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 '15 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 '15 at 15:25
  • $\begingroup$ @Augustin are you saying that $L$ is NOT a linear transformation? $\endgroup$ – qmd Jul 28 '15 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 '15 at 7:17

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 '15 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 '15 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.