I am given the following linear transformation $L$:

$A=\begin{bmatrix}1&2\\0&3\end{bmatrix} \in \Bbb R^{2 \times 2}$

$L: \space \Bbb R^{2 \times 2} \longrightarrow \Bbb R^{2 \times 2}; \space X \mapsto AX$

I want to find the transformation matrix with respect to the basis

$\mathcal B_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}, \space \mathcal B_2=\begin{bmatrix}0&0\\1&0\end{bmatrix}, \space \mathcal B_3=\begin{bmatrix}0&1\\0&0\end{bmatrix}, \space \mathcal B_4=\begin{bmatrix}0&0\\0&1\end{bmatrix}$

I know the answer is: $M_{\mathcal B}(L)=\begin{bmatrix}1&2&0&0\\0&3&0&0\\0&0&1&2\\ 0&0&0&3\end{bmatrix}$

but I don't know how to get to that matrix.

Usually I would find the new transformation $M$ with respect to a basis $\mathcal B$ by computing:


where $C$ is the matrix that has the alternate basis vectors $b_1,...,b_n$ as its columns. However, in this case, my matrix $C$ would look like this:


which makes no sense at all. What am I doing wrong here?

  • 1
    $\begingroup$ Be careful, $A$ is not the matrix of any linear map here, so you can't use the formula you said. One way to find the matrix of L is simply to calculate $L(M)$ for each $M$ is your base of matrix. $\endgroup$
    – Augustin
    Jul 28, 2015 at 15:21
  • $\begingroup$ @Augustin Thanks. Can you explain this sentence: "A is not the matrix of any linear map here, so you can't use the formula you said." I don't quite get why I can't use my formula here. $\endgroup$
    – qmd
    Jul 28, 2015 at 15:25
  • $\begingroup$ @Augustin are you saying that $L$ is NOT a linear transformation? $\endgroup$
    – qmd
    Jul 28, 2015 at 16:27
  • $\begingroup$ That's not what I'm saying. When you have a linear map $f$ with matrix $A$ in base $B$ and $C$ is the change of basis from $B$ to $B'$ then the matrix of $f$ in base $B'$ is $C^{-1}AC$. But here you're not in that case. $A$ is not the matrix of $L$ in base $B$, it's just a matrix used to define $L$. $\endgroup$
    – Augustin
    Jul 29, 2015 at 7:17

1 Answer 1


The first column of $M_\mathcal{B}(L) = (m_{ij})_{1\le i,j\le 4}$ is defined as the coordinates (with respect to the basis $\mathcal{B}=(\mathcal{B}_1,\mathcal{B}_2,\mathcal{B}_3,\mathcal{B}_4)$) of $A\mathcal{B}_1$.

Since $$\begin{array}{rcl} A\mathcal{B}_1 &=& \begin{bmatrix}1&2\\0&3\end{bmatrix} \begin{bmatrix}1&0\\0&0\end{bmatrix} \\ &=& \begin{bmatrix}1&0\\0&0\end{bmatrix} \\ &=& 1\times \mathcal{B}_1 + 0\times \mathcal{B}_2+ 0\times \mathcal{B}_3+ 0\times \mathcal{B}_4 \\ A\mathcal{B}_1&=& m_{11}\times \mathcal{B}_1 + m_{21}\times \mathcal{B}_2+ m_{31}\times \mathcal{B}_3+ m_{41}\times \mathcal{B}_4. \end{array}$$ This gives us the first column. I let you do the same for the rest of the matrix.

  • $\begingroup$ So I am basically plugging in the new basis matrices $\mathcal B_1,...,\mathcal B_n$ into the linear transformation? Why does it follow that $$A \mathcal B_1=m_11 \times \mathcal B_1+m_21 \times \mathcal B_2....$$? I don't understand that step. $\endgroup$
    – qmd
    Jul 28, 2015 at 15:31
  • $\begingroup$ I think you should rethink about what is a matrix (with respect to a basis) of a linear map. $\endgroup$
    – user37238
    Jul 29, 2015 at 7:36

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