Math 115a: Selected Solutions for HW 3
Paul Young
October 23, 2005
Exercise 2.1.3: Prove that T is a linear transformation, and find bases for
both N (T ) and R(T ). Then compute the nullity and rank of T , and verify the
dimension theorem. Finally, use the appropriate theorems in this section to
determine whether T is one-to-one or onto: Define T : R2 → R3 by
T (a1 , a2 ) = (a1 + a2 , 0, 2a1 − a2 )
Solution: We first prove that T is a linear transformation. Let x = (x1 , x2 ), y =
(y1 , y2 ) ∈ R2 and let c ∈ R.
T (cx + y) = T (c(x1 , x2 ) + (y1 , y2 ))
= T (cx1 + y1 , cx2 + y2 )
= ((cx1 + y1 ) + (cx2 + y2 ), 0, (2(cx1 + y1 ) − (cx2 + y2 ))
= c(x1 + x2 , 0, 2x1 − x2 ) + (y1 + y2 , 0, 2y1 − y2 )
= cT (x1 , x2 ) + T (y1 , y2 ),
and hence T is linear. Next we figure out what N (T ) and R(T ) look like. Let
x = (x1 , x2 ) ∈ N (T ). Then
0 = T (x1 , x2 ) = (x1 + x2 , 0, 2x1 − x2 ) →
x1 + x2 = 0, 2x1 − x2 = 0 →
x1 = 0, x2 = 0.
Therefore we conclude that N (T ) = {0}, so that the basis for N (T ) would be
{0}. We now look at the image space. Generally, what we do is take a basis
of the domain, and then transform each of these basis elements by T to see
what we get. More specifically, let β = be the canonical basis for R2 – that is,
β = {(1, 0), (0, 1)}. Then
T (1, 0) = (1, 0, 2)
T (0, 1) = (1, 0, −1)
and hence R(T ) = span({(1, 0, 2), (1, 0, −1)}). Since these two vectors are lin-
early independent, we conclude that this is actually a basis for R(T ). Therefore
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after computing N (T ) and R(T ), we conclude that nullity(T ) = dim(N (T )) =
0 and rank(T ) = dim(R(T )) = 2. This is clearly consistent with the dimension
formula:
dim(R2 ) = nullity(T ) + rank(T )
2 = 0 + 2.
Lastly, since N (T ) = {0}, by theorem 2.4 T is injective. Now since the range
space is R3 , which is larger in dimension than that of the domain space, we
conclude that T cannot be onto.
Exercise 2.1.9ace: In this exercise, T : R2 → R2 is a function. For each
of the following parts, state why T is not linear.
(a) T (a1 , a2 ) = (1, a2 )
(c) T (a1 , a2 ) = (sin a1 , 0)
(e) T (a1 , a2 ) = (a1 + 1, a2 )
Solution: (a) Fails under scalar multiplication when the scalar is 0. (c) Fail
addition for practically every pair of vectors. (e) Fails under scalar multipli-
cation when the scalar is 2. (Note: There may be other reasons why these
functions fail to be linear. I only picked out the most obvious reasons.)
Exercise 2.1.14c: Let V and W be vector spaces and T : V → W be lin-
ear. Suppose β = {v1 , v2 , ..., vn } is a basis for V and T is one-to-one and onto.
Prove that T (β) = {T (v1 ), T (v2 ), ..., T (vn )} is a basis for W .
Solution: In order to prove that T (β) is a basis, we need to show two things:
T (β) is a linearly independent set and span (T (β)) = W .
L.I.: Let a1 , a2 , ..., an ∈ F be scalars such that
n
X
ai T (vi ) = 0.
i=1
By linearity of T , we may rewrite the left hand side as:
n
X n
X
ai T (vi ) = T ai vi
i=1 i=1
Since T is injective, N (T ) = {0}. Therefore
n n
X X →
−
ai vi ∈ N (T ) → ai vi = 0 .
i=1 i=1
Since {v1 , v2 , ..., vn } is a basis for V , it is a linearly independent set. Therefore
the last equality we got implies that ai = 0 for all i. Therefore we’ve proven
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L.I. for T (β).
span (T (β)) = W : Let w ∈ W be arbitrary. Since T is surjective, there ex-
ists v ∈ V such that T (v) = w. We express this v in terms of the basis β:
v = c1 v1 + c2 v2 + · · · + cn vn , where the ci ’s are scalars. Therefore we can write
the following expressions:
X n
w = T (v) = T ci vi
i=1
n
?
X
= ci T (vi )
i=1
?
where = is given to us by linearity of T . Looking at what we’ve just done, we
have written w as a linear combination of elements from T (β). Therefore w ∈
span(T (β)). Since w was arbitrarily chosen, W ⊆ span(T (β)). We note that
this is enough to establish equality because span(T (β)) ⊆ R(T ) ⊆ W is given
to us for free. Therefore span (T (β).
This completes our proof.
Exercise 2.1.17: Let V and W be finite-dimensional vector spaces and T :
V → W be linear.
(a) Prove that if dim(V ) < dim(W ), then T cannot be onto.
(b) Prove that if dim(V ) > dim(W ), then T cannot be one-to-one.
Solution:
(a) Suppose for the sake of contradiction that T is onto. Then rank(T ) =
dim(W ). We are given the following chain of relations:
∗
dim(W ) > dim(V ) = nullity(T ) + rank(T )
= dim(V ) = nullity(T ) + dim(W ) →
dim(W ) > nullity(T ) + dim(W )
∗
where = is given to us by the Dimension Formula. But this means that nullity(T )
must be a negative number, which is nonsense. This is our contradiction and
therefore T cannot be onto.
(b) Let us suppose–for the sake of contradiction–that T is injective. Then by
Theorem 2.4, nullity(T ) = dim(N (T )) = 0. Then we are given the chain of
relations:
∗
dim(W ) < dim(V ) = nullity(T ) + rank(T )
= 0 + rank(T ) →
dim(W ) < rank(T ).
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But this is clearly impossible because R(T ) is a subspace of W and therefore
always has dimension less than or equal to the dimension of W . This is our
contradiction; therefore T is never injective.
Exercise 2.1.18: Give an example of a linear transformation T : R2 → R2
such that N (T ) = R(T ).
Solution: Define T : R2 → R2 such that
T (a1 , a2 ) = (0, a1 ).
I leave it to the students to verify that N (T ) = span({(0, 1)}) = R(T )
Exercise 2.1.24: Let T : R2 → R2 . Include figures for each of the follow-
ing parts.
(a) Find a formula for T (a, b), where T represents the projection on the y-axis
along the x-axis.
(b) Find a formula for T (a, b), where T represents the projection on the y-axis
along the line L = {(s, s) : s ∈ R}.
Solution: I leave it to the students to draw the figures. I’ll only construct
the formulas.
(a) Let v = (v1 , v2 ) ∈ R2 . Since R2 = W1 ⊕ W2 where W1 = span({(1, 0)}), W2
= span({(0, 1)}), we will write v with respect to this decomposition: v = vx +vy ,
such that vx = (v1 , 0), vy = (0, v2 ). Then by the definition of projection on the
y-axis along the x-axis,
T (v) = vy = (0, v2 )
(b) Let v = (v1 , v2 ) ∈ R2 . Since R2 = W1 ⊕ W2 where W1 = span({(1, 1)}), W2
= span({(0, 1)}), we will write v with respect to this decomposition: v = vL +vy ,
such that vL = (v1 , v1 ), vy = (0, v2 − v1 ). Then by the definition of projection
on the y-axis along L,
T (v) = vy = (0, v2 − v1 ).
