MATH20300: Linear Algebra 2Homework 5: Solutions

1. Let

B =

0 0 01 0 00 1 1

.

Show that there is no real or complex 3×3 matrix C such that CBC−1

is diagonal.

Solution: The characteristic polynomial of B is

χB(x) = det

−x 0 01 −x 00 1 1− x

= −x · (−x(1− x)) = x2(1− x).

Thus the eigenvalues of B are 0 and 1.

The corresponding eigenspaces are

E0 = null(

0 0 01 0 00 1 1

) = span{

01−1

} = span{w1}

and

E1 = null(

−1 0 01 −1 00 1 0

) = span{

001

} = span{w2}.

Since dim(E0) + dim(E1) = 1 + 1 = 2 < 3, the matrix B is notdiagonalizable by Theorem 5.1.

2. Find the eigenvalues and corresponding eigenspaces of the 6×6 matrix

M =

1 1 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 2 1 00 0 0 0 2 00 0 0 0 0 −1

.

What are the dimensions of the eigenspaces? Is this matrix diagonal-izable?

Solution: The characteristic polynomial of M is

χM(x) = det

1− x 1 0 0 0 0

0 1− x 0 0 0 00 0 1− x 0 0 00 0 0 2− x 1 00 0 0 0 2− x 00 0 0 0 0 −1− x

= −(1−x)3(2−x)2(1+x).

So the eigenvalues ofM are 1, 2 and−1. The corresponding eigenspacesare

E1 = null(

0 1 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 1 1 00 0 0 0 1 00 0 0 0 0 −2

) = span{e1, e3},

E2 = null(

−1 1 0 0 0 00 −1 0 0 0 00 0 −1 0 0 00 0 0 0 1 00 0 0 0 0 00 0 0 0 0 −3

) = span{e5}

and

E−1 = null(

2 1 0 0 0 00 2 0 0 0 00 0 2 0 0 00 0 0 3 1 00 0 0 0 3 00 0 0 0 0 0

) = span{e6}.

Thus dim(E1) = 2 and dim(E2) = dim(E−1) = 1.

Since the sum of the dimensions of the eigenspaces is 2+1+1 = 4 < 6,M is not diagonalizable by Theorem 5.1.

3. Let 〈 , 〉 be an inner product on the vector space V . Prove that〈v, 0〉 = 0 = 〈0, v〉 for all v ∈ V .

Solution:Let v ∈ V .

Then 〈0, v〉 = 〈0+0, v〉 = 〈0, v〉+〈0, v〉 =⇒ 〈0, v〉 = 0 after subtracting〈0, v〉 form both sides.

It follows that 〈v, 0〉 = 〈0, v〉 = 0̄ = 0.

4. Let 〈 , 〉 be an inner product on the vector space V . Let v1, . . . , vn, w1, . . . , wm ∈V and a1, . . . , an, b1, . . . , bm ∈ K. Prove that

〈n∑i=1

aivi,

m∑j=1

bjwj〉 =

n,m∑i,j=1

aibj〈vi, wj〉.

Solution: We’ll first prove by induction on n that

〈n∑i=1

aivi, w〉 =n∑i=1

ai〈vi, w〉

for all w ∈ W :

The case n = 1 holds by definition. Suppose now that n > 1 and thatthe result has been proved for n− 1. Then

〈n∑i=1

aivi, w〉 = 〈n−1∑i=1

aivi + anvn, w〉

= 〈n−1∑i=1

aivi, w〉+ 〈anvn, w〉

=n−1∑i=1

ai〈vi, w〉+ an〈vn, w〉 using the inductive hyp.

=n∑i=1

ai〈vi, w〉.

It follows that for any v ∈ V we have

〈v,m∑j=1

bjwj〉 = 〈m∑j=1

bjwj, v〉

=m∑j=1

bj〈wj, v〉

=m∑j=1

bj〈wj, v〉

=m∑j=1

b̄j〈wj, v〉

=m∑j=1

b̄j〈v, wj〉.

Putting these together we deduce that

〈n∑i=1

aivi,

m∑j=1

bjwj〉 =n∑i=1

ai〈vi,m∑j=1

bjwj〉

=n∑i=1

ai

(m∑j=1

bj〈vi, wj〉

)

=

n,m∑i,j=1

aibj〈vi, wj〉.

5. Let (V, 〈 , 〉) be an inner product space over K. Let W be a vectorspace over K and let T : W → V be an injective linear transformation.Define the pairing 〈 , 〉T on W by

〈w1, w2〉T := 〈T (w1), T (w2)〉.

Prove that 〈 , 〉T is an inner product on W . Where did you use thefact that T is injective?

Solution:

(a) For all w1, w2, w ∈ W

〈w1 + w2, w〉T = 〈T (w1 + w2), T (w)〉= 〈T (w1) + T (w2), T (w)〉 since T is linear

= 〈T (w1), T (w)〉+ 〈T (w2), T (w)〉= 〈w1, w〉T + 〈w2, w〉T .

(b) For all w1, w2 ∈ W,a ∈ K

〈aw1, w2〉T = 〈T (aw1), T (w2)〉= 〈aT (w1), T (w2)〉 since T is linear

= a〈T (w1), T (w2)〉= a〈w1, w2〉T .

(c) For all w1, w2 ∈ W

〈w2, w1〉T = 〈T (w2), T (w1)〉 = 〈T (w1), T (w2)〉 = 〈w1, w2〉T .

(d) For all 0 6= w ∈ W , we have T (w) 6= 0 in V since T is injectiveand hence

〈w,w〉T = 〈T (w), T (w)〉 > 0.

6. For non-zero vectors v, w in a real inner product space V , the Cauchy-Schwarz inequality tells us that

〈v, w〉‖v‖‖w‖

∈ [−1, 1].

We define the angle between v and w to be

θ(v, w) := cos−1

(〈v, w〉‖v‖‖w‖

)∈ [0, π].

Let V be the real vector space of all real-valued continuous functionsf : [0, 2π]→ R with the inner product

〈f, g〉 =

∫ 2π

0

f(t)g(t) dt.

Let f, g ∈ V be the functions f(t) = t, g(t) = t2. Calculate θ(f, g),the angle between these functions.

Solution: We have

‖f‖2 =

∫ 2π

0

t2 dt =(2π)3

3, ‖g‖2 =

∫ 2π

0

t4 dt =(2π)5

5.

Thus

‖f‖‖g‖ =

√(2π)8

15=

(2π)4√15

.

On the other hand

〈f, g〉 =

∫ 2π

0

t3 dt =(2π)4

4.

Thus〈f, g〉‖f‖‖g‖

=

√15

4

and hence

θ(f, g) = cos−1

(√15

4

)≈ 0.25268 radians.

7. Let V = R2 with the inner product

〈(x1, x2), (y1, y2)〉 = 2x1y1 + x1y2 + x2y1 + 2x2y2.

Apply the Gram-Schmidt process to the standard basis of R2 to findan orthonormal basis for this inner product.

Solution: We have ‖e1‖2 = 〈e1, e1〉 = 2. So ‖e1‖ =√

2 and so wecan take

u1 =1√2e1 =

(1√2, 0

).

Now

〈e2, u1〉 =1√2〈e2, e1〉 =

1√2.

Thus we let

w2 = e2 − 〈e2, u1〉u1 = e2 −1√2u1 = e2 −

1

2e1 =

(−1

2, 1

).

Note that

‖w2‖2 = 〈w2, w2〉 =1

2− 1

2− 1

2+ 3 =

3

2

and hence ‖w2‖ =√

32.

So we let

u2 =1

‖w2‖w2 =

√2

3

(−1

2, 1

)=

(− 1√

6,

√2

3

).

So

{u1, u2} =

{(1√2, 0

),

(− 1√

6,

√2

3

)}is an orthonormal basis of this inner product space.

8. Let V be the real vector space R[x]≤2. Equip V with the inner product

〈p(x), q(x)〉 :=

∫ 1

−1

p(x)q(x) dx.

Apply the Gram-Schmidt process to the basis {1, x, x2} to find anorthonormal basis {u1, u2, u3} of V for this inner product.

Solution:

We will start with the standard basis of V , {v1, v2, v3} = {1, x, x2},and use Gram-Schmidt to find an orthonormal basis.

Now

‖v1‖2 = ‖1‖2 = 〈1, 1〉 =

∫ 1

−1

12dx = 2

so that ‖1‖ =√

2 and we take u1 := 1/√

2.

We have

〈v2, 1〉 = 〈x, 1〉 =

∫ 1

−1

x dx =1

2− 1

2= 0.

So 〈v2, u1〉 = 0, p2 = 0 and w2 = v2 = x.

Now

‖x‖2 =

∫ 1

−1

x2 dx =x3

3|1−1 =

1

3−(−1

3

)=

2

3.

Thus ‖w2‖ = ‖x‖ =√

2/3 and hence

u2 :=

√3

2x.

〈x2, 1〉 =

∫ 1

−1

x2 dx =2

3.

Thus 〈v3, u1〉 = 〈x2, u1〉 =√

2/3.

〈x2, x〉 =

∫ 1

−1

x3 dx = 0.

Thus 〈v3, u2〉 = 0.

So

p3 =

√2

3u1 =

1

3and hence

w3 = x2 − 1

3.

Now

‖w3‖2 =

∫ 1

−1

(x2 − 1

3

)2

dx =

∫ 1

−1

x4 − 2

3x2 +

1

9dx =

8

45.

Thus ‖w3‖ =√

845

and thus

u3 :=

√45

8

(x2 − 1

3

).

So

{u1, u2, u3} =

{1√2,

√3

2x,

√45

8

(x2 − 1

3

)}is an orthonormal basis of this inner product space.