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I came across this exercise:

Let $\psi: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a linear map with

$$ \psi\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}1\\0\\-2\end{pmatrix} \quad\text{and}\quad \psi\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}0\\1\\1\end{pmatrix} $$

  • For which bases is the transformation matrix obvious?
  • Determine the transformation matrix regarding the standard bases.

For the first task I have chosen bases $$ B = \left( \begin{pmatrix}1\\1\end{pmatrix}, \begin{pmatrix}1\\2\end{pmatrix} \right) \quad\text{and}\quad C = \left(e_1, e_2, e_3\right) $$ where $e_i$ are the unit vectors.

I have concluded that the transformation matrix is $$ M^B_C (\psi) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ -2 & 1 \end{pmatrix} $$

For standard bases $S$ of $\mathbb{R}^2$ and $T$ of $\mathbb{R}^3$ I ended with this: $$ M^S_T(\psi) = \begin{pmatrix} 2 & -1 \\ -1 & 1 \\ -5 & 3 \end{pmatrix} $$

I am unsure if I am right, so it would be good if someone could check my results.

1 Answers 1

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Your solutions are correct. You can easily check them yourself. Multiply $M_C^B(\psi)$ with $\begin{pmatrix}1 \\0\end{pmatrix}$ and with $\begin{pmatrix}0 \\ 1\end{pmatrix}$ (which is how $\begin{pmatrix}1 \\1\end{pmatrix}$ and $\begin{pmatrix}1 \\2\end{pmatrix}$ are represented with base $B$), and multiply $M_T^S(\psi)$ with $\begin{pmatrix}1 \\1\end{pmatrix}$ and $\begin{pmatrix}1 \\2\end{pmatrix}$.