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I guess the answer is the $n^2\times n^2$ block matrix $diag(B, B, ..., B)$. Where shall I start to prove it? I tried to find the matrix by figuring out the image of the standard basis of $V$, but it gets complicated.

Any idea would be appreciated.

One more question

$\bullet$ Any idea about how to compute $det(L_B)$? I am sure that the answer is $det(B)^n$ if $B$ is invertible.

I found the answer of question $\bullet$ here :)

Majid
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  • Surely this will depend on your choice of ordered basis for V. For example one basis $\mathcal{E}$ in the $2\times 2$ matrices is $E_1=\begin{bmatrix}1 & 0 \ 0 & 0\end{bmatrix}, E_2=\begin{bmatrix}1 & 1 \ 0 & 0\end{bmatrix}, E_3=\begin{bmatrix}1 & 1 \ 1 & 0\end{bmatrix}, E_4=\begin{bmatrix}1 & 1 \ 1 & 1\end{bmatrix}$. Then you'd need to think of your vector space of matrices as column vectors $A=\sum x_iE$ then $[A]_\mathcal{E}=\begin{bmatrix}x_1 \ x_2\ x_3\ x_4\end{bmatrix}$. If you can post an example you're having trouble with I can be of more help. – N8tron May 20 '18 at 20:00

2 Answers2

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If we use a standard basis in which the matrix $$ A=\begin{pmatrix} a_{11}&a_{12}&\cdots &a_{1n}\\ a_{21}&a_{22}&\cdots &a_{2n}\\ \cdots\\ a_{n1}&a_{n2}&\cdots &a_{nn}\\ \end{pmatrix} $$ is represented as a vector of components $$ A=\begin{pmatrix} a_{11}\\\cdot\\\cdot \\a_{1n}\\ a_{21}\\\cdot\\\cdot \\a_{2n}\\ \cdots\\\cdots\\ a_{n1}\\\cdot \\\cdot\\a_{nn}\\ \end{pmatrix} $$ than the matrix that represents the transformation $L_B(A)=BA$ is a block matrix of the form $$ \begin{pmatrix} B_{11}&B_{12}&\cdots &B_{1n}\\ B_{21}&B_{22}&\cdots &B_{2n}\\ \cdots\\ B_{n1}&B_{n2}&\cdots &B_{nn}\\ \end{pmatrix} $$ where $B_{ij}$ is a diagonal matrix with as diagonal value the element $b_{ij}$ of the matrix $B$

You can prove this result starting from the case $n=2$ and using induction.

Emilio Novati
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Hint If you use the standard basis $E_{ij}$m, then $$L_B(E_{ij})=BE_{ij}=\begin{bmatrix} 0&0 &...& b_{1i} & 0 & ..0 \\ 0&0 &...& b_{1i} & 0 & ..&0 \\ 0&0 &...& b_{2i} & 0 & ..&0 \\ ...& ...&...&...&...&...&...\\ 0&0 &...& b_{ni} & 0 & ..0 \\ \end{bmatrix}$$ where the non-zero elements appear in column $j$. Write this as a linear combination of elementary matrices.

N. S.
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  • $E_{ij}=[\delta_{i,j}(m,n)]$ where $\delta_{i,j}(m,n)=\begin{cases} 1 &\text{ if } m=i \text{ and } n=j \ 0 & \text{otherwise} \end{cases}$. – N8tron May 20 '18 at 20:06
  • bases have to be matrices I don't think your multiplication is correct or you don't have a basis – N8tron May 20 '18 at 20:07
  • $E_{ij}$ is an $n\times n$ matrix with the only nonzero element in the row $i$ and column $j$ which is one. So, they form a basis. But, as I said, I tried this method, and I get confused every time! I am looking forward to another method. – Majid May 20 '18 at 20:09
  • @N. S., as I said in the question and comment above, I get confused with this straightforward technique. For example, could you please tell me where these entries will be set in the matrix of operator. They will be really sparse in the beginning and this makes the method complicated for me! – Majid May 20 '18 at 20:14
  • @Majid Look again at the last line in my answer. You need to write calculate the coefficients of $L_B(E_{ij})$ with respect to this basis, meaning write $L_B(E_{ij})$ as a linear combination of elementary matrices... Very important, you need to fix an order for this basis. Then, each $E_{ij}$ appears in some position in your list. The column with the same number as the position consists of its coefficients. – N. S. May 23 '18 at 13:27
  • @Majid For example, if $n=2$ and you pick the order $E_{11}, E_{12}, E_{21}, E_{22}$ you have $$L_B(E_{11})=\begin{bmatrix}b_{11} & 0 \b_{21} & 0 \end{bmatrix}=b_{11}E_{11}+0E_{12}+b_{21}E_{21}+0E_{22}$$ Thus, the first column is $\begin{bmatrix} b_{11} \ 0 \b_{21} \0 \end{bmatrix}$. – N. S. May 23 '18 at 13:29
  • @N.S. Thanks. I already figured it out after receiving the other answer here, but forgot to remove my comment on your answer. ;) – Majid May 23 '18 at 13:35