ı

].


Normal Matrices


Remark


example take A =

[0 11 0

]; B =

[1 00 2

].

(iv) Certainly Hermitian matrices are normal.

Of course, there are normal matrices


[ı 00 −ı

].


Normal Matrices


Remark


example take A =

[0 11 0

]; B =

[1 00 2

].


which are not Hermitian.

For example, take A =

[ı 00 −ı

].


Normal Matrices


Remark


example take A =

[0 11 0

]; B =

[1 00 2

].



[ı 00 −ı

].


Normal Matrices

Lemma

For a normal matrix A we have, ‖Ax‖2 = ‖A∗x‖2 for all x ∈ Cn;

in particular, foreach i , the norm of i th-row of A is equal to the norm of i th-column.

Proof: ‖Ax‖2 = (Ax)∗(Ax) = x∗A∗Ax = x∗AA∗x = (A∗x)∗(A∗x) = ‖A∗x‖2.Taking x = ei we get that norms of i th-column of A is equal to the norm ofi th-column of A∗= norm of the i th-row of A=norm of the i th-row of A. ♠

LemmaIf A is normal, then v is an eigenvector of A with eigenvalue µ iff v is aneigenvector of A∗ with eigenvalue µ.

Proof: Observe that if A is normal then A− µI is also normal. Now(A− µI )(v) = 0 iff ‖(A− µI )(v)‖ = 0 iff ‖(A− µI )∗v‖ = 0 iff (A∗ − µI )(v) = 0.♠


Normal Matrices

Lemma

For a normal matrix A we have, ‖Ax‖2 = ‖A∗x‖2 for all x ∈ Cn; in particular, for

each i , the norm of i th-row of A is equal to the norm of i th-column.





Normal Matrices

Lemma



Proof: ‖Ax‖2 = (Ax)∗(Ax) = x∗A∗Ax = x∗AA∗x = (A∗x)∗(A∗x) = ‖A∗x‖2.

Taking x = ei we get that norms of i th-column of A is equal to the norm ofi th-column of A∗= norm of the i th-row of A=norm of the i th-row of A. ♠




Normal Matrices

Lemma







Normal Matrices

Lemma







Normal Matrices

Lemma







Normal Matrices

Proposition

An upper triangular normal matrix is diagonal.

Proof: Let A be an upper triangular normal matrix. Inductively, we shallshow that aij = 0 for j > i . We have, Ae1 = a11e1. Hence ‖Ae1‖2 = |a11|2. Onthe other hand, this is equal to ‖A∗e1‖2 = |a11|2 + |a12|2 + · · ·+ |a1n|2. Hencea12 = a13 = · · · = a1n = 0. Inductively, suppose we have shown aij = 0 for j > ifor all 1 ≤ i ≤ k − 1. Then it follows that Aek = akkek . Exactly as in the firstcase, this implies that ‖A∗ek‖2 = |ak,k |2 + |ak,k+1|2 + · · ·+ |ak,n|2 = |ak,k |2.Hence ak,k+1 = · · · = ak,n = 0. ♠


Normal Matrices

Proposition


Proof: Let A be an upper triangular normal matrix.

Inductively, we shallshow that aij = 0 for j > i . We have, Ae1 = a11e1. Hence ‖Ae1‖2 = |a11|2. Onthe other hand, this is equal to ‖A∗e1‖2 = |a11|2 + |a12|2 + · · ·+ |a1n|2. Hencea12 = a13 = · · · = a1n = 0. Inductively, suppose we have shown aij = 0 for j > ifor all 1 ≤ i ≤ k − 1. Then it follows that Aek = akkek . Exactly as in the firstcase, this implies that ‖A∗ek‖2 = |ak,k |2 + |ak,k+1|2 + · · ·+ |ak,n|2 = |ak,k |2.Hence ak,k+1 = · · · = ak,n = 0. ♠


Normal Matrices

Proposition


Proof: Let A be an upper triangular normal matrix. Inductively, we shallshow that aij = 0 for j > i . We have, Ae1 = a11e1. Hence ‖Ae1‖2 = |a11|2.

Onthe other hand, this is equal to ‖A∗e1‖2 = |a11|2 + |a12|2 + · · ·+ |a1n|2. Hencea12 = a13 = · · · = a1n = 0. Inductively, suppose we have shown aij = 0 for j > ifor all 1 ≤ i ≤ k − 1. Then it follows that Aek = akkek . Exactly as in the firstcase, this implies that ‖A∗ek‖2 = |ak,k |2 + |ak,k+1|2 + · · ·+ |ak,n|2 = |ak,k |2.Hence ak,k+1 = · · · = ak,n = 0. ♠


Normal Matrices

Proposition


Proof: Let A be an upper triangular normal matrix. Inductively, we shallshow that aij = 0 for j > i . We have, Ae1 = a11e1. Hence ‖Ae1‖2 = |a11|2. Onthe other hand, this is equal to ‖A∗e1‖2 = |a11|2 + |a12|2 + · · ·+ |a1n|2.

Hencea12 = a13 = · · · = a1n = 0. Inductively, suppose we have shown aij = 0 for j > ifor all 1 ≤ i ≤ k − 1. Then it follows that Aek = akkek . Exactly as in the firstcase, this implies that ‖A∗ek‖2 = |ak,k |2 + |ak,k+1|2 + · · ·+ |ak,n|2 = |ak,k |2.Hence ak,k+1 = · · · = ak,n = 0. ♠


Normal Matrices

Proposition


Proof: Let A be an upper triangular normal matrix. Inductively, we shallshow that aij = 0 for j > i . We have, Ae1 = a11e1. Hence ‖Ae1‖2 = |a11|2. Onthe other hand, this is equal to ‖A∗e1‖2 = |a11|2 + |a12|2 + · · ·+ |a1n|2. Hencea12 = a13 = · · · = a1n = 0.

Inductively, suppose we have shown aij = 0 for j > ifor all 1 ≤ i ≤ k − 1. Then it follows that Aek = akkek . Exactly as in the firstcase, this implies that ‖A∗ek‖2 = |ak,k |2 + |ak,k+1|2 + · · ·+ |ak,n|2 = |ak,k |2.Hence ak,k+1 = · · · = ak,n = 0. ♠


Normal Matrices

Proposition


Proof: Let A be an upper triangular normal matrix. Inductively, we shallshow that aij = 0 for j > i . We have, Ae1 = a11e1. Hence ‖Ae1‖2 = |a11|2. Onthe other hand, this is equal to ‖A∗e1‖2 = |a11|2 + |a12|2 + · · ·+ |a1n|2. Hencea12 = a13 = · · · = a1n = 0. Inductively, suppose we have shown aij = 0 for j > ifor all 1 ≤ i ≤ k − 1.

Then it follows that Aek = akkek . Exactly as in the firstcase, this implies that ‖A∗ek‖2 = |ak,k |2 + |ak,k+1|2 + · · ·+ |ak,n|2 = |ak,k |2.Hence ak,k+1 = · · · = ak,n = 0. ♠


Normal Matrices

Proposition


Proof: Let A be an upper triangular normal matrix. Inductively, we shallshow that aij = 0 for j > i . We have, Ae1 = a11e1. Hence ‖Ae1‖2 = |a11|2. Onthe other hand, this is equal to ‖A∗e1‖2 = |a11|2 + |a12|2 + · · ·+ |a1n|2. Hencea12 = a13 = · · · = a1n = 0. Inductively, suppose we have shown aij = 0 for j > ifor all 1 ≤ i ≤ k − 1. Then it follows that Aek = akkek . Exactly as in the firstcase, this implies that ‖A∗ek‖2 = |ak,k |2 + |ak,k+1|2 + · · ·+ |ak,n|2 = |ak,k |2.Hence ak,k+1 = · · · = ak,n = 0. ♠


Normal Matrices

Why all this fuss? Well for one thing, we now have a big theorem. All we have todo is to combine propositions 13 and 15.

Theorem

Spectral Theorem Given any normal matrix A ∈ M(n,C), there exists a unitarymatrix C such that C∗AC is a diagonal matrix.

Corollary

(a) Every Hermitian matrix A is congruent to a diagonal matrix.(b) A real symmetric matrix is real-congruent to a diagonal matrix.

Proof: (a) We simply observe that a Hermitian matrix is normal and applythe above theorem.(b) We first recall that for a real symmetric matrix, all eigenvalues are real.Hence the proposition 14 is applicable and so we can choose C to be anorthogonal matrix. Along with proposition 15, this gives the result. ♠


Normal Matrices


Theorem


Corollary

(a) Every Hermitian matrix A is congruent to a diagonal matrix.

(b) A real symmetric matrix is real-congruent to a diagonal matrix.

Proof: (a) We simply observe that a Hermitian matrix is normal and applythe above theorem.(b) We first recall that for a real symmetric matrix, all eigenvalues are real.Hence the proposition 14 is applicable and so we can choose C to be anorthogonal matrix. Along with proposition 15, this gives the result. ♠


Normal Matrices


Theorem


Corollary


Proof: (a) We simply observe that a Hermitian matrix is normal and applythe above theorem.

(b) We first recall that for a real symmetric matrix, all eigenvalues are real.Hence the proposition 14 is applicable and so we can choose C to be anorthogonal matrix. Along with proposition 15, this gives the result. ♠


Normal Matrices


Theorem


Corollary


Proof: (a) We simply observe that a Hermitian matrix is normal and applythe above theorem.(b) We first recall that for a real symmetric matrix, all eigenvalues are real.

Hence the proposition 14 is applicable and so we can choose C to be anorthogonal matrix. Along with proposition 15, this gives the result. ♠


Quadratic forms and their diagonalization

Definition

Let A = (aij) be an n × n real matrix.

The function Q : Rn → R defined by :

QA(x) := QA((x1, x2, . . . , xn)t) :=n∑

i=1

n∑j=1

aijxixj

is called the quadratic form associated with A.If A = diag (λ1, λ2, . . . , λn) then Q(x) = λ1x2

1 + λ2x22 + · · ·+ λnx2

n is called adiagonal form.



Definition

Let A = (aij) be an n × n real matrix. The function Q : Rn → R defined by :

QA(x) := QA((x1, x2, . . . , xn)t) :=n∑

i=1

n∑j=1

aijxixj

is called the quadratic form associated with A.

If A = diag (λ1, λ2, . . . , λn) then Q(x) = λ1x21 + λ2x2

2 + · · ·+ λnx2n is called a

diagonal form.



Definition

Let A = (aij) be an n × n real matrix. The function Q : Rn → R defined by :

QA(x) := QA((x1, x2, . . . , xn)t) :=n∑

i=1

n∑j=1

aijxixj

is called the quadratic form associated with A.If A = diag (λ1, λ2, . . . , λn) then Q(x) = λ1x2

1 + λ2x22 + · · ·+ λnx2

n is called adiagonal form.


Quadratic forms

Proposition

Q(x) = [x1, x2, . . . , xn]A

x1x2...

xn

= xtAx where x = (x1, x2, . . . , xn)t .

Proof:

[x1, x2, . . . , xn]A

x1x2...

xn

= [x1, x2, . . . , xn]

n∑j=1

a1jxj

...n∑

j=1

anjxj

=

n∑j=1

a1jxjx1 + · · ·+n∑

j=1

anjxjxn =n∑

j=1

n∑i=1

aijxixj = Q(x).

♠


Quadratic forms

Proposition

Q(x) = [x1, x2, . . . , xn]A

x1x2...

xn


Proof:

[x1, x2, . . . , xn]A

x1x2...

xn

= [x1, x2, . . . , xn]

n∑j=1

a1jxj

...n∑

j=1

anjxj

=n∑

j=1

a1jxjx1 + · · ·+n∑

j=1

anjxjxn =n∑

j=1

n∑i=1

aijxixj = Q(x).

♠


Quadratic forms

Proposition

Q(x) = [x1, x2, . . . , xn]A

x1x2...

xn


Proof:

[x1, x2, . . . , xn]A

x1x2...

xn

= [x1, x2, . . . , xn]

n∑j=1

a1jxj

...n∑

j=1

anjxj

=

n∑j=1

a1jxjx1 + · · ·+n∑

j=1

anjxjxn

=n∑

j=1

n∑i=1

aijxixj = Q(x).

♠


Quadratic forms

Proposition

Q(x) = [x1, x2, . . . , xn]A

x1x2...

xn


Proof:

[x1, x2, . . . , xn]A

x1x2...

xn

= [x1, x2, . . . , xn]

n∑j=1

a1jxj

...n∑

j=1

anjxj

=

n∑j=1

a1jxjx1 + · · ·+n∑

j=1

anjxjxn =n∑

j=1

n∑i=1

aijxixj = Q(x).

♠ARS (IITB) MA106-Linear Algebra February 14, 2011 80 / 99

Quadratic forms

Example

(1) A =

[1 13 5

], X =

[xy

]. Then

X tAX = [x , y ]

[1 13 5

] [xy

]= [x , y ]

[x + y

3x + 5y

]= x2

1 + 4xy + 5y2.

(2) Let B =

[1 22 5

], X =

[xy

]. Then

X tBX = [x , y ]

[1 22 5

] [xy

]= [x , y ]

[x + 2y

2x + 5y

]= x2

1 + 4xy + 5x22 .

Notice that A and B give rise to same Q(x) and B = 12 (A + At) is a symmetric

matrix.


Quadratic forms

Example

(1) A =

[1 13 5

], X =

[xy

]. Then

X tAX = [x , y ]

[1 13 5

] [xy

]= [x , y ]

[x + y

3x + 5y

]= x2

1 + 4xy + 5y2.

(2) Let B =

[1 22 5

], X =

[xy

].

Then

X tBX = [x , y ]

[1 22 5

] [xy

]= [x , y ]

[x + 2y

2x + 5y

]= x2

1 + 4xy + 5x22 .


matrix.


Quadratic forms

Example

(1) A =

[1 13 5

], X =

[xy

]. Then

X tAX = [x , y ]

[1 13 5

] [xy

]= [x , y ]

[x + y

3x + 5y

]= x2

1 + 4xy + 5y2.

(2) Let B =

[1 22 5

], X =

[xy

]. Then

X tBX = [x , y ]

[1 22 5

] [xy

]= [x , y ]

[x + 2y

2x + 5y

]= x2

1 + 4xy + 5x22 .


matrix.


Quadratic forms

Example

(1) A =

[1 13 5

], X =

[xy

]. Then

X tAX = [x , y ]

[1 13 5

] [xy

]= [x , y ]

[x + y

3x + 5y

]= x2

1 + 4xy + 5y2.

(2) Let B =

[1 22 5

], X =

[xy

]. Then

X tBX = [x , y ]

[1 22 5

] [xy

]= [x , y ]

[x + 2y

2x + 5y

]= x2

1 + 4xy + 5x22 .


matrix.


PropositionFor any n × n matrix A and the column vector x = (x1, x2, . . . xn)t

xtAx = xtBx where B =12

(A + At ).

Hence every quadratic form is associated with a symmetric matrix.

Proof: xtAx is a 1× 1 matrix. Hence xtAx = (xtAx)t = xtAtx. Hence

xtAx =12

xtAx +12

xtAtx = xt 12

(A + At )x = xtBx.

□

59/59



(A + At ).



xtAx =12

xtAx +12

xtAtx = xt 12

(A + At )x = xtBx.

□

59/59



(A + At ).


Proof: xtAx is a 1× 1 matrix. Hence xtAx = (xtAx)t = xtAtx.

Hence

xtAx =12

xtAx +12

xtAtx = xt 12

(A + At )x = xtBx.

□

59/59



(A + At ).



xtAx =12

xtAx +12

xtAtx = xt 12

(A + At )x = xtBx.

□

59/59


We now show how the spectral theorem helps us in converting a quadratic forminto a diagonal form.

Theorem

Let xtAx be a quadratic form associated with a real symmetric matrix A. Let U bean orthogonal matrix such that U tAU = diag (λ1, λ2, . . . , λn). Then

xtAx = λ1y21 + λ2y2

2 + · · ·+ λny2n ,

where y = (y1, . . . , yn) are define by

x =

x1x2...

xn

= U

y1y2...

yn

= Uy.




Theorem

Let xtAx be a quadratic form associated with a real symmetric matrix A.

Let U bean orthogonal matrix such that U tAU = diag (λ1, λ2, . . . , λn). Then


2 + · · ·+ λny2n ,


x =

x1x2...

xn

= U

y1y2...

yn

= Uy.




Theorem

Let xtAx be a quadratic form associated with a real symmetric matrix A. Let U bean orthogonal matrix such that U tAU = diag (λ1, λ2, . . . , λn).

Then


2 + · · ·+ λny2n ,


x =

x1x2...

xn

= U

y1y2...

yn

= Uy.




Theorem

Let xtAx be a quadratic form associated with a real symmetric matrix A. Let U bean orthogonal matrix such that U tAU = diag (λ1, λ2, . . . , λn). Then


2 + · · ·+ λny2n ,


x =

x1x2...

xn

= U

y1y2...

yn

= Uy.



Proof: Since x = Uy,

xtAx = (Uy)tA(Uy) = yt(U tAU)y.

Since U tAU = diag (λ1, λ2, . . . , λn), we get

xtAx = [y1, y2, . . . , yn]

λ1

λ2. . .

λn

y1y2...

yn

= λ1y2

1 + λ2y22 + · · ·+ λny2

n .

♠



Proof: Since x = Uy,

xtAx = (Uy)tA(Uy) = yt(U tAU)y.

Since U tAU = diag (λ1, λ2, . . . , λn), we get

xtAx = [y1, y2, . . . , yn]

λ1

λ2. . .

λn

y1y2...

yn

= λ1y2

1 + λ2y22 + · · ·+ λny2

n .

♠


Quadratic forms: An example

Example

Let us determine the orthogonal matrix U which reduces the quadratic formQ(x) = 2x2

1 + 4xy + 5x22 to a diagonal form.

We write

Q(x) = [x , y ]

[2 22 5

] [xy

]= xtAx.

The symmetric matrix A can be diagonalized. The eigenvalues of A are λ1 = 1 andλ2 = 6. An orthonormal set of eigenvectors for λ1 and λ2 is

v1 =1√5

[2−1

]and v2 =

1√5

[12

].

Hence U = 1√5

[2 1−1 2

]. The change of variables equations are

[xy

]= U

[uv

].

The diagonal form is:

[u, v ]

[1 00 6

][u, v ]T = u2 + 6v2.

Check that UtAU = diag (1, 6).



Example


1 + 4xy + 5x22 to a diagonal form. We write

Q(x) = [x , y ]

[2 22 5

] [xy

]= xtAx.

The symmetric matrix A can be diagonalized.

The eigenvalues of A are λ1 = 1 andλ2 = 6. An orthonormal set of eigenvectors for λ1 and λ2 is

v1 =1√5

[2−1

]and v2 =

1√5

[12

].

Hence U = 1√5

[2 1−1 2


[xy

]= U

[uv

].


[u, v ]

[1 00 6

][u, v ]T = u2 + 6v2.




Example



Q(x) = [x , y ]

[2 22 5

] [xy

]= xtAx.

The symmetric matrix A can be diagonalized. The eigenvalues of A are λ1 = 1 andλ2 = 6.

An orthonormal set of eigenvectors for λ1 and λ2 is

v1 =1√5

[2−1

]and v2 =

1√5

[12

].

Hence U = 1√5

[2 1−1 2


[xy

]= U

[uv

].


[u, v ]

[1 00 6

][u, v ]T = u2 + 6v2.




Example



Q(x) = [x , y ]

[2 22 5

] [xy

]= xtAx.


v1 =1√5

[2−1

]and v2 =

1√5

[12

].

Hence U = 1√5

[2 1−1 2


[xy

]= U

[uv

].


[u, v ]

[1 00 6

][u, v ]T = u2 + 6v2.




Example



Q(x) = [x , y ]

[2 22 5

] [xy

]= xtAx.


v1 =1√5

[2−1

]and v2 =

1√5

[12

].

Hence U = 1√5

[2 1−1 2

].

The change of variables equations are

[xy

]= U

[uv

].


[u, v ]

[1 00 6

][u, v ]T = u2 + 6v2.




Example



Q(x) = [x , y ]

[2 22 5

] [xy

]= xtAx.


v1 =1√5

[2−1

]and v2 =

1√5

[12

].

Hence U = 1√5

[2 1−1 2


[xy

]= U

[uv

].


[u, v ]

[1 00 6

][u, v ]T = u2 + 6v2.




Example



Q(x) = [x , y ]

[2 22 5

] [xy

]= xtAx.


v1 =1√5

[2−1

]and v2 =

1√5

[12

].

Hence U = 1√5

[2 1−1 2


[xy

]= U

[uv

].


[u, v ]

[1 00 6

][u, v ]T = u2 + 6v2.



Conic Sections

A conic section is the locus in the Cartesian plane R2of an equation of the form

ax2 + bxy + cy2 + dx + ey + f = 0. (10)

It can be proved that this equation represents one of the following:(i) the empty set(ii) a single point(iii) one or two straight lines(iv) an ellipse(v) an hyperbola or(vi) a parabola.The second degree part of (10)

Q(x , y) = ax2 + bxy + cy2

is a quadratic form. This determines the type of the conic.


Conic Sections


ax2 + bxy + cy2 + dx + ey + f = 0. (10)

It can be proved that this equation represents one of the following:(i) the empty set

(ii) a single point(iii) one or two straight lines(iv) an ellipse(v) an hyperbola or(vi) a parabola.The second degree part of (10)

Q(x , y) = ax2 + bxy + cy2



Conic Sections


ax2 + bxy + cy2 + dx + ey + f = 0. (10)

It can be proved that this equation represents one of the following:(i) the empty set(ii) a single point

(iii) one or two straight lines(iv) an ellipse(v) an hyperbola or(vi) a parabola.The second degree part of (10)

Q(x , y) = ax2 + bxy + cy2



Conic Sections


ax2 + bxy + cy2 + dx + ey + f = 0. (10)

It can be proved that this equation represents one of the following:(i) the empty set(ii) a single point(iii) one or two straight lines

(iv) an ellipse(v) an hyperbola or(vi) a parabola.The second degree part of (10)

Q(x , y) = ax2 + bxy + cy2



Conic Sections


ax2 + bxy + cy2 + dx + ey + f = 0. (10)

It can be proved that this equation represents one of the following:(i) the empty set(ii) a single point(iii) one or two straight lines(iv) an ellipse

(v) an hyperbola or(vi) a parabola.The second degree part of (10)

Q(x , y) = ax2 + bxy + cy2



Conic Sections


ax2 + bxy + cy2 + dx + ey + f = 0. (10)

It can be proved that this equation represents one of the following:(i) the empty set(ii) a single point(iii) one or two straight lines(iv) an ellipse(v) an hyperbola or

(vi) a parabola.The second degree part of (10)

Q(x , y) = ax2 + bxy + cy2



Conic Sections


ax2 + bxy + cy2 + dx + ey + f = 0. (10)


Q(x , y) = ax2 + bxy + cy2

is a quadratic form.

This determines the type of the conic.


Conic Sections


ax2 + bxy + cy2 + dx + ey + f = 0. (10)


Q(x , y) = ax2 + bxy + cy2



Conic SectionsWe can write the equation (10) into matrix form:

[x , y ]

[a b/2

b/2 c

] [xy

]+ [d , e]

[xy

]+ f = 0 (11)

Write A =

[a b/2

b/2 c

]. Let U = [v1, v2] be an orthogonal matrix whose

column vectors v1 and v2 are eigenvectors of A with eigenvalues λ1 and λ2. Applythe change of variables

x =

[xy

]= U

[uv

]to diagonalize the quadratic form Q(x , y) to the diagonal form λ1u2 + λ2v2

2 . Theorthonormal basis {v1, v2} determines a new set of coordinate axes with respectto which the locus of the equation [x , y ]A[x , y ]T + B[x , y ]T + f = 0 withB = [d , e] is same as the locus of the equation

0 = [u, v ] diag (λ1, λ2)[u, v ]T + (BU)[u, v ]T + f

= λ1u2 + λ2v2 + [d , e][v1, v2][u, v ]T + f . (12)



[x , y ]

[a b/2

b/2 c

] [xy

]+ [d , e]

[xy

]+ f = 0 (11)

Write A =

[a b/2

b/2 c

].

Let U = [v1, v2] be an orthogonal matrix whose


x =

[xy

]= U

[uv




= λ1u2 + λ2v2 + [d , e][v1, v2][u, v ]T + f . (12)



[x , y ]

[a b/2

b/2 c

] [xy

]+ [d , e]

[xy

]+ f = 0 (11)

Write A =

[a b/2

b/2 c


column vectors v1 and v2 are eigenvectors of A with eigenvalues λ1 and λ2.

Applythe change of variables

x =

[xy

]= U

[uv




= λ1u2 + λ2v2 + [d , e][v1, v2][u, v ]T + f . (12)



[x , y ]

[a b/2

b/2 c

] [xy

]+ [d , e]

[xy

]+ f = 0 (11)

Write A =

[a b/2

b/2 c



x =

[xy

]= U

[uv


2 .

Theorthonormal basis {v1, v2} determines a new set of coordinate axes with respectto which the locus of the equation [x , y ]A[x , y ]T + B[x , y ]T + f = 0 withB = [d , e] is same as the locus of the equation


= λ1u2 + λ2v2 + [d , e][v1, v2][u, v ]T + f . (12)



[x , y ]

[a b/2

b/2 c

] [xy

]+ [d , e]

[xy

]+ f = 0 (11)

Write A =

[a b/2

b/2 c



x =

[xy

]= U

[uv




= λ1u2 + λ2v2 + [d , e][v1, v2][u, v ]T + f . (12)


Conic SectionsIf the conic determined by (12) is not degenerate i.e., not an emptyset, a point, nor line(s) then the signs of λ1 and λ2 determine whetherit is a parabola, an hyperbola or an ellipse.

The equation (10) willrepresent

(1) ellipse if λ1λ2 > 0

(2) hyperbola if λ1λ2 < 0

(3) parabola if λ1λ2 = 0.


Conic SectionsIf the conic determined by (12) is not degenerate i.e., not an emptyset, a point, nor line(s) then the signs of λ1 and λ2 determine whetherit is a parabola, an hyperbola or an ellipse. The equation (10) willrepresent

























Conic Sections: ExamplesExample

(1) 2x2 + 4xy + 5y2 + 4x + 13y − 1/4 = 0.We have earlier diagonalized the quadratic form 2x2 + 4xy + 5y2.

The associatedsymmetric matrix, the eigenvectors and eigenvalues are displayed in the equation ofdiagonalization :

U tAU =1√5

[2 −11 2

] [2 22 5

]1√5

[2 1−1 2

]=

[1 00 6

].

Put t = 1/√

5 for convenience. Then the change of coordinates equations are :[xy

]=

[2t t−t 2t

] [uv

],

i.e., x = t(2u + v) and y = t(−u + 2v). Substitute these into the original equationto get

u2 + 6v2 −√

5u + 6√

5v − 1

4= 0.

Complete the square to write this as

(u − 1

2

√5)2 + 6(v +

1

2

√5)2 = 9.

This is an equation of ellipse with center ( 12

√5,− 1

2

√5) in the uv-plane.



(1) 2x2 + 4xy + 5y2 + 4x + 13y − 1/4 = 0.We have earlier diagonalized the quadratic form 2x2 + 4xy + 5y2. The associatedsymmetric matrix, the eigenvectors and eigenvalues are displayed in the equation ofdiagonalization :

U tAU =1√5

[2 −11 2

] [2 22 5

]1√5

[2 1−1 2

]=

[1 00 6

].

Put t = 1/√


]=

[2t t−t 2t

] [uv

],


u2 + 6v2 −√

5u + 6√

5v − 1

4= 0.


(u − 1

2

√5)2 + 6(v +

1

2

√5)2 = 9.


√5,− 1

2





U tAU =1√5

[2 −11 2

] [2 22 5

]1√5

[2 1−1 2

]=

[1 00 6

].

Put t = 1/√


]=

[2t t−t 2t

] [uv

],


u2 + 6v2 −√

5u + 6√

5v − 1

4= 0.


(u − 1

2

√5)2 + 6(v +

1

2

√5)2 = 9.


√5,− 1

2





U tAU =1√5

[2 −11 2

] [2 22 5

]1√5

[2 1−1 2

]=

[1 00 6

].

Put t = 1/√

5 for convenience.

Then the change of coordinates equations are :[xy

]=

[2t t−t 2t

] [uv

],


u2 + 6v2 −√

5u + 6√

5v − 1

4= 0.


(u − 1

2

√5)2 + 6(v +

1

2

√5)2 = 9.


√5,− 1

2





U tAU =1√5

[2 −11 2

] [2 22 5

]1√5

[2 1−1 2

]=

[1 00 6

].

Put t = 1/√


]=

[2t t−t 2t

] [uv

],


u2 + 6v2 −√

5u + 6√

5v − 1

4= 0.


(u − 1

2

√5)2 + 6(v +

1

2

√5)2 = 9.


√5,− 1

2





U tAU =1√5

[2 −11 2

] [2 22 5

]1√5

[2 1−1 2

]=

[1 00 6

].

Put t = 1/√


]=

[2t t−t 2t

] [uv

],

i.e., x = t(2u + v) and y = t(−u + 2v).

Substitute these into the original equationto get

u2 + 6v2 −√

5u + 6√

5v − 1

4= 0.


(u − 1

2

√5)2 + 6(v +

1

2

√5)2 = 9.


√5,− 1

2





U tAU =1√5

[2 −11 2

] [2 22 5

]1√5

[2 1−1 2

]=

[1 00 6

].

Put t = 1/√


]=

[2t t−t 2t

] [uv

],


u2 + 6v2 −√

5u + 6√

5v − 1

4= 0.


(u − 1

2

√5)2 + 6(v +

1

2

√5)2 = 9.


√5,− 1

2



Conic Sections: Examples

The u-axis and v -axis are determined by the eigenvectors v1 and v2 as indicated inthe following figure :

y

uv



Example

(2) 2x2 − 4xy − y2 − 4x + 10y − 13 = 0.

Here, the matrix A =

[2 −2−2 −1

]gives the quadratic part of the equation. We write the equation in matrix form as

[x , y ]

[2 −2−2 −1

] [xy

]+ [−4, 10]

[xy

]− 13 = 0.

Let t = 1/√

5. The eigenvalues of A are λ1 = 3, λ2 = −2. An orthonormal set ofeigenvectors is v1 = t(2,−1)t and v2 = t(1, 2)t . Put

U = t

[2 1−1 2

]and

[xy

]= U

[uv

].

The transformed equation becomes

3u2 − 2v2 − 4t(2u + v) + 10t(−u + 2v)− 13 = 0



Example

(2) 2x2 − 4xy − y2 − 4x + 10y − 13 = 0. Here, the matrix A =

[2 −2−2 −1

]gives the quadratic part of the equation.

We write the equation in matrix form as

[x , y ]

[2 −2−2 −1

] [xy

]+ [−4, 10]

[xy

]− 13 = 0.

Let t = 1/√


U = t

[2 1−1 2

]and

[xy

]= U

[uv

].


3u2 − 2v2 − 4t(2u + v) + 10t(−u + 2v)− 13 = 0



Example


[2 −2−2 −1


[x , y ]

[2 −2−2 −1

] [xy

]+ [−4, 10]

[xy

]− 13 = 0.

Let t = 1/√


U = t

[2 1−1 2

]and

[xy

]= U

[uv

].


3u2 − 2v2 − 4t(2u + v) + 10t(−u + 2v)− 13 = 0



Example


[2 −2−2 −1


[x , y ]

[2 −2−2 −1

] [xy

]+ [−4, 10]

[xy

]− 13 = 0.

Let t = 1/√

5.

The eigenvalues of A are λ1 = 3, λ2 = −2. An orthonormal set ofeigenvectors is v1 = t(2,−1)t and v2 = t(1, 2)t . Put

U = t

[2 1−1 2

]and

[xy

]= U

[uv

].


3u2 − 2v2 − 4t(2u + v) + 10t(−u + 2v)− 13 = 0



Example


[2 −2−2 −1


[x , y ]

[2 −2−2 −1

] [xy

]+ [−4, 10]

[xy

]− 13 = 0.

Let t = 1/√

5. The eigenvalues of A are λ1 = 3, λ2 = −2.

An orthonormal set ofeigenvectors is v1 = t(2,−1)t and v2 = t(1, 2)t . Put

U = t

[2 1−1 2

]and

[xy

]= U

[uv

].


3u2 − 2v2 − 4t(2u + v) + 10t(−u + 2v)− 13 = 0



Example


[2 −2−2 −1


[x , y ]

[2 −2−2 −1

] [xy

]+ [−4, 10]

[xy

]− 13 = 0.

Let t = 1/√


U = t

[2 1−1 2

]and

[xy

]= U

[uv

].


3u2 − 2v2 − 4t(2u + v) + 10t(−u + 2v)− 13 = 0



Example


[2 −2−2 −1


[x , y ]

[2 −2−2 −1

] [xy

]+ [−4, 10]

[xy

]− 13 = 0.

Let t = 1/√


U = t

[2 1−1 2

]and

[xy

]= U

[uv

].


3u2 − 2v2 − 4t(2u + v) + 10t(−u + 2v)− 13 = 0



which is the same as

3u2 − 2v2 − 18tu + 16tv − 13 = 0.

Complete the square in u and v to get

3(u − 3t)2 − 2(v − 4t)2 = 12

or

(u − 3t)2

4− (v − 4t)2

6= 1.

This represents a hyperbola with center(3t, 4t) in the uv -plane. Theeigenvectors v1 and v2 determine thedirections of positive u and v axes.

u

v

x

y




3u2 − 2v2 − 18tu + 16tv − 13 = 0.


3(u − 3t)2 − 2(v − 4t)2 = 12

or

(u − 3t)2

4− (v − 4t)2

6= 1.


u

v

x

y




3u2 − 2v2 − 18tu + 16tv − 13 = 0.


3(u − 3t)2 − 2(v − 4t)2 = 12

or

(u − 3t)2

4− (v − 4t)2

6= 1.


u

v

x

y




3u2 − 2v2 − 18tu + 16tv − 13 = 0.


3(u − 3t)2 − 2(v − 4t)2 = 12

or

(u − 3t)2

4− (v − 4t)2

6= 1.

This represents a hyperbola with center(3t, 4t) in the uv -plane.

Theeigenvectors v1 and v2 determine thedirections of positive u and v axes.

u

v

x

y




3u2 − 2v2 − 18tu + 16tv − 13 = 0.


3(u − 3t)2 − 2(v − 4t)2 = 12

or

(u − 3t)2

4− (v − 4t)2

6= 1.


u

v

x

y



Example

(3) 9x2 + 24xy + 16xy2 − 20x + 15y = 0.

The symmetric matrix for the quadratic part is A =

[9 12

12 16

]. The eigenvalues

are λ1 = 25, λ2 = 0. An orthonormal set of eigenvectors isv1 = a(3, 4)t , v2 = a(−4, 3)t , where a = 1/5. An orthogonal diagonalizing matrix

is U = a

[3 −44 3

]. The equations of change of coordinates are

[xy

]= U

[uv

]i.e., x = a(3u − 4v), y = a(4u + 3v).

The equation in uv-plane is u2 + v = 0. This is an equation of parabola with itsvertex at the origin.



Example

(3) 9x2 + 24xy + 16xy2 − 20x + 15y = 0.


[9 12

12 16

].

The eigenvalues


is U = a

[3 −44 3


[xy

]= U

[uv

]i.e., x = a(3u − 4v), y = a(4u + 3v).




Example

(3) 9x2 + 24xy + 16xy2 − 20x + 15y = 0.


[9 12

12 16

]. The eigenvalues

are λ1 = 25, λ2 = 0.

An orthonormal set of eigenvectors isv1 = a(3, 4)t , v2 = a(−4, 3)t , where a = 1/5. An orthogonal diagonalizing matrix

is U = a

[3 −44 3


[xy

]= U

[uv

]i.e., x = a(3u − 4v), y = a(4u + 3v).




Example

(3) 9x2 + 24xy + 16xy2 − 20x + 15y = 0.


[9 12

12 16

]. The eigenvalues

are λ1 = 25, λ2 = 0. An orthonormal set of eigenvectors isv1 = a(3, 4)t , v2 = a(−4, 3)t , where a = 1/5.

An orthogonal diagonalizing matrix

is U = a

[3 −44 3


[xy

]= U

[uv

]i.e., x = a(3u − 4v), y = a(4u + 3v).




Example

(3) 9x2 + 24xy + 16xy2 − 20x + 15y = 0.


[9 12

12 16

]. The eigenvalues


is U = a

[3 −44 3

].

The equations of change of coordinates are

[xy

]= U

[uv

]i.e., x = a(3u − 4v), y = a(4u + 3v).




Example

(3) 9x2 + 24xy + 16xy2 − 20x + 15y = 0.


[9 12

12 16

]. The eigenvalues


is U = a

[3 −44 3


[xy

]= U

[uv

]i.e., x = a(3u − 4v), y = a(4u + 3v).




Example

(3) 9x2 + 24xy + 16xy2 − 20x + 15y = 0.


[9 12

12 16

]. The eigenvalues


is U = a

[3 −44 3


[xy

]= U

[uv

]i.e., x = a(3u − 4v), y = a(4u + 3v).

The equation in uv-plane is u2 + v = 0.

This is an equation of parabola with itsvertex at the origin.



Example

(3) 9x2 + 24xy + 16xy2 − 20x + 15y = 0.


[9 12

12 16

]. The eigenvalues


is U = a

[3 −44 3


[xy

]= U

[uv

]i.e., x = a(3u − 4v), y = a(4u + 3v).



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Linear Algebra - IIT Bombaydey/linearalg04.pdf · Triangularization Closely related to the problem of diagonalization is the problem of triangularization.We shall use this concept