Practice Exam 3
April 22, 2010
1. Let A be an n × n matrix.
a. If A is diagonalizable, then A3 is diagonalizable. Why?
Answer: A is diagonalizable means there’s an n × n invertible matrix P
and an n × n diagonal matrix D such that
A = P DP −1 .
Then
A3 = (P DP −1 )(P DP −1 )(P DP −1 ) = P D(P −1 P )D(P −1 P )DP −1 = P D3 P −1
You can easily check that if D is diagonal, then so is D3 . Therefore by definition
A3 is diagonalizable.
b. If A is invertible, then A3 is invertible. Why?
Answer: If A is invertible then det A 6= 0, so
det(A3 ) = det(AAA) = (det A)(det A)(det A) = (det A)3 6= 0.
Therefore A3 is invertible.
2. a. Find all h such that the following matrix not diagonalizable
1
h
1 −1
Answer: The characteristic polynomial of this matrix is
λ2 − (h + 1).
√
If h + 1 > 0, then this polynomial has two distinct roots, ± h + 1. The
matrix is therefore diagonalizable since it has distinct eigenvalues.
If h + 1 < 0, then the polynomial has complex roots. Therefore the matrix
is not diagonalizable (over R).
If h = −1, then λ = 0 is the only eigenvalue. Since the null space of
1 −1
1 −1
∼
1 −1
0
0
is the span of {(1, 1)}, there is only one linearly independent eigenvalue. Therefore the matrix is not diagonalizable.
b. Find all h such that the following matrix not diagonalizable


1 h 0
 0 1 0 
0 0 2
Practice Exam 3
April 22, 2010
Answer: If h = 0, then the matrix is diagonal, therefore diagonalizable.
So take the case h 6= 0. The eigenvalues of an upper triangular matrix are the
entries on the diagonal. Therefore, the eigenvalues of this matrix are λ = 1, 2.
We compute the eigenspaces for λ = 2




−1
h 0
1 0 0
Nul  0 −1 0  = Nul  0 1 0  = Span{(0, 0, 1)}
0
0 0
0 0 0
and for λ = 1

0
Nul  0
0


h 0
0
0 0  = Nul  0
0 1
0
1
0
0

0
1  = Span{(1, 0, 0)}
0
Where it’s very important that we’ve used the fact that h 6= 0 to scale the top
row. Thus if h 6= 0 then there are only two linearly independent eigenvectors,
so the matrix is not diagonalizable.
3. a. Define a linear transformation
T : P3 →
− P2
T : p(t) →
7− p0 (t)
For example, if p(t) = t3 − t, then
T (p(t)) = p0 (t) = 3t2 − 1
Find the matrix for T with respect to the bases B = {1, t, t2 , t3 } of P3 and
C = {1, t, t2 } of P2 .
Answer: Number the basis element of B by 1 = b1 , t = b2 , etc. Then the
matrix of T with respect to B and C is, by definition, the matrix
[T (b1 )]C [T (b2 )]C [T (b3 )]C [T (b4 )]C
d
d 3
t = 3t2 , and T (t) = dt
t = 1, etc., we get
Since T (t3 ) = dt
[T (1)]C [T (t)]C [T (t2 )]C [T (t3 )]C = [0]C [1]C [2t]C


0 1 0 0
= 0 0 2 0 
0 0 0 3
b. Multiplication by the matrix
A=
2
1
0
1
[3t2 ]C
Practice Exam 3
April 22, 2010
defines a linear transformation R2 →
− R2 . Find the matrix for this linear transformation with respect to the basis B = {(−1, 1), (1, 1)}.
Answer: Let T : x →
7− Ax be the linear transformation. Since the T maps
into Rn , there’s a short cut to this problem: take the matrix [ C | T (B) ] and
row reduce until you get the identity matrix on the left hand side:
C T (B) ∼ I M .
The matrix on the right hand side will be the answer. In this exercise, C = B =
{(−1, 1), (1, 1)}. We calculate
2 0
−1
−2
T (−1, 1) =
=
1 1
1
0
2 0
1
2
T (1, 1) =
=
1 1
1
2
So reducing the matrix,
C
T (B)
=
−1 1
1 1
−2
0
2
2
−1
0
1
0
1 0
0 1
∼
∼
∼
Therefore the answer is
1
−1
0
2
4. a. The transformation x →
7− Ax where
−3
A=
0
1
2
−2
−2
2
4
−1
2
1 −1
1
−1
0
2
−2
2
0
−3
is the composition of a scaling and rotation. Find the angle of rotation ϕ and
the scaling factor r.
Answer: This matrix is of the form ab −b
. Its eigenvalue is λ = −3; since
a
|λ| = 3 the matrix decomposes as
−3
0
3 0
−1
0
r 0
cos θ − sin θ
=
=
0 −3
0 3
0 −1
0 r
sin θ
cos θ
Thus r = 3, and θ has to satisfy
cos θ = −1
sin θ = 0
Practice Exam 3
April 22, 2010
The only −π < θ ≤ π that satisfies this is θ = π.
b. Find the eigenspaces of the matrix
1 3
B=
3 1
and prove they are orthogonal (Hint: just check the basis vectors).
Answer: The characteristic polynomial is
1−λ
3
det(B − λI) = det
3
1−λ
= (1 − λ)(1 − λ) − (3)(3) = λ2 − 2λ − 8
= (λ − 4)(λ + 2)
So the eigenvalues are λ = −2, 4. The corresponding eigenspaces are
3 3
1 1
Nul B + 2I = Nul
= Nul
= Span{(1, −1)}
3 3
0 0
and
Nul B − 4I = Nul
−3
3
3
−3
= Nul
−1
0
1
0
= Span{(1, 1)}
To show these two eigenspaces are orthogonal, we just have to check that the
basis vectors are orthogonal. This is true since
(1, −1) · (1, 1) = 1 · 1 + (−1) · 1 = 1 − 1 = 0.