Math 4326
Spring 2017
1. (20 points) Let T : M2×2
a
T
c
Exam 3
Solutions
→ P2 be defined by
b
= (a − b)t2 + (c + d)t + (a − b + c + d).
d
(a) Find a basis for the kernel of T.
Solution: We want matrices A for which T (A) = 0, the 0-polynomial. This
means we need a − b = 0, c + d = 0, and a − b + c + d = 0. The last equations
follows from the first two, so we need a = b and c = −d. If a and
c are our free
a a
variables, the kernel consists of all matrices of the form
. We can write
c −c 0 0
a a
1 1
0 0
1 1
this
=a
+c
, leading to basis
,
.
c −c
0 0
1 −1
0 0
1 −1
(b) Find a basis for the range of T.
Solution: Here are two approaches. Using first principles, we want all polynomials xt2 + yt + z that can occur on the right hand side of T so we need x, y, z
for which x = a − b, y = c + d, z = a − b + c + d for some a, b, c, d. The only condition for consistency here is that z = (a − b) + (c + d) = x + y. Thus, the range
consists of all polynomials of the form xt2 + yt + (x + y) = x(t2 + 1) + y(t + 1).
A basis is {t2 + 1, t + 1}.
A second approach: use the proof of the Rank-Nullity
for the kernel, we can extend it to a basis for M2×2 :
1 1
0 0
1 0
0
,
,
,
0 0
1 −1
0 0
1
Theorem. Given a basis
0
0
.
If we apply T to the vectors we added, we get a basis for the range. This gives
1 0
0 0
T
,T
= {t2 + 1, t + 1}.
0 0
1 0
In this case, we got the same answer both ways. Usually the first approach gives
a nice basis and the second gives a messy basis.
Page 2
2. (20 points) Two bases for P3 are
B = {t2 + t + 1, t2 + 1, t2 } and C = {t + 1, t2 + t, t2 + t + 1}.
(a) Find the transition matrix P .
B←C
Solution: If the vectors in B are b1 , b2 , b3 and the vectors in C are c1 , c2 , c3 ,
then we write the c-vectors as combinations of b-vectors, and the coefficients
in this combinations give us the columns of P . We have c1 = b1 − b3 , c2 =
b1 − b2 + b3 , c3 = b1 Thus,


1
1 1
P = ([c1 ]B | [c2 ]B | [c3 ]B ) =  0 −1 0 .
B←C
−1 1 0


1
(b) Use part (a) to find [p(t)]B given that [p(t)]C = −2
3
Solution:

   
1
1 1
1
2





0 −1 0
−2 =
2 .
[p(t)]B = P [p(t)]C =
B←C
−1 1 0
3
−3
Here is a way to check: Use
 thetwo coordinate vectors to calculate p(t) and see
1

if they agree. If [p(t)]C = −2 then
3
p(t) = c1 − 2c2 + 3c3 = (t + 1) − 2(t2 + t) + 3(t2 + t + 1) = t2 + 2t + 4.
 
2
If [p(t)]B =  2  then
−3
p(t) = 2b1 + 2b2 − 3b3 = 2(t2 + t + 1) + 2(t2 + 1) − 3t2 = t2 + 2t + 4.
Page 3

3
0
3. (20 points) Let A = 
0
1
0
4
0
0
0
0
4
0

1
0
.
0
3
(a) Find the characteristic polynomial of A. Explain any fancy reasoning.
Solution: We have
x − 3
0
0
−1
0
x−4
0
0 −1 2 x − 3
det(xI − A) = = (x − 4) 0
x−4
0 −1 x − 3
0
−1
0
0
x − 3
= (x − 4)2 ( (x − 3)2 − 1) = (x − 4)2 (x2 − 6x + 8) = (x − 4)3 (x − 2).
(b) For each eigenvalue of A, find a basis for its eigenspace. (There are only two
eigenvalues. You might be able to guess them if you have trouble with part (a)).

−1
0
Solution: For 4, A − 4I = 
0
1



0 1
1 0 0 −1


0 0
 ⇒ 0 0 0 0 . There is


0 0
0 0 0 0
0 −1 
0 0 0 0
     
1
0
0 


      

0 1 0

one pivot, three free variables leading to basis   ,   ,   .
0
0
1 





1
0
0




1 0 0 1
1 0 0 1
0 0 2 0
0 1 0 0



For 2, A − 2I = 
0 0 2 0 ⇒ 0 0 1 0. Now there are three pivots,
1 0 0 1
0 0 0 0


−1



  

0
 .
one free variables leading to basis 
 0 





1
0
0
0
0
Page 4
4. (20 points) A projections and reflections problem
3
(a) Find the matrix for the orthogonal projection of R2 onto the line y = x.
2
a
Solution: If we call the projection P then given a point
, we project onto
b
the line by finding theintersection
of y = 32 x with the line perpendicular to it
a
and passing through
. This second line has equation y − b = − 32 (x − a)
b
so 32 x − b = − 32 (x − a). Multiplying by 6, 9x − 6b = −4x + 4a, so x = 4a+6b
13
.
If
we
abuse
notation
and
now
let
P
be
the
matrix,
then
and y = 23 x=6a+9b
13 a
4a/13 + 6b/13
4/13 6/13
a
we need P
=
=
, so the matrix is
b
6a/13
+
9b/13
6/13
9/13
b
4/13 6/13
.
6/13 9/13
3
(b) Find the matrix for the orthogonal reflection of R2 through the line y = x.
2
Solution: Theeasiest method
is to use
the
formula R =2P − I to get the
4/13 6/13
1 0
−5/13 12/13
matrix to be 2
−
=
.
6/13 9/13
0 1
12/13 5/13
As a check, we can see if P 2 = P and R2 = I. If we factor
case,
2
P =
1
132
4 6
4 6
=
6 9
6 9
1
132
52 78
78 117
=
1
132
1
13
13 · 4 13 · 6
=
13 · 6 13 · 9
outside in each
1
13
4 6
= P.
6 9
and
1
R = 2
13
2
1 169 0
1 132 0
−5 12
−5 12
= 2
= 2
= I.
12 5
12 5
0 169
0 132
13
13
Page 5
5. (10 points) Suppose T is a
1
and suppose T (u) =
0
linear
transformation,
T
1
1
and T (v) =
0
1
:
V → M2×2 Let u and v be in V
0
.
0
(a) What is T (4u − 3v)? Explain.
Solution: By linearity, for any a and b, T (au + bv) =
case,
1 1
1
T (4u − 3v) = 4T (u) − 3T (v) = 4
−3
0 0
1
aT (u) + bT (v). In our
0
0
=
1 4
.
−3 0
(b) Show that u and v are independent. Hint: Try to use the fact that T (u) and T (v)
are independent, along with the definition of independence.
Solution: I would do the following: Suppose
that
0. Then by
xu+ yv = 1 1
1 0
x+y x
linearity, T (xu + yv) = xT (u) + yT (v) = x
+y
=
.
0 0
1 0
y
0
But xu + yv = 0 so T (xu + yv) = T (0) = 0. This forces the coefficients in the
matrix to be 0 so x = 0 and y = 0, showing u and v are independent.
6. (10 points) Define a linear transformation T : R3 → R2 as follows: given a vector v, the
top entry is the average the top two entires of v,and
 the bottom is the difference of the
1
1
(1 + 3)
2
2


second and third entries of v. For example, T 3 =
=
. Find the
4
(3 − 4)
−1
matrix of this transformation.
Solution: This can be done by any number of methods. If we use the formula for
the matrix,
      
1 0 0
1/2 1/2 0
   



A = (T (e1 ) | T (e2 ) | T (e3 )) = T 0 T 1 T 0
=
.
0
1 −1
0
0
1
Page 6
Some extra credit you can think about if you have time.
7. (3 points) With regard to problem 5, finda linear transformation
T : P2 → M2×2 and
1 1
1 0
vectors u, v in P2 for which T (u) =
and T (v) =
.
0 0
1 0
2
Solution: One fairly easy answer is T (at + bt + c) =
a+b a
b
c
with u = t2 and
v = t.
8. (4 points) With regard to problem 4, find rotation matrices that do the following:
(a) Rotate the line x = 0 onto the line y = 32 x.
(b) Rotate the line y = 23 x onto the y-axis.
9. (3 points) Give a formula for a linear transformation T : P3 → P3 with
Range = Span{1 + t, t + t2 } and Kernel = Span{t + t2 , t2 + t3 }.
10. (4 points) Suppose you have a 2n × 2n matrix A with 3’s on the diagonal
in
 and 1’s 
3 0 0 1
0 3 1 0

positions (1, 2n), (2, 2n − 1), . . . , (2n, 1). For example, when n = 2, A = 
0 1 3 0.
1 0 0 3
Find the characteristic polynomial of A, its eigenvalues, and a basis for each eigenspace.