Expressing a linear map $\phi: \>{\mathbb R}^4\to{\mathbb R}^4$ in terms of a new base

Question

In terms of the standard coordinates $x$, $y$, $z$, $t$ a linear map $\phi: \>{\mathbb R}^4\to{\mathbb R}^4$ appears as $$\phi{(x,y,z,t)} = (t,x,y,z)\ .$$ Now we are given a new base $$R = \{ (1,0,1,0),(0,1,0,1),(0,1,1,0),(0,1,1,1) \}$$ of $\mathbb{R}^4$ and should find the matrix $M_R(\phi)$.

Is my thinking correct? As I understand I need to find the matrix created by the linear transormation: $$ \left( \begin{matrix} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0& 1& 0 & 0\\ 0& 0& 1 & 0\\ \end{matrix} \right) $$ then, solve $$ \left[ \begin{array}{cccc|c} 1&0&0&0&a\\ 0&1&1&1&b\\ 1&0&1&1&c\\ 0&1&0&1&d\\ \end{array} \right] $$

by turning it to row echelon form to find out how the vector will look in the R basis. Is it the correct way to do a question like this?

Answer 1

The most straight-forward way (as I see it) would be to follow this:

The columns of the matrix representation of a linear transformation are the images of the basis vectors.

So, since the first basis vector in $R$ is $(1,0,1,0)$, the first column of $M_R(\phi)$ is $\phi(1, 0, 1, 0) = (0,1,0,1)$, but expressed using the basis $R$ rather than the standard basis. And so on.

Answer 2

The matrix $P = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ \end{bmatrix}$ transforms the basis $R$ to the canonical basis $E$.

Therefore

$$M_R(\phi) = P^{-1}M_E(\phi) P = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ \end{bmatrix}^{-1} \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 2 & 1 \\ \end{bmatrix}$$