Question

1. The figure shows two bases, consisting of unit vectors, for ll?2: S —— (i,j) and 0 = (ut,ut).

(a) Find the transition matrix s B

(b) Find the transition matrix PB S

Answer 1

Kindly Note

a) The figure referred to was not provided

b) There are typing errors.

Adjusting the conditions of the questions, and rewriting it as;

The figure show two basis for $\reals^2:S=(i,j)$ and $B=(u_1',u_2')$

a) Find the transition matrix $P_S\to\>_B$

b) Find the transition matrix $P_B\to\>_S$

Solution:

In reference to the figure, first basis $S=(i,j)$

Where $i=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ ,$\quad\>j=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$

Second basis $B(u'_1,u'_2)$

Where $u'_1=\begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}$ , $u'_2= \begin{pmatrix} -1 \\ 2 \end{pmatrix}$

a) Expressing elements of $B$ in terms of elements of $S$

$\begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}$ $=a\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $+b\begin{pmatrix} 0 \\ 1 \end{pmatrix}$

$\implies a=\frac{1}{2},b=0$

$\begin{pmatrix} -1 \\ 2 \end{pmatrix}$ $=a'\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $+b'\begin{pmatrix} 0\\ 1 \end{pmatrix}$

$\implies\>a'=-1,b'=2$

$\therefore$ Transition matrix is $\begin{pmatrix} a & a' \\ b& b' \end{pmatrix}$

$\therefore$ Transition matrix $P_S\>\to_B=\begin{pmatrix} \frac{1}{2} & -1\\ 0& 2 \end{pmatrix}$

b) Transition matrix $P_B\>\to\>S$ $=(P_S\to\>_B)^{-1}$

$\begin{pmatrix} \frac{1}{2} & -1 \\ 0& 2 \end{pmatrix}$ $^{-1}$ $= \frac{1}{1}\begin{pmatrix} 2 & 1 \\ 0 & \frac{1}{2} \end{pmatrix}$

$\therefore P_B\to\>_S$ $= \begin{pmatrix} 2 & 1 \\ 0 & \frac{1}{2} \end{pmatrix}$