Math 416 Homework 4. Solutions.
1. Let T : R2 → R2 be any function of the form
T (x, y) = (ax + by, cx + dy),
a, b, c, d ∈ R.
Prove that T is linear.
Solution: We need to show that if v, w ∈ R2 and a ∈ R, then
T (v + w) = T v + T w,
T (av) = aT v.
Let us write v = (v1 , v2 ) and w = (w1 , w2 ). Then we have
T (v + w) = T (v1 + w1 , v2 + w2 ) = (a(v1 + w1 ) + b(v2 + w2 ), c(v1 + w1 ) + d(v2 + w2 )).
On the other hand, we have
T v + T w = T (v1 , v2 ) + T (w1 , w2 )
= (av1 + bv2 , cv1 + dv2 ) + (aw1 + bw2 , cw1 + dw2 )
= (a(v1 + w1 ) + b(v2 + w2 ), c(v1 + w1 ) + d(v2 + w2 )).
The proof for the scalar multiple is very similar.
2. Show that T : V → W is linear if and only if for any vi ∈ V and αi ∈ F, we have
!
n
n
X
X
T
αi vi =
αi T (vi ).
i=1
i=1
Solution: We use induction on n.
First, let us check that it is true for n = 1: we have
T (a1 v1 ) = a1 T (v1 ),
which is one of the conditions of linearity. So this checks.
Now, let us assume that the identity is true for n. Then we have
!
!
n+1
n
X
X
T
αi v i = T
αi vi + αn+1 vn+1
i=1
=T
=T
i=1
n
X
i=1
n
X
i=1
!
αi vi
+ T (αn+1 vn+1 )
!
αi vi
+ αn+1 T (vn+1 ).
(1)
The last two lines are just using the definition of linearity. Now, using the induction hypothesis, we have
!
n+1
n
n+1
X
X
X
T
αi vi =
αi T (vi ) + αn+1 T (vn+1 ) =
αi T (vi ),
i=1
i=1
i=1
and we are done.
3. Show that if T : V → W is linear, then T (0V ) = 0W .
Solution: Let v ∈ V be any vector. Then 0 · v = 0V . So we have
T (0V ) = T (0 · v) = 0 · T (v) = 0W .
4. (q.v. Freidberg, Insel, Spence 2.1.2).
Solution: We have T : R3 → R2 defined by T (a1 , a2 , a3 ) = (a1 − a2 , 2a3 ).
Let us first consider N (T ). These are the points that satisfy
a1 − a2 = 0,
2a3 = 0,
or a1 = a2 and a3 = 0. We can think of a2 as free, and thus the solution set can be written
(a2 , a2 , 0) = a2 (1, 1, 0)
so that a basis is clearly {(1, 1, 0)}, and the nullity is one.
The quickest way to compute the range is to write down a set that spans it, and we know we
can obtain this by choosing a basis v1 , v2 , v3 for R3 and then computing
T v1 = (1, 0),
T v2 = (−1, 0),
T v3 = (0, 2).
Since R(T ) contains both (1, 0) and (0, 2), and this is a basis for R2 , then this is a basis for
R(T ). Thus T has rank 2.
Rank–Nullity gives 1 + 2 = 3, which is true.
Also, we see it is not one-to-one, but it is onto.
5. (q.v. Freidberg, Insel, Spence 2.1.3).
Solution: We have T : R2 → R3 defined by T (a1 , a2 ) = (a1 + a2 , 0, 2a1 − a2 ).
Let us first consider N (T ). These are the points that satisfy
a1 + a2 = 0,
2a1 − a2 = 0.
This is a homogeneous system, and it is not hard to see that the equations are independent,
so this system has only the trivial solution. Thus N (T ) = {(0, 0)} and the nullity is zero. T
is one-to-one.
If we choose v1 = (1, 0) and v2 = (0, 1), then
T v1 = (1, 0, 2),
T v2 = (1, 0, −1).
Since these vectors are not scalar multiples of each other, they are independent and thus form
a basis for R(T ). From this we see that the rank of T is 2.
Rank–Nullity gives 0 + 2 = 2 which is true.
Finally, it is one-to-one, but not onto.
6. (q.v. Freidberg, Insel, Spence 2.1.5).We have T : P2 (R) → P3 (R) given by
T (f (x)) = xf (x) + f 0 (x).
We can choose {1, x, x2 } as a basis for P2 (R), so we have
T 1 = x,
T x = x2 + 1,
T x2 = x3 + 2x.
Note that these polynomials are independent — one way to see this is that they are all different
degrees, so there is no way in which we can take a linear combination of some of them and
obtain another one. A more direct proof is: let us write
ax + b(x2 + 1) + c(x3 + 2x) = 0,
then we see that the coefficient of x3 is c, the coefficient of x2 is b, and the coefficient of x is
a, and all of these must be zero. Since this set is independent, it spans R(T ) and therefore
the rank of the transformation is 3.
To compute the nullspace, we need to find a polynomial that satisfies
xf (x) + f 0 (x) = 0.
But notice that if f (x) 6= 0, then xf (x) and f 0 (x) are different degrees (the first is one more
than f , the second is one less) and thus they cannot cancel. So N (T ) = {0}.
Rank–Nullity gives 0 + 3 = 3, which is true.
Finally, we see that T is one-to-one but not onto.
7. (q.v. Freidberg, Insel, Spence 2.1.6).
Solution: We have T : Mn×n (R) → R defined by T (A) = Tr(A).
Let us first compute R(T ). Notice that the target vector space is one-dimensional, and therefore
R(T ) = R or R(T ) = {0}. If we can show that, for any A ∈ Mn×n (R), T (A) 6= 0, then it
must be that R(T ) = {0}. But it’s not hard to see that T (I) = n 6= 0, and so the rank of this
transformation is 1.
If we could use Rank–Nullity now, then we would know that the nullity is n−1, but the problem
says not to use it here.
So, ok, we need to find a basis for the nullspace of T explicitly. First, let us write down a
basis for Mn×n (R) as follows: for any 1 ≤ i, j ≤ n, let us define A(i,j) as the matrix with a 1
in the (i, j)th spot and zeros elsewhere. These are clearly linearly independent, and there are
n2 of them, so they form a basis for Mn×n (R).
(A compact way of writing this definition is
(
1, i = k ∧ j = l,
(i,j)
Ak,l =
0, else,
but that’s not necessary.)
So we want to compute T (A(i,j) ). Notice that if i 6= j, then A(i,j) has all zeros on the diagonal,
so T (A(i,j) ) = 0. But,if i = j, then there is a single one on the diagonal, and we will have
T (A(i,i) ) = 1.
But then this means that
T A(i,i) − A(n,n) = T (A(i,i) ) − T (A(n,n) ) = 1 − 1 = 0.
Therefore, for any i, A(i,i) − A(n,n) ∈ N (T ), and if i 6= n, then this is a nonzero matrix in
N (T ).
If we count these up, we see that we have them all: we have n(n − 1) = n2 − n matrices of
the form A(i,j) with i 6= j, and n − 1 matrices of the form A(i,i) − A(n,n) with i 6= n, giving a
grand total of n2 − n + n − 1 = n2 − 1 matrices. It is not hard to see that these matrices are
independent (e.g. every one of them has a 1 in a spot where no other matrix has a nonzero
number), and this then forms a basis.
Since (n2 − 1) + 1 = n2 , Rank–Nullity is again established.
The transformation is onto, but it is extremely not one-to-one.
8. (q.v. Freidberg, Insel, Spence 2.1.9).
Solution:
(a) T (0, 0) = (1, 0) 6= (0, 0).
(b) T (1, 1) = (1, 1). T (2, 2) = (2, 4) 6= 2(1, 1).
(c) T (1, 0) = (sin 1, 0). T (2, 0) = (sin 2, 0). Since sin 2 6= 2 sin 1, this is not linear.
(d) T (1, 0) = (1, 0). T (−1, 0) = (1, 0). T ((−1, 0) + (1, 0)) = T (0, 0) = (0, 0) 6= T (−1, 0) +
T (1, 0).
(e) T (0, 0) = (1, 0) 6= (0, 0).
9. (q.v. Freidberg, Insel, Spence 2.1.15).
Solution: This is a tricky problem in some ways. Note that since the vector space is infinitedimensional we cannot use Rank–Nullity and we have to do things “by hand”.
First, let us show that T is linear. This uses some results from calculus. We have
x
Z
Z
x
Z
0
0
0
Also,
Z
x
Z
cf (t) dt = c
T (cf ) =
g(t) dt = T (f ) + T (g).
f (t) dt +
(f (t) + g(t)) dt =
T (f + g) =
x
x
f (t) dt = cT (f ).
0
0
So T is linear!
Now, we want to show that it is one-to-one, i.e. that N (T ) = {0}. Let us assume that
T (f ) = 0; this would mean that
Z x
f (t) dt = 0 for all x.
0
Let us write
x
Z
g(x) =
f (t) dt.
0
Since g(x) = 0 for all x, then g 0 (x) = 0 for all x, so we have, by the Fundamental Theorem
of Calculus,
Z x
d
0=
f (t) dt = f (x) for all x,
dx 0
and thus f = 0.
Now we want to show that it is not onto. We will show that R(T ) contains no nonzero
constant. For example, if T f = 1, then the same argument as before would give that g(x) = 1,
so that g 0 (x) = 0, and this implies that f (x) = 0. Thus, the only way we can have a constant
come out of the transformation T would be if we plug in 0, and in that case we could get the
zero constant.
10. (q.v. Freidberg, Insel, Spence 2.1.16).Like the previous problem, we cannot use Rank–
Nullity and must do things by hand.
First, let us note that T 1 = 0, so T is not one-to-one.
Now we want to show that T is onto. In particular, we need to show that for any polynomial
g, there is another polynomial f such that T f = g, then we are done.
There are two ways to do this. The first is to use the Fundamental Theorem of Calculus: if
we define
Z x
f (x) =
g(t) dt,
0
Then
d
d
Tf =
f=
dx
dx
Z
x
g(t) dt = g(x),
0
so we have T f = g.
Another way is to construct f explicitly. Let
g(x) =
n
X
ak x k .
k=0
Then if we define
n
X
ak k+1
x ,
f (x) =
k+1
k=0
then we know from calculus that T f = g.
11. (q.v. Freidberg, Insel, Spence 2.1.18).Since N (T ) = R(T ), T must have the same rank
and nullity. By Rank–Nullity, we have
dim(N (T )) + dim(R(T )) = 2,
so we then must have dim(N (T )) = dim(R(T )) = 1.
Let us pick a one-dimensional subspace of R2 : let W = Span((1, 0)). So we want to write
down a transformation T such that T (1, 0) = (0, 0), but for any x, y ∈ R2 , T (x, y) = (a, 0) for
some a.
What if we have a transformation with T (0, 1) = (1, 0)? Then if we let v1 = (1, 0), v2 = (0, 1),
then T v1 , T v2 span R(T ), and we have
T (1, 0) = (0, 0),
T (0, 1) = (1, 0).
Thus {(0, 0), (1, 0)} spans R(T ). But this is clearly a dependent set since it has 0, but we can
just as clearly make it independent by discarding the zero vector, so we have that {(1, 0)} is
a basis for R(T ).
The formula for T is then
T (x, y) = T (x(1, 0) + y(0, 1)) = xT (1, 0) + yT (0, 1) = x(0, 0) + y(1, 0) = (y, 0).
If we want to represent T as a matrix, we see that by using the standard basis, we can write
the matrix
0 1
.
0 0