Linear Algebra II

Course No. 100 222

Spring 2007

Michael Stoll

Contents

1. Review of Eigenvalues, Eigenvectors and Characteristic Polynomial 2

2. The Cayley-Hamilton Theorem and the Minimal Polynomial 2

3. The Structure of Nilpotent Endomorphisms 7

4. Direct Sums of Subspaces 9

5. The Jordan Normal Form Theorem 11

6. The Dual Vector Space 16

7. Norms on Real Vector Spaces 22

8. Bilinear Forms 24

9. Inner Product Spaces 32

10. Orthogonal Diagonalization 39

11. External Direct Sums 43

12. The Tensor Product 44

13. Symmetric and Alternating Products 53

References 59

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1. Review of Eigenvalues, Eigenvectors and CharacteristicPolynomial

Recall the topics we finished Linear Algebra I with. We were discussing eigenvaluesand eigenvectors of endomorphisms and square matrices, and the question whenthey are diagonalizable. For your convenience, I will repeat here the most relevantdefinitions and results.

Let V be a finite-dimensional F -vector space, dimV = n, and let f : V V bean endomorphism. Then for F , the -eigenspace of f was defined to be

E(f) = {v V : f(v) = v} = ker(f idV ) . is an eigenvalue of f if E(f) 6= {0}, i.e., if there is 0 6= v V such thatf(v) = v. Such a vector v is called an eigenvector of f for the eigenvalue .

The eigenvalues are exactly the roots (in F ) of the characteristic polynomial of f ,

Pf (x) = det(x idV f) ,which is a monic polynomial of degree n with coefficients in F .

The geometric multiplicity of as an eigenvalue of f is defined to be the dimensionof the -eigenspace, whereas the algebraic multiplicity of as an eigenvalue of fis defined to be its multiplicity as a root of the characteristic polynomial.

The endomorphism f is said to be diagonalizable if there exists a basis of Vconsisting of eigenvectors of f . The matrix representing f relative to this basis isthen a diagonal matrix, with the various eigenvalues appearing on the diagonal.

Since n n matrices can be identified with endomorphisms F n F n, all notionsand results makes sense for square matrices, too. A matrix A Mat(n, F ) isdiagonalizable if and only if it is similar to a diagonal matrix, i.e., if there is aninvertible matrix P Mat(n, F ) such that P1AP is diagonal.It is an important fact that the geometric multiplicity of an eigenvalue cannotexceed its algebraic multiplicity. An endomorphism or square matrix is diagonal-izable if and only if the sum of the geometric multiplicities of all eigenvalues equalsthe dimension of the space. This in turn is equivalent to the two conditions (a)the characteristic polynomial is a product of linear factors, and (b) for each eigen-value, algebraic and geometric multiplicities agree. For example, both conditionsare satisfied if Pf is the product of n distinct monic linear factors.

2. The Cayley-Hamilton Theorem and the Minimal Polynomial

Let A Mat(n, F ). We know that Mat(n, F ) is an F -vector space of dimension n2.Therefore, the elements I, A,A2, . . . , An

2cannot be linearly independent (because

their number exceeds the dimension). If we define p(A) in the obvious way for pa polynomial with coefficients in F , then we can deduce that there is a (non-zero)polynomial p of degree at most n2 such that p(A) = 0 (0 here is the zero matrix).In fact, much more is true.

Consider a diagonal matrix D = diag(1, 2, . . . , n). (This notation is supposedto mean that j is the (j, j) entry of D; the off-diagonal entries are zero, of course.)Its characteristic polynomial is

PD(x) = (x 1)(x 2) (x n) .Since the diagonal entries are roots of PD, we also have PD(D) = 0. More generally,consider a diagonalizable matrix A. Then there is an invertible matrix Q such

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that D = Q1AQ is diagonal. Since (Exercise!) p(Q1AQ) = Q1p(A)Q for p apolynomial, we find

0 = PD(D) = Q1PD(A)Q = Q

1PA(A)Q = PA(A) = 0 .(Recall that PA = PD similar matrices have the same characteristic polynomial.)

The following theorem states that this is true for all square matrices (or endomor-phisms of finite-dimensional vector spaces).

2.1. Theorem (Cayley-Hamilton). Let A Mat(n, F ). Then PA(A) = 0.

Proof. Here is a simple, but wrong proof. By definition, PA(x) = det(xI A),so, plugging in A for x, we have PA(A) = det(AIA) = det(AA) = det(0) = 0.(Exercise: find the mistake!)

For the correct proof, we need to consider matrices whose entries are polynomials.Since polynomials satisfy the field axioms except for the existence of inverses, wecan perform all operations that do not require divisions. This includes addition,multiplication and determinants; in particular, we can use the adjugate matrix.

Let B = xIA, then det(B) = PA(x). Let B be the adjugate matrix; then we stillhave BB = det(B)I. The entries of B come from determinants of (n1) (n1)submatrices of B, therefore they are polynomials of degree at most n 1. We canthen write

B = xn1Bn1 + xn2Bn2 + + xB1 +B0 ,

and we have the equality (of matrices with polynomial entries)

(xn1Bn1 +xn2Bn2 + +B0)(xIA) = PA(x)I = (xn+bn1xn1 + +b0)I ,

where we have set PA(x) = xn + bn1x

n1 + + b0. Expanding the left hand sideand comparing coefficients of like powers of x, we find the relations

Bn1 = I, Bn2 Bn1A = bn1I, . . . , B0 B1A = b1I, B0A = b0I .We multiply these from the right by An, An1, . . . , A, I, and add:

Bn1An = An

Bn2An1 Bn1An = bn1An1

......

...B0A B1A2 = b1A

B0A = b0I0 = PA(A)

2.2. Remarks.

(1) The reason why we cannot simply plug in A for x in B(xI A) = PA(x)Iis that whereas x (as a scalar) commutes with the matrices occurring ascoefficients of powers of x, it is not a priori clear that A does so, too. Wewill discuss this in more detail in the Introductory Algebra course, wherepolynomial rings will be studied in some detail.

(2) Another idea of proof (and maybe easier to grasp) is to say that a genericmatrix is diagonalizable (if we assume F to be algebraically closed. . . ),hence the statement holds for most matrices. Since it is just a bunch ofpolynomial relations between the matrix entries, it then must hold for allmatrices. This can indeed be turned into a proof, but unfortunately, thisrequires rather advanced tools from algebra.

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(3) Of course, the statement of the theorem remains true for endomorphisms.Let f : V V be an endomorphism of the finite-dimensional F -vectorspace V , then Pf (f) = 0 (which is the zero endomorphism in this case).For evaluating the polynomial at f , we have to interpret fn as the n-foldcomposition f f f , and f 0 = idV .

Our next goal is to define the minimal polynomial of a matrix or endomorphism,as the monic polynomial of smallest degree that has the matrix or endomorphismas a root. However, we need to know a few more facts about polynomials inorder to see that this definition makes sense.

2.3. Lemma (Polynomial Division). Let f and g be polynomials, with g monic.Then there are unique polynomials q and r such that r = 0 or deg(r) < deg(g)and such that

f = qg + r .

Proof. We first prove existence, by induction on the degree of f . If deg(f) 1 (in fact, m = 3), henceA is not diagonalizable.

On the other hand, the matrix (for F = R, say)

B =

1 2 30 4 50 0 6

has MB(x) = PB(x) = (x 1)(x 4)(x 6); B therefore is diagonalizable.Exercise: what happens for fields F of small characteristic?

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3. The Structure of Nilpotent Endomorphisms

3.1. Definition. A matrix A Mat(n, F ) is said to be nilpotent, if Am = 0 forsome m 1. Similarly, if V is a finite-dimensional vector space and f : V Vis an endomorphism, then f is said to be nilpotent if fm = f f f

m times

= 0 for

some m 1.It follows that the minimal polynomial of A or f is of the form xm, where m isthe smallest number that has the property required in the definition.

3.2. Corollary. A nilpotent matrix or endomorphism is diagonalizable if and onlyif it is zero.

Proof. The minimal polynomial is xm. Prop. 2.6 then implies that the matrixor endomorphism is diagonalizable if and only if m = 1. But then the minimalpolynomial is x, which means that the matrix or endomorphism is zero.

The following result tells us more about the structure of nilpotent endomorphisms.

3.3. Theorem. Let V be an F -vector space, dimV = n, and let f : V V bea nilpotent endomorphism. Then V has a basis v1, v2, . . . , vn such that f(vj) iseither zero or vj+1.

Proof. Let Mf (x) = xm. We will do induction on m. When m = 0 (then V = {0})

or m = 1, the claim is trivial, so we assume m 2. Then {0} ( ker(f) ( V ,and we can consider the quotient space W = V/ ker(f). Then f induces an

endomorphism f : W W that satisfies fm1 = 0. (Note that fm1(V ) ker(f).) So by induction, there is a basis w1, . . . , wk of W such that f(wj) = 0or wj+1. We can then write the basis in the form

(w1, . . . , wk) =(w1, f(w

1), . . . , f

e1(w1), w2, f(w

2), . . . , f

e2(w2), . . . , w`, . . . , f

e`(w`))

where f ej+1(wj) = 0 for j = 1, . . . , `. We now lift wj to an element v

j V (i.e.,

we pick vj V that maps to wj under the canonical epimorphism V W ). Thenv1, f(v

1), . . . , f

e1(v1), v2, f(v

2), . . . , f

e2(v2), . . . , v`, f(v

`), . . . , f

e`(v`)

are linearly independent in V (since their images in W are linearly independent;

note that the image of f i(vj) is fi(wj)). We must have f

ej+1(vj) ker(f) (sinceits image in W is zero). Note that their linear hull L is a complementary subspaceof ker(f). I claim that the extended sequence

v1, f(v1), . . . , f

e1+1(v1), v2, f(v

2),