Question

9. Consider a basis B = {bi, b2} of a sulspoo, W of R4 where -3 (a) Determine the coordinates of x(3,-1,-2,1) in the basis B (i.e. fnd x). (b) Suppose that bl el-C2 and b2 2c1 +c2. Determine the coordinates of x = (3.-1,-21) in the basis C = {c,,c) (i.e. find [x le) (e) Suppose t dbb an d2b 3b s D- di da a basis of W Why or why not?

I need all details. Thx

Answer 1

Solution:

$\small 9.\left ( a \right )$

Let $\small x=\alpha b_{1}+\beta b_{2}$

(3,一1,-2.1)=0 (5,1,-3.3)+3(-1,3.1, 1)

$\small \therefore 5\alpha -\beta =3,\: \: \: \alpha +3\beta =-1,\: \: \: -3\alpha +\beta =-2,\: \: \: 3\alpha +\beta =1$

Solving these equations we get ,

1-2 2

b2 r)

$\small \left [ x \right ]_{B}=\begin{bmatrix} \frac{1}{2}\\ \\ -\frac{1}{2}\\ \end{bmatrix}$

$\small \left ( b \right )$

Let  $\small P$ be the change of basis matrix from $\small B$ to  $\small C$

2 B+C

$\small \left [ x \right ]_{C}=P_{B\rightarrow C}\left [ x \right ]_{B}$

$\small =\begin{bmatrix} 1 & 2\\ -1& 1 \end{bmatrix}\begin{bmatrix} \frac{1}{2}\\ \\ -\frac{1}{2}\\ \end{bmatrix}=\begin{bmatrix} -\frac{1}{2}\\ \\ -1\\ \end{bmatrix}$

$\small \left ( c \right )$

$\small d_{2}=-\left ( 2b_{1}-3b_{2} \right )$

$\small d_{1}=\frac{2b_{1}-3b_{2}}{2}=-\frac{d_{2}}{2}$

Since  $\small d_{1}$ can be written as a linear combination of  $\small d_{2}$ ,  $\small \left \{ d_{1} ,d_{2}\right \}$ is linearly dependent.

$\small \therefore \left \{ d_{1} ,d_{2}\right \}$ is not a basis of  $\small W$ .