1. (5 points each) State whether the following are true or false, and justify all of your answers.

a. Let A be a 3×3 matrix with eigenvectors v1 = , v2 = , v3 = . A is

diagonalizable.

Solution:

TRUE. These vectors are linearly independent since:

~ ~ the columns are linearly independent.

Thus we have 3 linearly independent eigenvectors for a 3×3 matrix, which implies that A is diagonalizable by The Diagonalization Theorem.

b. det (ATA) ≥ 0.

Solution:

TRUE.

det (ATA) = det (AT)(det A) since det(AB) = (det A)(det B) = (det A)(det A) since det A = det (AT) = (det A)2 ≥ 0 since det A R

c. It is possible for a 3×3 matrix with real entries to have the eigenvalues 2, -3, and i.

Solution:

FALSE.

Complex eigenvalues of a real-valued matrix come in conjugate pairs, so -i would also need to be an eigenvalue.

1. (continued)

d. If J is the Jordan canonical form of a matrix A, then A is similar to J.

Solution:

TRUE.

If J is the Jordan canonical form of a matrix A, then A = PJP-1, which implies that A is similar to J.

e. Let A = . The vector v = is in Nul(A).

Solution:

FALSE.

A vector v is in Nul(A) if Av = 0.

Here, Av = = 0. Thus v Nul(A).

f. Each eigenvalue of A is also an eigenvalue of A2.

Solution:

FALSE.

If A = , then the eigenvalues of A are 1 and 2. But A2 = , and its

eigenvalues are 1 and 4. Thus the eigenvalue 2 of A is not an eigenvalue of A2.

In general, if is an eigenvalue of A, then 2 is an eigenvalue of A2.

2. Let u = , v = , w = be vectors in R3.

a. (5 points) What is the distance from u to w?

Solution:

dist(u, w) = ║u – w║ = ║ ║ = = =

b. (10 points) Find the component of u orthogonal to v.

Solution:

Let L = Span{v}.

u – projLu = – = – =

c. (10 points) Find the projection of u onto the subspace of R3 spanned by v and w. [NOTE: To use the nice dot product formula, what kind of set/basis must you have??]

Solution:

We must have an orthogonal basis for Span{v, w} – we use Gram-Schmidt!

Let v1 = v; v2 = w – = – = .

Then, if W = Span{v, w}:

projWu = + = + = =

d. (10 points) Find the distance from u to the subspace of R3 spanned by v and w.

Solution:

Let W = Span{v, w}.

dist(u, W) = ║u – projWu║ = ║ – ║ = ║ ║ =

= =

3. (20 points) Let u1, …, up be an orthogonal basis for a subspace W of Rn, and let T: Rn→ Rn be defined by T(x) = projW x. Show that T is a linear transformation.

Solution:

Since u1, …, up is an orthogonal basis, we can write:

projW x = + … +

To show that T is a linear transformation, we must show 2 things:

(i) T(x + y) = T(x) + T(y):

T(x + y) = + … +

= + … + (prop’s of dot product)

= + … + + + … + (prop’s of vectors)

= T(x) + T(y)

(ii) T(cx) = cT(x):

T(cx) = + … +

= + … + (prop’s of dot product)

= (prop’s of vectors)

= cT(x)

Thus, T is a linear transformation.

4. Let a subspace U of R5 be defined by: U = {(x1, x2, x3, x4, x5): x1 = 3x2, x3 = 7x4}

a. (15 points) Find a basis for U.

Solution:

Vectors in U look like:

= x2 + x4 + x5 . Therefore a basis is:

b. (5 points) What is the dimension of U?

Solution:

The dimension of U is the number of elements in the basis, which is 3.

5. Given the quadratic form on R3: Q(x) = x12 + 25x2

2 – 10x1x2

a. (5 points) What is the matrix of this quadratic form?

Solution:

A =

b. (15 points) Make a change of variable that transforms this quadratic form into one with no cross- product terms. Tell me what the P matrix is, and write what the new quadratic form looks like in the new variable.

Solution:

To make a change of variable to satisfy the conditions that we want, we find a P such that x = Py, and with yT(PTAP)y a quadratic form with no cross product terms (i.e. with PTAP a diagonal matrix).

So, we want to orthogonally diagonalize A! Meaning we find an orthonormal basis for R2 such that PTAP is diagonal, where P is the matrix whose columns are the orthonormal basis.

The eigenvalues of A are the roots of the polynomial:(1 – )(25 – ) – 25 = 2 – 26 = ( – 26)

eigenvalues are 0 and 26

Now we find bases for the eigenspaces:

For = 0: A – 0I = ~ basis is

For = 26: A – 26I = ~ basis is

Since these eigenvectors correspond to different eigenvalues, they are orthogonal. So to get an orthonormal basis we just normalize.

u1 = , u2 =

P = and the new quadratic form is 0y12 + 26y2

2.

6. (20 points) Suppose that T1, …, Tn are injective linear transformations such that T1◦…◦ Tn makes sense (i.e. the codomain of Ti+1 is the domain of Ti.) Prove that T1◦…◦ Tn is injective. [Recall: T1◦…◦ Tn(x) = T1(…(Tn(x))…) = composition of functions]

Solution:

We can prove this in a few ways… I’ll do 2:

(i) A transformation T is one-to-one if an only if ker(T) = {0}

Since T1, …, Tn are injective, ker(T1) = {0}, …, ker(Tn) = {0}.

So now, what is in the kernel of T1◦…◦ Tn?

Assume T1◦…◦ Tn(x) = T1(…(Tn(x))…) = 0. We show that x = 0.

T1 injective T2(…(Tn(x))…) = 0.T2 injective T3(…(Tn(x))…) = 0.

………….Tn-1 injective Tn(x) = 0.Tn injective x = 0.

Thus, T1◦…◦ Tn(x) = 0 x = 0, which means that ker(T1◦…◦ Tn) = {0}, and T1◦…◦ Tn is injective.

(ii) A transformation T is one-to-one if and only if [T(x) = T(y) x = y]

So Ti(x) = Ti(y) x = y for all i = 1, …, n.

Assume T1◦…◦ Tn(x) = T1◦…◦ Tn(y). We show that x = y.

T1 injective T2(…(Tn(x))…) = T2(…(Tn(y))…)T2 injective T3(…(Tn(x))…) = T3(…(Tn(y))…)

………….Tn-1 injective Tn(x) = Tn(y)Tn injective x = y

Thus T1◦…◦ Tn(x) = T1◦…◦ Tn(y) x = y , which means that T1◦…◦ Tn is injective.

7.a. (5 points) What is the definition of orthogonally diagonalizable?

Solution:

A matrix A is orthogonally diagonalizable if there exist an orthogonal matrix P and a diagonal matrix D such that A = PDP-1 = PDPT.

b. (20 points) Orthogonally diagonalize the matrix A = , given that its

eigenvalues are 7 and -2.

Solution:

We want to find an orthonormal basis for R3.

Since we have the eigenvalues already, we find bases for the eigenspaces:

For = 7: A – 7I = ~ ~

basis is

For = -2: A + 2I = ~ ~ ~

basis is OR

The basis that we found for the eigenspace of = 7 is not orthogonal, so we use Gram-Schmidt to make it orthogonal:

v1 = ; v2 = – =

So {v1, v2} is an orthogonal basis for the eigenspace of = 7, and {v1, v2, } is an

orthogonal basis for R3 since corresponds to a different eigenvalue than v1 and v2,

which implies that it is orthogonal to both v1 and v2.

Now we just have to normalize:

u1 = , u2 = , u3 = .

Thus A = PDPT for

P = and D = .

8. (15 points) Show that two vectors u and v of an inner product space V are orthogonal if and only if ║u – v║2 = ║u║2 + ║v║2.

Solution:

First, we note that:

║u – v║2 = <u – v, u – v > (by the def of ║u║2) = <u, u> – <u, v> – <v, u> + <v, v> (by prop’s of inner product) = <u, u> – 2<u, v> + <v, v> (by prop’s of inner product) = ║u║2 – 2<u, v> + ║v║2 (by the def of ║u║2)

Then ║u – v║2 = ║u║2 + ║v║2 -2<u, v> = 0 <u, v> = 0 u and v are orthogonal.

9.a. (5 points) State the Cauchy-Schwarz Inequality.

Solution:

For all u, v in a vector space V, <u, v> ║u║║v║.

b. (5 points) Circle one: “Cauchy” is pronounced cow-shee coh-shee cow-chee

c. (10 points) Let u = and v =

11

. Show that .

Solution:

We use Cauchy-Schwarz!

<u, v> = a + b ║u║ = ║v║ =

Then Cauchy-Schwarz a + b ( )( )

(a + b)2 (a2 + b2)(2)

a2 + b2

.