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Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a linear map and consider the matrix representation $$M_{B}^{B} =\begin{pmatrix} 4 & 6\\ 6 & 3 \end{pmatrix}$$ with respect to the basis $B=\bigg\{ \begin{pmatrix} 2 \\ 2 \end{pmatrix} , \begin{pmatrix} 3\\ 2 \end{pmatrix}\bigg\}$. Find the matrix representation of $f$ with respect to the canonical basis.

MY ATTEMPT: Let $b_1=\begin{pmatrix} 2 \\ 2 \end{pmatrix}$, $b_2=\begin{pmatrix} 3\\ 2 \end{pmatrix}$ and $e_j$ the j-th vector of the canonical basis. Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix} \Longrightarrow f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) =4e_1 + 6e_2$

$f(b_2)=\begin{pmatrix} 6\\ 3 \end{pmatrix} \Longrightarrow f(3e_1 +2e_2)=3 f(e_1) +2f(e_2) =6e_1 + 3e_2$

Solving this system I obtain that $f(e_1)=2e_1 -3e_2=\begin{pmatrix} 2\\ -3 \end{pmatrix}$ and $f(e_2)= 6e_2=\begin{pmatrix} 0\\ 6 \end{pmatrix}$. So the matrix rappresentation is $M_{E}^{E}=\begin{pmatrix} 2 & 0\\ -3 & 6 \end{pmatrix}$.

BUT my theacher says that the solution is $M_{E}^{E}=\begin{pmatrix} 13 & 10\\ 0 & 0 \end{pmatrix}$. Where I am wrong?

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$$f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) \color{red}{=4e_1 + 6e_2}$$

This is where you are wrong, $4$ and $6$ are coordinates in the basis $B$, so there is one more step to get $e_1$ and $e_2$ into the game. Otherwise your approach is nice.

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  • $\begingroup$ How can I do that? $\endgroup$ – fcoulomb Jun 1 '18 at 18:47
  • $\begingroup$ @Besh00 write $\color{red}{=4b_1+6b_2}$ instead of the red part, and then write both $b_1$ and $b_2$ in terms of $e_1$, $e_2$. $\endgroup$ – Arnaud Mortier Jun 1 '18 at 18:50
  • $\begingroup$ Why $f(b_1)=4b_1+6b_2$? $\endgroup$ – fcoulomb Jun 1 '18 at 21:10
  • $\begingroup$ You wrote yourself "Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix}$". These coordinates are with respect to the basis $B$ since they are taken from a matrix wrt $B,B$. $\endgroup$ – Arnaud Mortier Jun 1 '18 at 21:14

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