Download MA 237-102 Linear Algebra I Homework 5 Solutions 3/3/10 1. Which

MA 237-102
Linear Algebra I
Homework 5 Solutions
3/3/10
1. Which of the following transformations are linear? Justify your answers.
(i) T : P2 → P3 defined by T (p(t)) = t3 p′ (0) + t2 p(0);
For any polynomials p(t) and q(t) we have
T ((p + q)(t)) = t3 ((p + q)′ (0))+t2 (p+q)(0) = t3 p′ (0) + t2 p(0) + t3 q ′ (0) + t2 q(0) .
Similarly for any real number a we have
T (ap(t)) = t3 ((ap)′ (0)) + t2 (ap)(0) = a t3 p′ (0) + t2 p(0) .
It follows that T is linear.
(ii) T : P1 → P2 defined by T (p(t)) = tp(t) + p(0);
For any polynomials p(t) and q(t) we have
T ((p + q)(t)) = t ((p + q)(t)) + (p + q)(0) = (tp(t) + p(0)) + (tq(t) + q(0)) .
Similarly for any real real number a we have
T (ap(t)) = t ((ap)(t)) + (ap)(0) = a (tp(t) + p(0)) .
Thus T is linear.
(iii) T : P1 → P2 defined by T (p(t)) = tp(t) + 1.
If T (p(t)) = tp(t) + 1, then T (3p(t)) = t(3p(t)) + 1 = 3tp(t) + 1, which is
not the same as 3T (p(t)) = 3tp(t) + 3. Thus T is not linear.
2. Let V and W be vector spaces and suppose T : V → W is a linear transformation.
(i) Show that T (0V ) = 0W ;
For any vector X in V we have
T (0V ) = T (0 · X) = 0 · T (X) = 0W .
(ii) Show that the set of vectors X in V with the property that T (X) = 0 is a subspace of
V (this subspace is called the kernel of T );
If T (X) = 0 and T (Y ) = 0, then for any real numbers s and t we have
T (sX + tY ) = sT (X) + tT (Y ) = s0 + t0 = 0.
Thus the kernel of T is a subspace of V.
(iii) Show that the set of vectors Y in W such that the equation T (X) = Y is solvable is a
subspace of W (this subspace is called the image of T ).
Assume that Y1 and Y2 are in the image of T . Then there are (not necessarily
unique) vectors X1 and X2 so that T (X1 ) = Y1 and T (X2 ) = Y2 . Then
for any real numbers s and t we have
T (sX1 + tX2 ) = sT (X1 ) + tT (X2 ) = sY1 + tY2 ,
so that sY1 +tY2 is also in the image of T .
a subspace of W.
Thus the image of T is
3. (i) Find the matrix representation for the linear map T : R3 → R2 given by
 
x
2y
−
z
T y  =
.
3x − y
z
We apply T to the standard basis vectors in R3 :
 
1
0




T
,
=
0
3
0
 
0
2




,
T
=
1
−1
0
 
0
−1




.
T
=
0
0
1
It follws that the matrix representation of T is
0 2 −1
.
AT =
3 −1 0
(ii) Find a basis for the kernel of T .
The kernel of T is exactly the nullspace of AT , so we row-reduce AT
as follows:
3 −1 0
0 2 −1
.
0 2 −1
3 −1 0
Solving the corresponding homogeneous system gives
z = t,
2y − z = 0 ⇒ y = t/2
3x − y = 0 ⇒ x = t/6,
so the solution is

 
t/6
1/6
t/2 = t 1/2 .

t
1
It follows that this vector is a basis for the nullspace of AT , and
hence for the kernel of T .
(iii) Find a basis for the image of T .
The image of T is the column space of AT . From the row-reduction above,
we see that the first two columns of AT are its pivot columns. Thus
a basis for the column space of AT , and hence also for the image of
T , is
0
2
,
.
3
−1