Transition matrix problems

Question

Recently learned about Transition matrix and other related stuff, was trying to do some practice questions and came across this:

Consider the ordered bases: $S$ = ($1$, $x$, $x^2$, $x^3$)

$B$ = ($1$, $2$ $-$ $x^2$, $x + x^3$, $-3 + 2x^3$)

I wish to find a transition matrix $P$($S,B$), but I am confused about this one.

I attempted to do something like this:

[($1$, $2$ $-$ $x^2$, $x + x^3$, $-3 + 2x^3$)$s$] but then I'm stuck, I don't know how to separate those and find the matrix..

Hopefully someone can offer some suggestions!

Thank you!

Answer 1

Just write the new basis' elements as linear combinations of the old basis:

$$\begin{align*}&1=\color{red}1\cdot1+\color{red}0\cdot x+\color{red}0\cdot x^2+\color{red}0\cdot x^3\\{}\\ &2-x^2=\color{red}2\cdot1+\color{red}0\cdot x+\color{red}{(-1)}x^2+\color{red}0\cdot x^3\\{}\\ &x+x^3=\color{red}0\cdot 1+\color{red}1\cdot x+\color{red}0\cdot x^2+\color{red}1\cdot x^3\\{}\\ &-3+2x^3=\color{red}{(-3)}\cdot1+\color{red}0\cdot x+\color{red}0\cdot x^2+\color{red}2\cdot x^3\end{align*}$$

and now just take the (transpose, usually) matrix of coefficients:

$$\begin{pmatrix}1&2&0&\!-3\\ 0&0&1&0\\ 0&\!-1&0&0\\ 0&0&1&2\end{pmatrix}$$

Sometimes, though, the question can be understood as writing the old basis elements as linear combinations of the new ones. The method, of course, is the same.

Answer 2

$$ \left[\begin{array}{cccc} 1 & -\frac{3}{2} & 2 & \frac{3}{2} \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & \frac{1}{2} \end{array}\right] $$

There's no other way. For instance, consider the first column. If the second entry is non-zero, then the resulting linear combination will contain $x^2$. If the third entry is non-zero, then the resulting linear combination will contain $x$. If the third entry is non-zero, then, since we've established that the third entry must be zero, the resulting linear combination will contain $x^3$. Similar considerations apply to each of the other three columns.