Winter 2014
Math325
Solutions to Problems 8
1) By Problems 6, 1) the map
5 1
(v, w) < v, w := v t
1 1
w
is an inner product on R2 . Hence if the columns of a matrix S are an
orthonormal basis v1 , v2 of R2 with respect to this inner product we have
S t A
Winter 2014
Math 325
Problems 1
01/15/2014
1) Let A =
1 4
3 1
and A : R2 R2 , v A v be the corresponding R-linear map.
Give the matrix of the linear map A with respect to the basis
2) Let A =
2 1
0 3
2
2
,
2
3
of R2 .
. What are the A-invariant subspaces
Winter 2016
Math 325
Problems 4
02/24/2016
1) Compute the Jordan normal form of the following 3 3-matrix
3
1 2
A = 1 1 1 ,
2
1 1
and give a Jordan basis for A. (Hint: 1 is an eigenvalue of A.)
2) Compute the Jordan normal form of the following 3 3-matrix
Winter 2016
Math325
Solutions to Problems 3
0 1
1) Set A :=
1
0
!
and B :=
2 1
1 4
!
. We first compute the characteristic
polynomials:
PA (T ) = T 2 + 1 = (T ) (T + ) ,
where =
1 is square root of 1, and
PB (T ) = T 2 6T + 9 = (T 3)2 .
Therefore A is dia
Winter 2016
Math 325
Problems 3
01/27/2016
1) Determine the eigenvalues and the eigenspaces of the following two matrices
!
!
0 1
2 1
and
,
1 0
1 4
and if one of them is not diagonalizable give a Jordan basis for it.
3 2
0
2) Let A = 0 1 1 .
1 1
1
(i) Sho
Winter 2016
Math 325
Problems 5
03/02/2016
1) Compute a square root of the following 2 2-matrix:
!
3 1
A =
.
1 1
2) Compute exp(A) for
1 1
A =
2
4
!
.
(3) Give a Jordan basis and the Jordan
4 4-matrix:
3 1
1 1
A =
0 0
normal form for the following comp
Winter 2016
Math325
Solutions to Problems 2
1) The characteristic polynomial of the matrix A is
PA (T ) = T 3 5T 2 + 8T 6 .
By the Cayley-Hamilton Theorem we have therefore
0 = PA (A) = A3 5 A2 + 8 A 6 I3 ,
and this is equivalent to the equation A
A1 =
1
Winter 2016
Math 325
Praxis Midterm Exam
02/03/2016
1) Let
1 1
A =
1
!
,
3
and A : C2 C2 , v 7 A v be the corresponding C-linear map. Give a Jordan
basis for A .
(4 credits)
1 1 3
2) Let A = 0 1 1 . Use the Cayley-Hamilton Theorem to compute A1 .
1 0 1
(4
Winter 2016
Math 325
Problems 1
01/13/2016
1 4
!
and A : R2 R2 , v 7 A v be the corresponding R-linear map.
3 1
!
!
2
1
Give the matrix of the linear map A with respect to the basis
,
of R2 .
1
3
!
2 2
2) Let A =
. What are the A-invariant subspaces in R2
Winter 2014
Math 325
Problems 3
01/29/2014
1) Compute the eigenvalues and give Jordan bases of the generalized eigenspaces of the
following complex 2 2-matrices:
2 4
1
2
3 2
0
and
3
2 1
1 4
.
2) Let A = 0 1 1 .
1 1
1
(i) Show that A has two dierent eigenv
Winter 2014
Math325
Solutions to Problems 3
1) Let A =
mials:
2 4
1
2
3
and B =
2 1
. We rst compute the characteristic polyno-
1 4
PA (T ) = T 2 5T + 4 = (T 4) (T 1) ,
and
PB (T ) = T 2 6T + 9 = (T 3)2 .
Therefore A is diagonalizable with eigenvalues 1 a
Winter 2014
Math325
Solutions to Problems 6
a 1
1) Let A =
. It is straightforward to check that the map in question
1 b
v, w : R2 R2 R , (v, w) v T A w
satises axioms (L1) and (H) of the denition of an inner product for all
real numbers a, b, see 5.1 of
Winter 2014
Math 325
Problems 7
03/12/2014
1)
(i) Show that the map
: C2 C2 C ,
(
a1
a2
,
b1
b2
) 3a11 + a12 + a21 + 2a22
b
b
b
b
is an inner product on C2 .
(ii) Give an orthogonal basis of C2 with this inner product.
2) Let V be a C-vector space with i
Winter 2014
Math325
Solutions to Problems 7
1) Except for axiom (P) part (i) is a straightforward verication of the axioms.
To check (P) we observe rst that we can re-write this product as follows:
3 1
x, y = xt
for all x, y C2 . Let A :=
y
1 2
3 1
. The
Winter 2014
Math 325
Practice Final Exam
03/28/2014
1) Compute the Jordan normal form of the two following complex 3 3-matrices:
1 0 1
11 16 8
(a) 0 1 1 and (b) 9 13 6 .
0 0 1
0
0
1
2) Is the map
(
a1
a2
,
b1
b2
b
b
b
b
) a11 + 2(a12 + a21 ) + 3a22
an inn
Winter 2014
Math 325
Problems 5
02/26/2014
1) Compute exp(A) for
23 11
A =
11
1
.
2) Compute exp(A) for
1 1 1
A = 0
1
2
1
1 .
2
Due on 03/05/2014 handed in class.
All answers have to be justified.
Every problem is worth 4 points.
8 points = 100%
1
Winter
Winter 2014
Math325
Solutions to Problems 5
1) We have
PA (T ) = (T 23) (T 1) + 121 = T 2 24T + 144 = (T 12)2
1
is an
1
eigenvector and spans the eigenspace of the eigenvalue = 12. We extend
1
it to a basis by z =
. We have
0
and so = 12 is the only eigen
Winter 2014
Math325
Solutions to Problems 2
1) The characteristic polynomial of the matrix A is
PA (T ) = T 3 4T 2 T + 2 .
By the Cayley-Hamilton Theorem we have therefore
0 = PA (A) = A3 4 A2 A + 2 I3 ,
and this is equivalent to the equation A
A1 =
1
2
Winter 2014
Math325
Solutions to Problems 1
1) We have
A
2
=
2
10
and A
8
2
=
3
14
9
.
We express this images as linear combinations of the basis vectors:
10
8
2
= 7
2
+ (2)
2
3
and
14
9
2
= 12
2
+ (5)
2
3
.
Hence the matrix of A with respect to this
Winter 2014
Math325
Solutions to Problems 4
1) We have
PA (T ) = T 3 3T 2 + 3T 1 = (T 1)3 ,
and so = 1 is the only eigenvalue of A. The corresponding generalized eigenspace
is therefore the whole space C3 .
1
The eigenspace of this eigenvalue is 1-dimensi
Winter 2016
Math 325
Problems 2
01/20/2016
2 1 0
1) Let A = 0 2 2 . Compute A1 using the Cayley-Hamilton theorem.
1 0 1
2) Let A =
2 9
1 8
!
and A : C2 C2 , v 7 A v, the corresponding linear map.
(i) Show that A has only one eigenvalue and that A is not d