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I've been struggling with this question for two days now. I usually don't have a problem with change of basis but this particular question I couldn't solve, I really hope someone can help.

We have $(e_1, e_2, e_3)$ a basis of $\mathbb R^3$ and $(v_1, v_2, v_3)$ another basis defined as:

$v_1 = 3 e_1 - e_2$
$v_2 = e_1$
$v_3 = 2e_2 - e_3$

We consider the endomorphism f of $\mathbb R^3$ with a matrix in the basis $(e_1, e_2, e_3)$ as:

$$ \begin{pmatrix} -2 & -2 & -1 \\ 1 & -1 & 1 \\ -2 & 2 & 1 \\ \end{pmatrix} $$

What is its matrix C in the basis $(v_1, v_2, v_3)$ ?

The answer given is:

$$ \begin{pmatrix} 12 & 3 & -3 \\ -40 & -11 & 6 \\ 8 & 2 & -3 \\ \end{pmatrix} $$

I don't know how we got to this answer. I didn't know whether $(e_1, e_2, e_3)$ were the Basis Vectors so I thought to compute what's the Basis of the Linear Transformation associated with the matrix given (since it comes from $(e_1, e_2, e_3)$) then plugging the resulting vectors in their respective places in each one of those three Linear Combinations given to get $(v_1, v_2, v_3)$ I then supposed (most likely mistakenly) that the resulting matrix made off the column vectors $(v_1, v_2, v_3)$ would be the Change of Basis Matrix, wich I multiplied by the matrix given in the question and of course it didn't work.

I tried many other approaches but in vain. I would really appreaciate it if someone tells me the steps to solve this question, and it would be even better if someone shows me the problem in my line of reasoning so I can make the necessary adjustments. Much appreciated.

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  • $\begingroup$ What was your change-of-basis matrix? Also, have you calculated its inverse? $\endgroup$ – Berci Apr 24 '18 at 23:34
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If$$M=\begin{pmatrix}3 & 1 & 0 \\ -1 & 0 & 2 \\ 0 & 0 & -1\end{pmatrix},$$then $M$ is the change of bases matrix from the basis $(v_1,v_2,v_3)$ to the basis $(e_1,e_2,e_3)$. So, the matrix that you're after is\begin{align}M^{-1}.\begin{pmatrix}2 & -2 & -1 \\ 1 & -1 & 1 \\ -2 & 2 & 1\end{pmatrix}.M&=\begin{pmatrix} 0 & -1 & -2 \\ 1 & 3 & 6 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}2 & -2 & -1 \\ 1 & -1 & 1 \\ -2 & 2 & 1\end{pmatrix}.\begin{pmatrix}3 & 1 & 0 \\ -1 & 0 & 2 \\ 0 & 0 & -1\end{pmatrix}\\&=\begin{pmatrix}12 & 3 & -3 \\ -40 & -11 & 6 \\ 8 & 2 & -3\end{pmatrix}\end{align}

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  • $\begingroup$ Thank you so much ! $\endgroup$ – Metrician Apr 25 '18 at 0:32
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In order to keep things straight in problems like these I like to think in terms of input and output bases. You have a matrix for $f$ that has the $e$-basis as inputs and outputs and want one that has the $v$-basis instead. This means that you have to convert the input from the $v$-basis to the $e$-basis and vice-versa for the output. The equations that you have relating the two bases tell you how to convert from $v$ to $e$, i.e., if you form the matrix of coefficients you get the $v$-to-$e$ change of basis matrix. That goes on the right, on the input side, and its inverse goes on the left, on the output side.

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  • $\begingroup$ Thank a lot for your comment. It helped correct some misconceptions I had when viewing this problem. $\endgroup$ – Metrician Apr 25 '18 at 0:32
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Hint: You will need the matrix $P=[v_1 v_2 v_3]$. It converts from the basis $\{v_1,v_2,v_3\}$ to the standard basis.

You will then need to compute $P^{-1}$.

Your matrix will be $C=P^{-1}AP$, where $A$ is the (given) matrix representing $f$...

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