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Mathematics

MAT224H1

Sean Uppal

Fall

Description

University of Toronto
Department of Mathematics
MAT224H1F
Linear Algebra II
Midterm Examination
October 13, 2010
S. Uppal
Duration: 1 hour 20 minutes
Last Name:
Given Name:
Student Number:
Tutorial Code:
No calculators or other aids are allowed.
FOR MARKER USE ONLY
Question Mark
1 /10
2 /10
3 /10
4 /10
5 /5
6 /5
TOTAL /50
1 of 10 1. Find a basis for the kernel and the range of the linear transformation T: C ! C
3
whose matrix with respect to the standard basis of C is
0 1
2 i 1 + i
@ A
1 ▯ i 1 2 :
0 1 ▯ i 2
Solution: We ▯nd the row reduced echelon form of the given matrix.
0 1 0 1 0 1
2 i 1 + i 2 !2 2▯1 2 i 1 + i r3$2 2 i 1 + i
@ 1 ▯ i 1 2 A ▯▯▯▯▯▯▯! @0 0 0 A ▯▯▯! @ 0 1 ▯ i 2 A
0 1 ▯ i 2 0 1 ▯ i 2 0 0
0 1 0 1
i 2 0 2 1 0 1
1 !1 1▯i @ A @ A
▯▯▯▯▯▯▯▯▯! 0 1 ▯ i 2 ▯▯▯▯! 0 1 i + 1
0 0 0 0 0 0
Hence the kernel is spanned by f(1;i + 1;▯1)g and a basis for the range is the set the
▯rst two columns of the matrix for T, that is the basis
0 1 0 1
( 2 i )
@ 1 ▯ i; 1 A
0 1 ▯ i
2 of 10 EXTRA PAGE FOR QUESTION 1 - do not remove.
3 of 10 2. Let T: P2(R) ! P 2R) be de▯ned by T(p(x)) = p(ax+b), where a;b 2 R are constants,
a 6= 0;1. Find a basis ▯ for P (R) such that [T]is diagonal.
2 ▯▯
Solution: We begin by ▯nding the eigenvalues and eigenvectors for the matrix of T with respect
to the standard basis ▯. We have that T(1) = 1, T(x) = ax+b and T(x ) = (ax+b) =
2 2 2
a x + 2abx + b hence 0 1
1 b b2
@ A
[T]▯▯ = 0 a 2ab
0 0 a 2
2
We have that det([T▯▯ ▯▯I) = (1▯▯)(a▯▯)(a ▯▯) so the eigenvalues are ▯ 1 1, ▯ 2
a and ▯ = a : To get the eigenvectors we solve the system of equations corresponding
3
to the following matrices (corresponding to1▯ 2▯ 3▯ respectively)
0 2 1 0 2 1 0 2 2 1
0 b b 1 ▯ a b b 1 ▯ a b b
@0 a ▯ 1 2ab A ; 0 0 2ab A ; 0 a ▯ a2 2abA
2 2
0 0 a ▯ 1 0 0 a ▯ a 0 0 0
b ▯b2 2b
we get the eigenvectors 1 = (0;0;1), 2 = (a▯1;1;0) and ((1▯a); 1▯a;▯1). If we take
P to be the matrix with vector i as column i then we have that
▯1 2
P [T]▯▯ = diagf1;a;a g:
Hence we can take ▯ to be the basis
2 b ▯b2 2b 2
▯ = fx ; + x; 2 + x ▯ x g
a ▯ 1 (1 ▯ a) 1 ▯ a
4 of 10 EXTRA PAGE FOR QUESTION 2 - do not remove.
5 of 10 3 2
3. Let V = f(x ;x1;x2) 23R j x +x +x1= 0g2and 3 = fa+bx+cx 2 P (R) j a▯b = 2
2cg. Show that V and W are isomorphic and construct an isomorphism T: V ! W.
Justify your answer.
Solution: Suppose that x = (x ;x ;x1) 2 V 3 Then
x = (x 1x ;2 )3
= (x 1x ;2(x + 1 )) 2
= x 11;0;▯1) + x (0;2;▯1)
showing that a basis for V is f(1;0;▯1);(0;1;▯1)g and so dim(V ) = 2 Similarly if
p(x) = a + bx + cx 2 W then
2
p(x) = a + bx + cx
a ▯ b 2
= a + bx + x
2
1 2 1 2
= a(1 + x ) + b(x ▯ x )
2 2
showing that a basis for W is f1 + x2;x ▯ x2g and so dim(W) = 2. Since V and W
2 2
have the same dimension and are ▯nite dimensional they are isomorphic. An explicit
1 2 1 2
isomorphism is given by T(x ;x ;1 ) 2 x3(1 + 1 2x ) + x 2x ▯ x 2. We need to show
that this is a linear function which is one -to -one and onto. To see the function is
linear note that if ▯ 2 R and x;y 2 V then
1

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