EXAM
Exam 1
Math 5316, Fall 2012
December 2, 2012
• Write all of your answers on separate sheets of paper.
You can keep the exam questions. This is a
takehome exam, to be worked individually. You can
use your notes. You may use Maple as indicated and
turn in hardcopy Maple output with your written
answers. There is a Maple worksheet on my website
with the matrices already entered.
• The exam is due Saturday, December 8 by
5pm.
• You must show enough work to justify your answers.
Unless otherwise instructed,
give exact answers, not
√
approximations (e.g., 2, not 1.414).
• This exam has 8 problems. There are 370 points
total.
Good luck!
60 pts.
Problem 1. In each part, find a matrix P so that P −1 AP is in Jordan Canonical Form.
I would not undertake this by hand. There is a maple worksheet with the
matrices already entered.
A.

1400 1570
 561 631

A=
294 330
495 555
−9793
−3927
−2057
−3465

86
34 

18 
32
B.






A=





40 pts.
−4899
3100
1110
1765
−14269
−16920
−36857
29120

72
48 −124
1222
−58
20 −236
−106
276 −543
−790
−203
115
195
22 −358
275 −265
294 −150
34
−9 −216
144 −435
182
−90
77
261 −432
−40
3575
294 −180
−713
0
828 −3405
4155 −688
345 −600
−51
2556 −7499
9033 −2116
1067 −1297
−171 −3096
2384 −7185
2619 −1290
1272











Problem 2.
Find a matrix P so that P −1 AP = J, where J is in Jordan Canonical Form.
Find a real matrix S so that S −1 AS = R, where R is in real Jordan form.
Again, not a problem for hand computation! See the Maple worksheet.
A=












−49
−69
−501
−33
−2283
−489
−1686
−1971
5092766
−372123 1084 158927 5091364
−1697132
−70821
5288
151
−2223
−71006
23652
7754744
−564518 2197 241674 7752164
−2584179
1254085
−91603
318
39132 1253677
−417900
38991800 −2838877 10495 1215215 38979500 −12993735
3376128
−245787 1759 105244 3373954
−1124769
11189119 −814529 6002 348795 11181691
−3727636
14532072 −1051262 4913 451845 14527028
−4842842
1
−79220
1094
−120712
−19510
−606928
−52571
−174236
−226453












40 pts.
40 pts.
Problem 3.
Again, not a problem for hand computation.
The following matrix A is diagonalizable, so we have P −1 AP = D, where D
is diagonal. Find a matrix S so that At = S −1 AS.


14 23
−73 −39
 117 80 −273 −351 
.
A=
 39 26
−89 −117 
0 5
−15
1
Problem 4.
Again, not a problem for hand computation. You can use the Maple GramSchmidt command to check yourself. Read the help page for that command
first.
A Use the Gram-Schmidt process on the vectors v1 , v2 , v3 , v4 to find an orthonormal basis of R4 . Show the Gram-Schmidt process one step at a time.
Find an orthogonal matrix Q and an upper triangular matrix R so that
A = QR.


1
 1

v1 = 
 0 ,
−1

1
 1
A=
 0
−1
 
0
1

v2 = 
0 ,
1
0
1
0
1

0
0 
,
1 
1
 
1
1

v3 = 
1 ,
1
1
1
1
1
 
0
0

v4 = 
1 .
1
B Use the Gram-Schmidt process on the vectors v1 , v2 , v3 , v4 to find an orthonormal basis of C4 . Show the Gram-Schmidt process one step at a time.
Find an unitary matrix U and an upper triangular matrix R so that A = U R.


i
0 1
i
 i
0 0
i 

A=
 0
1 1 −2 i 
1 1 + 2i 1
1
 


 


i
0
1
i
 i

0
 i
0

,


v1 = 
v2 = 
v3 = 
v4 = 
0 ,

1 ,
−2i
1
1
1 + 2i
1
1
2
50 pts.
40 pts.
Problem 5.
Again, not a problem for hand computation. You can use the Maple GramSchmidt command to check yourself. Read the help page for that command
first.
The following matrix A is symmetric. Find an orthonormal basis for R5
consisting of eigenvectors of A.
Find an orthogonal matrix Q so that Qt AQ is diagonal.


−5 0
0 −10
5

0 20
0
0
0 



0 0 −10
0
0 
A=

 −10 0
0
10 −10 
5 0
0 −10
−5
Problem 6. Let A be an m × n matrix over K. Multiplication by A is a linear transformation Kn → Km and multiplication by A∗ is linear
transformation Km → Kn .
We use the standard scalar product on Kn and Km .
Show that
Km = im(A) ⊕ ker(A∗ )
Kn = im(A∗ ) ⊕ ker(A),
and that these are orthogonal direct sums, e.g., the subspaces im(A) and ker(A∗ )
are orthogonal. Hint: Do the first equation first, suppose that v ∈ im(A)⊥ , see
if you can get it in ker(A∗ ).
50 pts.
Problem 7. Let V be a vector space over K with a K-inner product.
Recall that a linear transformation V → V is symmetric if
hT (u), V i = hu, T (v)i,
∀u, v ∈ V.
Recall that an operator P : V → V is a projection operator if P 2 = P (which
is the same as saying P acts as the identity on im(P )). We know from previous
work that in this case
V = im(P ) ⊕ ker(P ).
Show that this is an orthogonal direct sum if and only if P is symmetric.
A word on Matrix Norms
Let V be a vector space over K. Recall that a norm on V is a function V →
K : v 7→ kvk that has the following properties.
3
1. kvk ≥ 0 for all v ∈ V and kvk = 0 ⇐⇒ v = 0.
2. kλvk = |λ|kvk for all λ ∈ K and v ∈ V .
3. The triangle inequality holds: for all v, w ∈ V ,
kv + wk ≤ kvk + kwk.
In the following problem it is convenient to use the following norm on Kd :
kxk = max{|xi | | i = 1, 2, . . . , d},
where


x1
 x2 
 
x =  .  ∈ Kd .
 .. 
xd
On the vector space Kd×d of d × d matrices, we use the following norm. If
A = [aij ], then
X
d
|aij | | i = 1, 2, . . . , d
kAk = max
j=1
In other words, for each row we form the sum of the absolute values of all the
entries in the row, and then we take the largest of these row sums.
It’s easy to see that if kAk is small, the entries in A must be small. Indeed,
we can form the norm of Kd×d defined by
kAkmax = max{|ai,j | | i, j = 1, 2, . . . , d},
where we take the maximum size of an entry. It’s easy to check that
kAkmax ≤ kAk ≤ dkAkmax ,
so if one of these two norms is small, the other is forced to be small.
The advantage of the norm k·k is the following two inequalities
kAxk ≤ kAk kxk,
x ∈ Kd , A ∈ Kd×d ,
kABk ≤ kAk kBk,
A, B ∈ Kd×d .
4
50 pts.
Problem 8.
Let A be a d × d matrix over C. Let λ1 , λ2 , . . . , λk be the eigenvalues of A.
Strictly for the purposes of this problem, we say that the matrix A is stable
if |λj | < 1 for j = 1, 2, . . . , k, i.e., all the eigenvalues have modulus strictly less
that one.
In this problem, we’re working over C.
A. Show that if D is a stable d × d diagonal matrix, then
kDn k → 0,
n → ∞.
B. If S is a stable d × d matrix which is diagonalizable, then
kS n k → 0,
n → ∞.
C. Let A be a stable d × d matrix which is not diagonalizable. Show that
kAn k → 0,
n → ∞.
To do this, use the Jordan Decomposition A = S + N . Since S and N
commute, you can compute (S + N )n by the Binomial Theorem.
5