Solutions for Math 225 Assignment #6
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(1) Let T1 and T2 be two linear transformations from R2 → R2
given by
T1 (x, y) = (2x + y, x + 2y) and T2 (x, y) = (x − y, y − x).
(a) Find T1 + T2 , T1 ◦ T2 and T2 ◦ T1 .
(b) Find the matrices
[T1 + T2 ]B←B , [T1 ◦ T2 ]B←B and [T2 ◦ T1 ]B←B
representing T1 + T2 , T1 ◦ T2 and T2 ◦ T1 under the standard
basis B of R2 .
(c) Find K(T1 + T2 ), K(T1 ◦ T2 ) and K(T2 ◦ T1 ).
(d) Find R(T1 + T2 ), R(T1 ◦ T2 ) and R(T2 ◦ T1 ). Verify Rank
Theorem for T1 + T2 , T1 ◦ T2 and T2 ◦ T1 .
Solution. We have
(T1 + T2 )(x, y) = T1 (x, y) + T2 (x, y)
= (2x + y, x + 2y) + (x − y, y − x) = (3x, 3y)
T1 ◦ T2 (x, y) = T1 (T2 (x, y)) = T1 (x − y, y − x)
= (2(x − y) + (y − x), (x − y) + 2(y − x))
= (x − y, y − x)
T2 ◦ T1 (x, y) = T2 (T1 (x, y)) = T1 (2x + y, x + 2y)
= ((2x + y) − (x + 2y), (x + 2y) − (2x + y))
= (x − y, y − x)
3
[T1 + T2 ] =
3
1 −1
[T1 ◦ T2 ] = [T2 ◦ T1 ] =
−1 1
K(T1 + T2 ) = {(x, y) : (T1 + T2 )(x, y) = 0}
= {(x, y) : 3x = 3y = 0} = {(0, 0)}
K(T1 ◦ T2 ) = K(T2 ◦ T1 ) = {(x, y) : x − y = y − x = 0}
= Span{(1, 1)}
R(T1 + T2 ) = Col[T1 + T2 ] = Span{(3, 0), (0, 3)} = R2
R(T1 ◦ T2 ) = R(T2 ◦ T1 ) = Span{(1, −1), (−1, 1)} = Span{(1, −1)}.
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For T1 + T2 , dim K(T1 + T2 ) = 0 and
rank(T1 + T2 ) = dim R(T1 + T2 ) = 2.
Hence
dim K(T1 + T2 ) + rank(T1 + T2 ) = 2 = dim R2 .
For T1 ◦ T2 , dim K(T1 ◦ T2 ) = 1 and
rank(T1 ◦ T2 ) = dim R(T1 ◦ T2 ) = 1.
Hence
dim K(T1 ◦ T2 ) + rank(T1 ◦ T2 ) = 2 = dim R2 .
For T2 ◦ T1 , dim K(T2 ◦ T1 ) = 1 and
rank(T2 ◦ T1 ) = dim R(T2 ◦ T1 ) = 1.
Hence
dim K(T2 ◦ T1 ) + rank(T2 ◦ T1 ) = 2 = dim R2 .
(2) Find a polynomial f (x) ∈ R[x] of degree ≤ 3 such that
f (−1) = 0, f (0) = 1, f (1) = 2 and f (2) = 9.
Solution. Let
(x − 0)(x − 1)(x − 2)
1
f1 (x) =
= − x(x − 1)(x − 2)
(−1 − 0)(−1 − 1)(−1 − 2)
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(x + 1)(x − 1)(x − 2)
1
f2 (x) =
= (x + 1)(x − 1)(x − 2)
(0 + 1)(0 − 1)(0 − 2)
2
(x + 1)(x − 0)(x − 2)
1
f3 (x) =
= − (x + 1)x(x − 2)
(1 + 1)(1 − 0)(1 − 2)
2
(x + 1)(x − 0)(x − 1)
1
f4 (x) =
= (x + 1)x(x − 1).
(2 + 1)(2 − 0)(2 − 1)
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Then
f (x) = 0 · f1 (x) + 1 · f2 (x) + 2 · f3 (x) + 9 · f4 (x)
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1
= (x + 1)(x − 1)(x − 2) − (x + 1)x(x − 2) + (x + 1)x(x − 1)
2
2
3
=x +1
by Lagrange Interpolation.
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(3) Which of the following statements are true and which are false?
Justify your answer.
(a) The composition of two injective linear transformations is
injective. That is, if T1 : V → W and T2 : U → V are
two injective linear transformations, then T1 ◦ T2 is also
injective.
Proof. True. For all u1 6= u2 ∈ U , T2 (u1 ) 6= T2 (u2 ) since
T2 is injective; since T2 (u1 ) 6= T2 (u2 ) and T1 is injective,
T1 (T2 (u1 )) 6= T1 (T2 (u2 )). Therefore, T1 ◦T2 is injective. (b) The sum of two surjective linear transformations is surjective. That is, if T1 : V → W and T2 : V → W are
two surjective linear transformations, then T1 + T2 is also
surjective.
Proof. False. For example, let T1 : R → R and T2 : R → R
be given by T1 (x) = x and T2 (x) = −x. Then
(T1 + T2 )(x) = T1 (x) + T2 (x) ≡ 0
and it is not onto.
(c) There is no injective linear transformation from R3 to R5 .
Proof. False. For example, let T : R3 → R5 be given by
T (x1 , x2 , x3 ) = (x1 , x2 , x3 , 0, 0). Then T is injective since
K(T ) = {(0, 0, 0)}.
(d) There is no surjective linear transformation from R5 to R3 .
Proof. False. For example, let T : R5 → R3 be given by
T (x1 , x2 , x3 , x4 , x5 ) = (x1 , x2 , x3 ). Then T is onto since
R(T ) = Span{T (e1 ), T (e2 ), T (e3 ), T (e4 ), T (e5 )}
= Span{e1 , e2 , e3 , 0, 0} = R3 .
(4) Determine whether the following linear transformations are 1-1,
onto and/or bijective. Justify your answer:
(a) T : R2 → R3 given by T (x, y) = (2x + y, x + 2y, x − y).
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Solution. Since
K(T ) = {(x, y) : T (x, y) = 0}
= {(x, y) : 2x + y = x + 2y = x − y = 0}
= {(0, 0)},
T is 1-1.
Since dim(R2 ) < dim(R3 ), T is not onto by Rank Theorem.
Consequently, T is not bijective.
(b) T : R3 → R2 given by T (x, y, z) = (x + y + z, x + 2y + 4z).
Solution. Since
1 1 1
rank(T ) = rank([T ]B2 ←B3 ) = rank
= 2 = dim(R2 ),
1 2 4
T is onto, where B2 and B3 are the standard basis of R2
and R3 .
Since dim(R3 ) > dim(R2 ), T is not 1-1. Consequently, T
is not bijective.
(c) T : R[x] → R[x] given by T (f (x)) = f (1 − x).
Solution. Since
T ◦ T (f (x)) = T (T (f (x))) = T (f (1 − x)) = f (1 − (1 − x)) = f (x),
T −1 exists and T −1 = T . So T is 1-1, onto and bijective.
(d) T : M3×3 (R) → R given by T (A) being the sum of all
entries of A.
Solution. Since
R(T ) = Span{T (Eij ) : 1 ≤ i, j ≤ 3} = Span{1, 1, ..., 1} = R,
T is onto, where {Eij : 1 ≤ i, j ≤ 3} is a standard basis of
M3×3 (R).
Since dim(M3×3 (R)) = 9 > dim(R), T is not 1-1 by Rank
Theorem. Consequently, T is not bijective.
(5) Let T1 : V → W and T2 : V → W be two linear transformations
between two finite-dimensional vector spaces V and W . Do the
following:
(a) Prove that R(T1 + T2 ) ⊂ R(T1 ) + R(T2 ).
(b) Prove that rank(T1 + T2 ) ≤ rank(T1 ) + rank(T2 ).
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(c) Prove that
dim K(T1 ) + dim K(T2 ) ≤ dim K(T1 + T2 ) + dim V.
(d) Is it true that K(T1 + T2 ) ⊂ K(T1 ) + K(T2 )? Justify your
answer.
Proof. For every vector v ∈ V ,
(T1 + T2 )(v) = T1 (v) + T2 (v) ∈ R(T1 ) + R(T2 ),
since T1 (v) ∈ R(T1 ) and T2 (v) ∈ R(T2 ). Therefore,
R(T1 + T2 ) ⊂ R(T1 ) + R(T2 )
and hence
rank(T1 + T2 ) = dim R(T1 + T2 ) ≤ dim(R(T1 ) + R(T2 ))
≤ dim R(T1 ) + dim R(T2 ) = rank(T1 ) + rank(T2 ).
By Rank Theorem,
rank(T1 + T2 ) = dim V − dim K(T1 + T2 )
rank(T1 ) = dim V − dim K(T1 )
rank(T2 ) = dim V − dim K(T2 ).
Therefore,
dim V − dim K(T1 + T2 )
≤ (dim V − dim K(T1 )) + (dim V − dim K(T2 ))
⇒ dim K(T1 ) + dim K(T2 ) ≤ dim K(T1 + T2 ) + dim V.
The statement
K(T1 + T2 ) ⊂ K(T1 ) + K(T2 )
is false. For example, let T1 : R → R and T2 : R → R be
given by T1 (x) = x and T2 (x) = −x. Then K(T1 ) = {0},
K(T2 ) = {0} and K(T1 ) + K(T2 ) = {0}. But (T1 + T2 )(x) ≡ 0
and hence K(T1 + T2 ) = R.