Question

o 1 0 -1 Exercise 2. Let A= in M3,R, and ✓ = 0 in R3. -1 0 For every vector W E R3, set g(W) = WT AT ER. (i) Show that g: R3 → R defines a linear transformation. What is the matrix [g]C,B in the - 1 bases C = {1} and B { 8.00 } ? (ii) Let f : R3 → R be the function defined by f() = 7T Aw E R. Show that f(w) = g(w) for every w E R3. (iii) Find a vector v such that g() = 0;" w for every ✓ € R3.

Answer 1

given that,

А 1 0 -1 0 1 0 -1 0 1
and
1 ū=

Now, g is defined by $g: \mathbb{R}^3 \to \mathbb{R}$

g() = " Av
for every
WER

(i) We now show that, g defines a linear transformation.

Let, , then,

w1, 2 E R

g(W1 + 2) = (1 + 2)" A✓ = (w1 + w ) A✓ = 1 A✓ + 1 AŬ = g(W1) + 9(2)

and let, , then,

QER

glaw) = (aw" Av = a(w)TAT = ag()

Hence,  g defines a linear transformation.

Now,

1 B = { 0 0 1 0
and   
C = {1}
are the bases of
R3, R
respectively.

Now,

1 1 g(0) ) = [1 0 0.0 1 0 -1 0 -1 0 1 1 0 1 = [1 0 -1].0 =1-1=0 0

1 0 1 0 -1 9( 1 ) = [o 1 0 0 0 0 -1 0 1 1 0 1 [o 1 0].0 0

0 1 1 g(0) = [o o 1]. O 1 0 -1 0 -1 0 1 1 0 = [-1 0 1].0 = 0 1 1

Then,

The matrix

19]C,B= [0 0 0

(ii) f is defined by

f(w) = " Aw
for every
WER

Let,

T w Y

then,

T 1 T f(y) = [1 0 1].0 0 -1 0 -1 0 1 y = [0 0 0]y = 0 1

. 1 1 9 y) = [2 y ) = [r y ] .0 0 -1 0 -1 0 1 1 0 = [x – y 2 y +1 +2] . O = x - 2-2 +2 = 0

Hence, f=g

(iii) Since g is zero linear transformation we choose

1 0 0 0

for which  
9(W) = 7T.
holds   for every
WER