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MATH 415, SECTION D
REVIEW SHEET 3
SOLUTIONS
BERT GUILLOU
True or False
- The symmetric matrix A is positive definite if and only if det(A) > 0.
False. If A is positive definite, then det(A) must be positive, but there are examples of
symmetric matrices with positive determinant that are not positive definite. For instance,(
− 1 0 0 − 1
has determinant equal to 1, but it is negative definite.
- If A is negative definite, then det(A) < 0.
False. The example
from above is negative definite and has determinant 1.
- If A is positive definite, then A is invertible.
True. A is a square matrix with det(A) > 0. Since det(A) 6 = 0, A is invertible.
- If A is positive semi-definite, then A is invertible.
False. The zero matrix is positive semi-definite but not invertible.
In problems 5 through 9, A and B denote square matrices of the same size.
- If λ is an eigenvalue for A and σ is an eigenvalue for B, then λσ is an eigenvalue for
AB.
False. Take A =
and B =
. Then AB =
. So 3 is an
eigenvalue for A, 2 is an eigenvalue for B, but 6 is not an eigenvalue for AB.
The statement is true if the eigenvectors for λ and σ are the same.
- If λ is an eigenvalue for A and σ is an eigenvalue for B, then λ + σ is an eigenvalue for
A + B.
Date: November 17, 2009.
2 BERT GUILLOU
False. Take A and B as in the previous problem. Then A + B =
. Again, 3 is an
eigenvalue for A, 2 is an eigenvalue for B, but 5 is not an eigenvalue for A + B.
The statement is true if the eigenvectors for λ and σ are the same.
- If v is an eigenvector for A and w is an eigenvector for B, then v + w is an eigenvalue
for A + B.
False. Take A and B as above. Then
is an eigenvector for A and
is an
eigenvector for B. But
is not an eigenvector for A + B.
The statement is true if the eigenvalues for v and w are the same.
- If v is an eigenvector for A and also for B, then v is an eigenvalue for A + B.
True. If Av = λv and Bv = σv, then (A + B)v = (λ + σ)v.
- If v is an eigenvector for A and also for B, then v is an eigenvalue for AB.
True. If Av = λv and Bv = σv, then (AB)v = (λσ)v.
- If A and B are similar, they have the same eigenvalues.
True. If Bv = λv and A = SBS−^1 , then A(Sv) = SBv = Sλv = λSv, so λ is an
eigenvalue for A.
- If A and B are similar, they have the same eigenvectors.
False. If Bv = λv and A = SBS−^1 , then A(Sv) = SBv = Sλv = λSv, so Sv is an
eigenvector for A, but there is no reason for v to be an eigenvector. For instance, take
B =
and A =
. Then B
but
A
, so
is not an eigenvector for A.
- If A and B are similar, they have the same rank.
True. If A = SBS−^1 and b 1 ,... , br is a basis for the column space of B, then Sb 1 ,... , Sbr
is a basis for the column space of A, so they have the same rank,
4 BERT GUILLOU
(ii) The characteristic polynomial is
λ 2 − 10 λ + 25 = (λ − 5) 2 ,
so the only eigenvalue is 5 (with multiplicity 2).
(iii) The characteristic polynomial is
λ 2 − 3 λ + 3.
The quadratic formula then gives the eigenvalues as λ = 3 2
√ 3 i 2
(iv) The characteristic polynomial is
(− 5 − λ)(−λ)(5 − λ) + 12(− 2 λ) = λ(25 − λ 2 − 24) = λ(1 − λ)(1 + λ),
so the eigenvalues are −1, 1, and 0.
- Find the eigenspaces for each of the eigenvalues in Problem 2. Which matrices are
diagonalizable?
Solution. (i) The eigenspace for 4 is N
= span
and the eigenspace
for −1 is N
= span
. The matrix is automatically diagonalizable since
it has 2 distinct eigenvalues.
(ii) The eigenspace for 5 is N
= span
. Since this eigenvalue has
multiplicity 2, the matrix is not diagonalizable.
(iii) The eigenspace for 3 2 +^
√ 3 i 2 is
N
3 i)/ 2 − 1 1 (1 −
3 i)/ 2
= span
− 1 −
√ 3 i 2
It follows that the eigenspace for the conjugate eigenvalue is the complex conjugate sub-
space, span
−1+
√ 3 i 2
. The matrix is automatically diagonalizable since it has 2 distinct
eigenvalues.
(iv) The eigenspace for −1 is
N
= N
(^) = span
MATH 415, SECTION D1 REVIEW SHEET 3 SOLUTIONS 5
The eigenspace for 1 is
N
= N
(^) = span
Finally, the eigenspace for 0 is
N
= N
(^) = span
The matrix is automatically diagonalizable since it has three distinct eigenvalues.
- Diagonalize the matices in Problem 2 that are diagonalizable (that is, write A = SDS−^1
for appropriate matrices S and D).
Solution. According to what we found in Problems 2 and 3,
(i)
(iii)
− 1 −
√ 3 i 2
−1+
√ 3 i 2
3+
√ 3 i 2 0 0 3 −
√ 3 i 2
− 1 −
√ 3 i 2
−1+
√ 3 i 2
− 1 −
√ 3 i 2
−1+
√ 3 i 2
3+
√ 3 i 2 0 0 3 −
√ 3 i 2
−i √ 3
−1+
√ 3 i 2 −^1 1+
√ 3 i 2 1
(iv)
− 1
- Find the spectral decomposition of the following symmetric matrices:
(i)
(ii)
(iii)
MATH 415, SECTION D1 REVIEW SHEET 3 SOLUTIONS 7
(iv) The characteristic polynomial is
−λ^3 + 7λ + 6 = (λ + 1)(−λ^2 + λ + 6) = −(λ + 1)(λ + 2)(λ − 3),
so the eigenvalues are 3, −1 and −2. We find corresponding eigenvectors v 3 =
v− 1 =
, and v− 2 =
. These do not have norm 1, so we normalize to get
(^) and
. The spectral decomposition is then
- Which of the matrices in Problem 5 are positive-definite (or positive semi-definite)?
Solution. For this, we only have to look at the eigenvalues. (i) Both eigenvalues are positive, so the matrix is positive-definite.
(ii) We have one positive eigenvalue and one eigenvalue of 0, so the matrix is only positive
semi-definite.
(iii) Both eigenvalues are positive, so the matrix is positive definite.
(iv) There are both positive and negative eigenvalues, so the matrix is not even positive
semi-definite.
- Determine whether the following functions can take on negative values by expressing
them as sums or differences of squares.
(i) 9x^2 + 12xy + 4y^2.
(ii) 2x 2 − 8 xy + 11y 2 .
Solution. (i) We first rewrite the function as
9 x 2
x y
x y
The spectral decomposition from problem 5(ii) allows us to write the function as
9 x 2
3 x + 2y √ 13
= (3x + 2y) 2 .
8 BERT GUILLOU
This function never takes on negative values, but it does produce a value of 0 for some
nonzero vectors
x y
. Plugging in x = 2 and y = −3, for instance, gives a value of 0.
(ii) We first rewrite the function as
2 x 2 − 8 xy + 11y 2 =
x y
x y
We could then use the spectral decomposition of this matrix found in problem 5(iii). Another
method is to use the LDV factorization of the matrix, which is quite a bit simpler. Here we
have (^) (
2 − 4 − 4 11
so that (^) (
2 − 4 − 4 11
From this we can see that
2 x 2 − 8 xy + 11y 2 = 2(x − 2 y) 2
This shows that the function can never be negative, and we can also deduce from this
factorization that the function only attains the value of 0 if x = 0 and y = 0.
- Construct a symmetric matrix with the following eigenvectors and eigenvalues, or
explain why none exists:
(i) λ 1 = −2, λ 2 = 1, v 1 =
, v 2 =
(ii) λ 1 = 3, λ 2 = 0, λ 3 = −1, v 1 =
, v 2 =
, v 3 =
(iii) λ 1 =
3, λ 2 = −2, λ 3 = 5, v 1 =
, v 2 =
, v 3 =
Solution. (i) The given vectors are orthogonal, so if we normalize them we will get an
orthogonal matrix, and
A =
(ii) v 1 is not orthogonal to v 3 , so no symmetric matrix can have these as eigenvectors.
10 BERT GUILLOU
The solution to the differential equation is then given by
u(t) = e At
e^4 t^0 0 e −t
e^4 t^0 0 e−t
− 9 e^4 t 7 e−t
− 9 e^4 t^ + 7e−t − 6 e^4 t^ + 21e−t
- Give a Jordan form for each of the following matrices:
(i)
(^) (ii)
(iii)
Solution. (i) We must find the eigenvalues and the dimensions of the eigenspaces. The
characteristic polynomial is
(4 − λ)[(11 − λ)(− 9 − λ) + 96] − 12[(− 9 − λ) + 12] − 18[8 − (11 − λ)]
= (4 − λ)[λ 2 − 2 λ − 3] + 12λ − 36 − 18 λ + 54
= −λ 3
- 6λ 2 − 5 λ − 12 − 6 λ + 18 = −λ 3
- 6λ 2 − 11 λ + 6.
Since 1 is a root of this polynomial, we can factor out a λ − 1 to get
−λ 3
- 6λ 2 − 11 λ + 6 = (λ − 1)(−λ 2
- 5λ − 6) = −(λ − 1)(λ − 2)(λ − 3),
so the eigenvalues are 1, 2, and 3. Since this matrix has three distinct eigenvalues, it follows
that it is diagonalizable and so a Jordan form is
(ii) We must find the eigenvalues and the dimensions of the eigenspaces. The characteristic
polynomial is
(7 − λ)[(− 3 − λ)(− 3 − λ) − 18] − 4[−6(− 3 − λ) − 27] − [− 108 − 27(− 3 − λ)]
= (7 − λ)[λ 2
- 6λ − 9] − 24 λ + 36 − 27 λ + 27
= −λ 3
- λ 2
- 51λ − 63 − 51 λ + 63 = −lambda 3
- λ 2 = −λ 2 (λ − 1).
So the eigenvalues are 0, 0, and 1. To determine the Jordan form, we need to know if there
are two eigenvectors for 0 or just 1. In other words, we want to know the dimension of the
MATH 415, SECTION D1 REVIEW SHEET 3 SOLUTIONS 11
null space of the matrix. Since
we find that the rank of the matrix is 2 and the dimension of the null space is just 1. This
means that the Jordan form of the matrix is
(iii) We must find the eigenvalues and the dimensions of the eigenspaces. The characteristic
polynomial is
(− 9 − λ)[(10 − λ)(− 1 − λ) + 8] − 23[−4(− 1 − λ) − 4] + 9[8 − (10 − λ)]
= (− 9 − λ)[λ 2 − 9 λ − 2] − 92 λ + 9λ − 18
= −λ 3
- 83λ + 18 − 83 λ − 18 = −λ 3 .
So 0 is a triple eigenvalue. To determine the Jordan form, we need to find the dimension of
the null space.
Since the null space is only 1-dimensional, this means the Jordan form is
- Find the Singular Value Decomposition of the following matrices.
(i)
(ii)
Solution. (i) We normally begin by finding the v’s, which are an orthonormal basis for
R^3 consisting of eigenvectors of the 3 × 3 matrix AT^ A. On the other hand, the vectors u
will be an orthonormal basis for R^2 consisting of eigenvectors for the 2 × 2 matrix AAT^. So
MATH 415, SECTION D1 REVIEW SHEET 3 SOLUTIONS 13
(ii) We begin by finding a basis for R^3 consisting of eigenvectors for
A
T A =
The characteristic polynomial is
(6 − λ)[(6 − λ)(3 − λ) − 9] + 3[−3(6 − λ)] = (6 − λ)[λ 2 − 9 λ + 9 − 9] = (6 − λ)λ(λ − 9).
The eigenvalues are λ 1 = 9, λ 2 = 6 and λ 3 = 0. From this, we find orthonormal eigenvectors
of
v 1 =
(^) , v 2 =
(^) and v 3 =
The singular values are σ 1 = 3 and σ 2 =
6 (0 does not count as a singular value). Then
we have
u 1 = Av 1 /σ 1 =
and^ u^2 =^ Av^2 /σ^2 =^
Again, we can complete this to an orthonormal basis for R 4 by finding an orthonormal basis
for N (AA T ) = N (A T ). We have
A
T
An orthonormal basis for the null space is given by
u 3 =
and
The singular value decomposition of the matrix is then
14 BERT GUILLOU
- In the previous problem, give the rank one approximation of each of the matrices
coming from the first (largest) singular value and corresponding singular vectors.
Solution. (i) The rank one approximation is ( 1 /
(ii) The rank one approximation is
