What happens to a transformation matrix when you switch the order of the basis of a linear transformation matrix

Question

I had a question on an exam that asked what the matrix of a linear transformation T. In this case, T was \begin{bmatrix} 2 & 1\\ 4 & 3 \end{bmatrix} with respect to a basis (f,g). If the new basis of T was (g,f), what would the transformation matrix be? Neither g nor f are specified, and I've tried looking everywhere to figure out what would happen if the order of the basis was switched. I honestly have no idea how to go about approaching this question.

Answer 1

For this matrix, let call it $A_{(f,g)}=\begin{pmatrix} 2 & 1 \\ 4 & 3\end{pmatrix}$ in the base $(f,g)$.

We are reading it like this $\begin{cases} Af=2f+4g\\ Ag=1f+3g \end{cases}$

So if we swap the order of the vectors in the base we have $\begin{cases} Ag=3g+1f\\ Af=4g+2f \end{cases}$

And the matrix $A$ in the base $(g,f)$ is simply $A_{(g,f)}=\begin{pmatrix} 3 & 4 \\ 1 & 2\end{pmatrix}$

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Let's have a look at a $3\times 3$ matrix in base $(e_1,e_2,e_3)$

$A_{(e_1,e_2,e_3)}=\begin{pmatrix} 1&2&4 \\-3&5&0\\2&-2&7\end{pmatrix}$

It reads as below :

$\begin{array} {ccc|c} A(e_1) & A(e_2) & A(e_3)\\ \hline 1 & 2 & 4 & e_1 \\ -3 & 5 & 0 & e_2 \\ 2 & -2 & 7 & e_3 \\ \end{array}\qquad$ meaning for instance $A(e_1)=e_1-3e_2+2e_3$

In the shuffled base $(e_2,e_3,e_1)$ let shuffle first the columns,

$\begin{array} {ccc|c} A(e_2) & A(e_3) & A(e_1)\\ \hline 2 & 4 & 1 & e_1 \\ 5 & 0 & -3 & e_2 \\ -2 & 7 & 2 & e_3 \\ \end{array}$

And then the rows accordingly

$\begin{array} {ccc|c} A(e_2) & A(e_3) & A(e_1)\\ \hline 5 & 0 & -3 & e_2 \\ -2 & 7 & 2 & e_3 \\ 2 & 4 & 1 & e_1 \\ \end{array}$

And the new matrix is $A_{(e_2,e_3,e_1)}=\begin{pmatrix} 5&0&-3 \\ -2&7&2\\ 2&4&1\end{pmatrix}$

Is the manual processus clearer now ?

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Let's now look at the theory

We define $P=\begin{pmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{pmatrix}$ the matrix of change of basis.

Notice how $e_2$ is in first column, $e_3$ in the second and $e_1$ in the third column: $P=(e_2,e_3,e_1)$.

Its inverse is $P^{-1}=\begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{pmatrix}$ and if you carry on the calculation, you can observe the equality $$P^{-1}A_{(e_1,e_2,e_3)}P=A_{(e_2,e_3,e_1)}$$

This how a change of base is operated : by writing $P^{-1}AP$ with a suitable matrix $P$ of change of basis.