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$A$ is an $n\times n$ matrix and $L_A$ is the left-multiplication operator on the $n\times n$ matrices.

I have seen this question and the given answer. But I could not understand the answer. I have commented on the answer, but did not get any reply for more details.

So, could any one please give me some more details.

Majid
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  • Do you understand that if $V, W$ are vector spaces, and $A, B$ are linear operators on those spaces respectively, then for $A \oplus B: V \times W \to V \times W$ defined by $(A \oplus B)(v, w) = (Av, Bw)$ we have $\det(A \oplus B) = \det(A)\det(B)$? – Mees de Vries May 22 '18 at 15:20
  • Which matrix is multiplied from the left ? Apparently, not $A$ itself, otherwise the result would only hold for $n=2$. – Peter May 22 '18 at 15:20
  • This duplicate has an answer, which is quite understandable, I think. – Dietrich Burde May 22 '18 at 15:21
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    @Peter: I think it is $A$ itself -- left multiplication by $A$ is viewed as a linear operator on the $n^2$-dimensional vector space of $n\times n$ matrices. – hmakholm left over Monica May 22 '18 at 15:21
  • @MeesdeVries Thansk for your help. Is this a known result? Based on what you said, I think now I can understand the other answer. Could you please let me know where shall I find the proof of the result you said? – Majid May 22 '18 at 17:47

1 Answers1

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We're viewing left multiplication of $A$ as a linear operator on the $n^2$-dimensional vector space of $n\times n$ matrices.

One convenient basis for the $n^2$-dimensional space consists of all matrices that have $0$ in every entry except $1$ in one of them. $L_A$ maps each of these basis elements to a matrix that contains one column from $A$ and zeroes in all the other columns. If we write down the matrix of $L_A$ with respect to this basis (assuming we have ordered the basis in column-major order), we get this $n^2\times n^2$ block matrix:

$$ \begin{pmatrix} A & 0 & \cdots & 0 \\ 0 & A &\cdots & 0 \\ \vdots &\vdots & \ddots & \vdots \\ 0 & 0 &\cdots & A \end{pmatrix} $$

It is an elementary fact that the determinant of a block diagonal matrix is the product of the determinants of the blocks on the diagonal -- in this case, $(\det A)^n$.

(This can be seen by considering the full expansion of the determinant, for example, or by writing each of the blocks as a product of elementary matrices).

hmakholm left over Monica
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  • Thanks Henning. Could you please have a look at the first answer to this question https://math.stackexchange.com/questions/2789186/let-v-be-the-space-of-n-times-n-matrices-b-a-fix-matrix-then-what-is-mat There the matrix of $L_B$ is presented different from what you said. Also, I computed it for the case with $n=2$ and it was not like what you said but like the one I addressed you to see. Could you please explain what is my mistake? – Majid May 22 '18 at 17:37
  • You may also see here the matrix of $L_B$ is computed different from what you said and the same as the other one I mentioned in previous comment. https://math.stackexchange.com/questions/339263/finding-trace-and-determinant-of-linear-operator?rq=1 – Majid May 22 '18 at 17:43
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    @Majid: That other answer represents matrices as vectors in the order $(a_{11},a_{12},\ldots, a_{1n}, a_{21}, a_{22}, \ldots)^T$. In order to get a nice block diagonal matrix for $L_B$ you need instead to use $(a_{11},a_{21},\ldots, a_{n1}, a_{12}, a_{22}, \ldots)^T$ as a representation of the input matrix. In other words, keep each column of the input matrix together and just stack them on top of each other, instead of taking it row-by-row. (That's what I meant by "column-major order", but that was probably a bit too obscure -- mea culpa!) – hmakholm left over Monica May 22 '18 at 17:49
  • It was not "sua culpa", I did not read your answer enough carefully! Now, I understood! Thanks! – Majid May 22 '18 at 18:10