Download Problem 1. Let R 2×2 denote the vector space of 2 × 2 real matrices

Problem 1. Let R2×2 denote the vector space of 2 × 2 real matrices, and define a transformation
S : R2×2 → R2×2 ,
S(A) = AT − A ,
where AT is the transpose of the matrix A.
(i) Show that the transformation S is linear.
(ii) Let S 2 = S ◦ S denote the composition of the map S with itself. Show that S 2 = −S.
(iii) Find a basis of the image of S and a basis of the kernel of S. Explain how the dimensions of
these subspaces are related.
(iv) Describe a map T : R2×2 → R2×2 such that the image of S is the kernel of T , and the kernel of
S is the image of T .
Solution:
(i) The cheap way to show linearity is to use indicial notation: if A = (aij ), then S(A) = (aji −aij ).
Given any two matrices A, B ∈ R2×2 and a scalar x ∈ R, one then has
S(A + xB) = (aij + xbij )T − (aij + xbij ) = (aji − aij ) + x(bji − bij ) = S(A) + xS(B) .
Alternatively, one can use that (A + B)T = AT + B T , and (xA)T = x(AT ).
(ii)
S(A)T = (AT − A)T = (AT )T − AT = A − AT = −S(A)
S 2 (A) = S(AT − A) = −(A − AT ) − (AT − A) = −2S(A) .
(iii) The kernel is characterized by the symmetry of its elements: A ∈ ker S ⇐⇒ AT − A = 0 ⇐⇒
AT = A, thus
1 0
0 0
0 1
ker S = span
,
,
.
0 0
0 1
1 0
To understand the image, note that since S 2 = −2S and also S(A)T = −S(A) every matrix in
the image is skew symmetric: A ∈ S(R2×2 ) =⇒ AT = −A. Moreover, every skew symmetric
matrix is in the image: if B T = −B, then
1
1
S − 21 B = − (B T − B) = − (−B − B) = B .
2
2
Thus the image can be expressed as
0 −1
S(R2×2 ) = span
.
1 0
Note that the kernel has dimension 3 and the image has dimension 1. The relevant relationship is the rank and nullity theorem:
4 = dim R2×2 = dim S(R2×2 ) + dim ker S = 3 + 1 .
(iv) Let T : R2×2 → R2×2 be the map defined by T (A) = AT + A. Then
ker A = {A ∈ R2×2 | AT = −AT } = S(R2×2 )
T (R2×2 ) = {A ∈ R2×2 | AT = A} = ker S .
1
2
Problem 2. Find all eigenvalues of the matrix


1
0 6
A = −5 −1 2
1
0 0
Solution.


1−λ
0
6
det  −5 −1 − λ 2  = (−1 − λ)(−λ(1 − λ) − 6) = −(1 + λ)(λ − 3)(λ + 2)
1
0
−λ
So the eigenvalues are 3, −2, and −1.
Problem 3. Let A be the matrix


5 0 0
A =  2 1 1
−2 2 0
Determine if an eigenbasis for A exists. If one does exists, find an eigenbasis for A. If an eigenbasis for
A does not exist, explain why not.
Solution. Yes, an eigenbasis exists. The eigenvalues of A are 5, 2, and −1, and the corresponding
eigenvectors form the eigenbasis
     
0
0 
 −9
−4 , 1 , −1


2
2
1
Problem 4. True or false.
(i) If A is a square matrix and ~v is an eigenvector of A with eigenvalue λ, then −~v is an eigenvector
of A with eigenvalue −λ.
(ii) If A is a square matrix and an eigenbasis for A exists, then A has n distinct eigenvalues.
(iii) If A is a 2 × 2 matrix such that both standard basis vectors are eigenvectors of A, then A is a
diagonal matrix.
(iv) If A and B are n × n matrices, ~v is an eigenvector of A, and w
~ is an eigenvector of B, then
~v + w
~ is an eigenvector of the matrix A + B.
(v) If A is not invertible, then 0 is an eigenvalue of A.
Solution.
(i) False. (It has eigenvalue λ. Av = λv =⇒ A(−v) = −Av = −λv = λ(−v))
(ii) False. (Consider the identity matrix)
(iii) True. (This is the definition of diagonal. Or, A~e1 = k~e1 =⇒ the first column of A is k~e1 .
Similarly for second column.)
1 1
1
1 0
1
(iv) False. (Consider A =
, ~v =
,B=
, and w
~=
.)
0 0
0
1 0
1
(v) True. (A not invertible means there’s a nonzero ~v in the kernel of A, so A~v = 0~v .)
3
Problem 5.
(i) Write the definition of what it means for λ to be an eigenvalue of a linear transformation T .
(ii) Let S be the set of all sequences (x1 , x2 , x3 , . . .) of real numbers. Consider the linear left-shift
map:
L : S → S : L(x1 , x2 , x3 , . . .) = (x2 , x3 , x4 , . . .)
Show that every real number λ is an eigenvalue of L.
Solution:
(i) λ is an eigenvalue of T if there exists a nonzero vector ~v so that T (~v ) = λ~v .
(ii) ~v = (1, λ, λ2 , λ3 , λ4 , . . . ) is an eigenvector: S(~v ) = λ~v .
Problem 6.
(i) Compute the determinant of the matrix


2 3 1
A = 1 c 2
4 6 5
where c ∈ R is a constant.
(ii) For what values of c is the matrix A invertible?
Solution
(i) 6c − 9.
(ii) c 6= 3/2.
Problem 7.
(i) Recall that M2 (R) denotes the vector space of 2 × 2 matrices with real number entries. Let W
be the subset of matrices which equal their own transpose, i.e.
W = {A ∈ M2 (R)|A = AT }.
(Such matrices are called symmetric.)
(a) Show that W is a subspace.
(b) Find a basis for W . What is the
dimension of W ?
2 −7
(c) Write the matrix A = −7 5 in coordinates using the basis you found in part (b).
Solution.
(i) We check that W is a subspace using the definition.
(a) Certainly the zero matrix is in W .
(b) If A = AT and B = B T , then (A + B)T = AT + B T = A + B, so W is closed under
addition.
(c) If A = AT then for any scalar c we have (cA)T = c(AT ) = cA, so W is closed under scalar
multiplication.
4
(ii) Let A =
then
a b
c d
. Then AT = ( ab dc ), so if A = AT then b = c, while a and d are free. So if A ∈ W
A=
a b
b d
= a ( 10 00 ) + b ( 01 10 ) + d ( 00 01 ) .
So a basis is given by
1 0
0 1
0 0
,
,
0 0
1 0
0 1
and this subspace has dimension
 3.
2
(iii) In coordinates, this matrix is −7.
5
Problem 8
Let P2 denote the vector space of polynomials of degree at most 2. Consider the linear transformation
T : P2 → P2
given by
T (p(x)) = p(x) − xp0 (x)
(you do not need to justify that this is a linear transformation). Consider the basis
B = {x2 , x, 1}
of P2
(you do not need to justify that this is a basis of P2 . Also, pay attention to the order of elements.)
(i) Find the matrix A of the linear transformation T with respect to the basis B, i.e. find the
matrix of TB .
(ii) Find bases for the kernel and the image of the matrix A.
(iii) Use your answer in part (b) to find bases for the kernel and the image of T .
Solution
(i) We have
T (x2 ) = x2 − x(x2 )0 = −x2 ,
T (x) = x − x(x)0 = 0,
T (1) = 1 − x(1)0 = 1.
We can write these elements in terms of the basis B as


 
 
−1
0
0
[T (x2 )]B =  0  , [T (x)]B =  0  , [T (1)]B =  0  .
0
0
1
Then the matrix of T with respect to the basis

−1 0
A= 0 0
0 0
B is

0
0 
1
(ii) The reduced row-echelon form of A is


1 0 0
rref A =  0 0 1 
0 0 0
5
The general solution of the system A~x = 0 is
  
 

0
0
x1
~x =  x2  =  t  = t  1  .
0
0
x3
The kernel and the image of A have bases
 

  
0 
 0 
 −1
 1 
 0 , 0 
and




0
0
1
respectively.
(iii) The vector


0
 1 ,
0
is the coordinate vector of the polynomial x, i.e.
polynomial x forms a basis for ker T .
Similarly, the vectors



−1
 0  and 
0
[x]B . Since it forms a basis for ker A, the

0
0 
1
are the coordinate vectors of the polynomials −x2 and 1, respectively. Since they form a basis
for im A, the polynomials −x2 and 1 form a basis for im T .
Problem 9. Compute A100 for
Solution
7 −8
.
4 −5
100
2(3 ) − 1 2 − 2(3100 )
3100 − 1
2 − 3100
Problem 10. True or False, justify your answer
(i) If A and B are diagonalizable then so is A + B.
(ii) If A is diagonalizable so is AT
Solution
(i) False
(ii) True