Math 544–Practice Exam 2B
" !!1
$ !!5
1. A = $
$ !!4
$
# !!2
!2
!!10
!!8
!4
!4
!9
!9
!5
!3
!7
!2
!!0
!!3 % "
!!8 ' $
'!$
!!7 ' $
' $
!6 & #
1
0
0
0
2
0
0
0
!4 !3 !!3 %
!!1 !2 !!0 '
'.
!!0 !0 !5 '
'
!!0 !0 !!0 &
(a). Find a basis for Col(A).
How did you determine this basis?
Solution:
( " !1 % " !4 % " !3 % ,
*$ ' $ ' $ ' *
* $ !5 ' $ !9 ' $ !8 ' *
,!
,!
!! - The pivot columns of A form a basis for Col(A).
)!
* $ !4 ' $ !9 ' $ !7 ' *
*+ $# !2 '& $# !5 '& $# !6 '& *.
(b). Find a basis for Row(A).
How did you determine this basis?
Solution:
{![1!!2!!!4!!3!!3],!![ 0!!0!!1!!!2!!0 ],!![ 0!!0!!0!!0!!!5 ]!}
The pivot rows of the echelon form of A form a basis for Row(A).
2. Find the eigenspace of the eigenvalue ! = 4 for the matrix A below.
!5 2 3$
A = ## 1 6 3 && .
#" 0 0 4 &%
" 1 2 3% " 1 2 3%
Solution: A ! 2I 3 = $$ 1 2 3 '' ~ $$ 0 0 0 '' .
$# 0 0 0 '& $# 0 0 0 '&
( " !2 % " !3% ,
*
*
So, a basis for the eigenspace is ) ! $$ 1 '' ,! $$ 0 '' ! - .
*$0' $1' *
+# & # &.
3. Suppose that T : M 2 ! R 3 is a linear transformation such that
' ! 1 1 $* ! 2 $
' ! 1 0 $* ! 1 $
' ! 1 0 $* ! 2 $ ' ! 0 1 $* !1 $
#
&
#
&
#
&
#
&
&, = # 4 & ,!T ) #
&, = #1 & .
)
,
)
,
T
= 1 ,!!T
= 5 ,!!T ) #
&, # &
&, # &
&, # & ) #
&, # &
)#
)#
)#
)( # 0 0 &,+ # 4 &
)( # 0 0 &,+ # 1 &
)( # 0 1 &,+ # 1 & )( # 1 0 &,+ # 3&
"
%
"
%
"
%
"
%
" %
" %
" %
" %
' ! 7 4 $*
&, ?
What is the value of T ) ##
&,
)
)( # 1 2 &,+
"
%
Solution:
!7 4 $
!1 1 $
!1 0 $
!1 0 $ !0 1 $
#
&
#
&
#
&
&+#
& , we get that
Since
=3
+2
+ 2#
#
&
#
&
#
&
#
& #
&
#" 1 2 &%
#" 0 0 &%
#" 0 0 &%
#" 0 1 &% #" 1 0 &%
' ! 7 4 $*
' ! 1 1 $*
' ! 1 0 $*
' ! 1 0 $*
' ! 0 1 $*
#
&
#
&
#
&
#
&
&,
)
,
)
,
)
,
)
,
T
= 3T
+ 2T
+ 2T
+T )#
&,
&,
&,
&,
&,
)#
)#
)#
)#
)#
)( # 1 2 &,+
)( # 0 0 &,+
)( # 0 0 &,+
)( # 0 1 &,+
)( # 1 0 &,+
"
%
"
%
"
%
"
%
"
%
!2 $
!1 $
! 2 $ !1 $ ! 13 $
#
&
#
&
= 3 # 1 & + 2 # 5 & + 2 ## 4 && + ##1 && = ## 22 && .
#" 4 &%
#" 1 &%
#" 1 &% #" 3&% #"19 &%
"1 2 %
'.
4. Let W = { A !M 2 : AB = BA} ,!where B = $
$
'
$# 0 1 '&
Then W is a subspace of M 2 (you need not verify that). Find a basis for W.
!a b $
Solution: Let A = #
& 'W . Then,
"c d %
! a b $ ! 1 2 $ ! 1 2 $ ! a b $ ! a 2a + b $ ! a + 2c b + 2d $
.
# c d & # 0 1 & = # 0 1 & # c d & ' # c 2c + d & = # c
d &%
"
%"
% "
%"
% "
% "
Hence, a = a + 2c ! c = 0!and 2a + b = b + 2d ! a = d .
!a b $
!1 0 $
!0 1 $
So a typical A looks like A = #
= a#
+ b#
&
&
&.
"0 a %
"0 1 %
"0 0 %
' !1 0 $ !0 1 $ *
So, a basis for A is ( ! #
& ,! #
&! + .
) "0 1 % "0 0 % ,
5. (a). Suppose that A is a 5 ! 12 matrix whose columns span R 5 .
What is the value of the nullity of A? Explain your answer.
Solution: Since the columns of A span R 5 , the rank of A is 5. Thus the nullity of
A is 12 – 5 = 7.
(b). The dimension of a vector space V is _______________________________
(Provide a definition.)
Solution: The dimension of a vector space is the number of vectors in any basis.
0
' ! a b $* ! a + b - c
$
#
&
#
&
)
,
=
6. (a). Let T : M 2 ! M 2 be defined by. T #
&, #
&
)
)( # c d &,+ # 0
&%
b
+
2c
d
"
%
"
Determine a basis for the kernel of T. Answer: _________________________
!a b $
Solution: If A = #
& 'ker(T ) , then a + b ! c = 0 and b + 2c ! d = 0 .
"c d %
So, writing a = !b + c,!!!d = b + 2c , we get that
b %
" !b + c
" !1 1%
"1 0 %
A=$
= b$
+ c$
'
'
'.
b + 2c &
# c
# !!0 1&
#1 2 &
( " !1 1% "1 0 % +
So a basis for the kernel is ) ! $
' ,! $1 2 ' ! , .
0
1
#
& #
&*
(b). Let T : P4 ! P4 be defined by T ( p(t)) = p ''(t) .
Give two independent elements of the kernel of T.
Solution: 3, 2t + 1 would work – or infinitely many others e.g., 3t + 1,!2t + 1 .
7. Let S = {v1 ,!v 2 ,!v 3 ,…,!v n } be a spanning set for the vector space V and let
T :V ! U be a linear transformation such that T is onto. We want to show that
T (S) = {u1 ,!u 2 ,…,!u n } where ui = T (v i ) for each i =1,2,…, n spans U.
Finish the argument below.
Solution:
Let b be a vector in U. We must show that b is a linear combination of the ui ' s .
Since T is onto there exists some
v
in V
such that b = T (v) .
Since S spans V, v = a1v1 + a2 v 2 + ! + an v n , for some real numbers a1 ,!a2 ,!…,!an .
But now using the fact that T is a linear transformation,
b = T (v) = T ( a1v1 + a2 v 2 + ! + an v n ) = a1T (v1 ) + a2T (v2 ) + ! + anT (vn ) .
Hence, b = a1u1 + a2 u 2 + ! + an u n .
8.
'!a b $
+
)#
)
&
:!a,b, c, d are integers , . Is W a subspace of M 2 ?
(a). Let W = ( #
&
)# c d &
)
"
%
*
If so verify it. If not, give a specific example to show that it is not a subspace.
!1
$
1&
1 !1 2 $ # 2
!1 2 $
Solution: No, #
& 'W .
& 'W , but 2 # 3 4 & = # 3
"
% #
"3 4 %
2&
#" 2
&%
'! a $
+
)# &
)
)# b &
)
(b). Let W = (
:!ab = cd,!a,b, c, d! are real numbers , . Is W a subspace of R 4 ?
)# c &
)
)* #" d &%
)If so verify it. If not, give a specific example to show that it is not a subspace.
!1 $
!3$
#6 &
#6 &
#
&
Solution: No,
and # & are both in W, since 1 ! 6 = 2 ! 3 and 3 ! 6 = 2 ! 9 ,
#2 &
#2 &
# &
# &
"3%
"9 %
!1 $ ! 2 $ ! 3 $
# 6 & # 6 & #12 &
but # & + # & = # & since 3 ! 12 " 5 ! 7 .
#2 & # 3& # 5 &
# & # & # &
" 3% " 4 % " 7 %
9. (a). Find a 2 ! 2 matrix A with no zero entries such that 3, 6 are the
eigenvalues of A.
" 2 !1%
Solution: A = $
' will do.
#4 7 &
(b). Determine all invertible 2 ! 2 matrices A such that A and A !1 have
the same eigenvalues. Express your answer in terms of det(A) and explain.
Solution: These are precisely those matrices A with Det(A) = 1.
!a b $
Suppose that A = #
& and let D = det(A) .
"c d %
Then the characteristic polynomial for A is p ( ! ) = ! 2 " (a + d)! + D and the
(
)
characteristic polynomial for A !1 is q(t) = ! 2 " (a + b) D ! + 1 D .
For A and A to have the same eigenvalues, it must be that p ( ! ) = q ( ! ) and this
means that D = 1 .
!1
( " a + b + 2c %
,
*$
*
'
10. Let W = ) $ 2a + b + 4c ' : a,b, c are real numbers - .
* $ !a ! 3b ! 2c '
*
&
+#
.
3
Explain how you know that W is a subspace of R , and find a basis for W.
( " !!1 % " !!1 % " !!2 % ,
*
*
Solution: W is a subspace since it is the span of ) ! $$ !!2 '' ,! $$ !!1 '' ,! $$ !!4 '' ! - .
* $ !1 ' $ !3' $ !2 ' *
+# & # & # &.
" !!2 %
" !!1 %
$
'
Since ! $ !!4 ' = 2 $$ !!2 '' , the vectors in
$# !2 '&
$# !1 '&
( " !!1 % " !!1 % ,
*$ ' $ '*
) ! $ !!2 ' ,! $ !!1 ' ! - form a basis.
* $ !1 ' $ !3' *
+# & # &.