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Given a matrix block diagonal matrix as follows

${\bf A} = \left[ \begin{array}{cccc} {\bf a}_1 & 0 & 0 & 0 & \\ 0 &{\bf a}_2 & 0 & 0 \\ 0 & 0 & {\bf a}_3& 0 \\ 0 & 0 & 0& {\bf a}_4 \end{array} \right]$

where each non zero entry ${\bf a_i}$is $M\times 1$ vector and the zero are also $M\times 1$ vectors. Is it correct to say that the product of this matrix and any other matrix for example ${\bf D}$ ${\bf P = AD} = \left[ \begin{array}{cccc} {\bf a}_1 & 0 & 0 & 0 & \\ 0 &{\bf a}_2 & 0 & 0 \\ 0 & 0 & {\bf a}_3& 0 \\ 0 & 0 & 0& {\bf a}_4 \\ \end{array} \right]\left[ \begin{array}{cccc} d_1 & d_2 & d_3 & d_4 & \\ d_5 &d_6 &d_7 &d_8\\ d_9 &d_{10} &d_{11} &d_{12} \\ d_{13} & d_{14} &d_{15} & d_{16} & \end{array} \right]=\\$$ \stackrel{=}{?????} \left[ \begin{array}{cccc} {\bf a}_1 & 0 & 0 & 0 & \\ 0 &{\bf a}_2 & 0 & 0 \\ 0 & 0 & {\bf a}_3& 0 \\ 0 & 0 & 0& {\bf a}_4 \end{array} \right] \times \operatorname{diag} ({\bf D})$

where $d_i$ are scalars, where $\operatorname{diag} ({\bf D})$ is a matrix with only the diagonal entries of matrix ${\bf D}$?

Thanks

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  • $\begingroup$ What do you mean by an "$M\times 1$ matrix"? What is $M$? Take a trivial example where every $a_i$ is a $1\times 1$ matrix equal to $[1]$ (i.e. $A$ is the $4\times 4$ identity matrix). What do you know about $AD$ in that case? Is $AD = A~diag(D)$? $\endgroup$ – JMoravitz Sep 9 '15 at 2:47
  • $\begingroup$ Sorry I meant $M\times 1$ vector $\endgroup$ – Henry Sep 9 '15 at 2:54
  • $\begingroup$ Then the product you describe makes little to no sense if $M$ is not equal to one. Can you give an explicit example for what $A$ looks like if $M$ isn't one? Something like $\begin{bmatrix}\left[\begin{smallmatrix} a_{1,1}\\a_{1,2}\end{smallmatrix}\right]&0&0&0\\ 0&\left[\begin{smallmatrix} a_{2,1}\\a_{2,2}\end{smallmatrix}\right]&0&0\\0&0&\left[\begin{smallmatrix} a_{3,1}\\a_{3,2}\end{smallmatrix}\right]&0\\0&0&0&\left[\begin{smallmatrix} a_{4,1}\\a_{4,2}\end{smallmatrix}\right]\end{bmatrix}$? Are those zeroes actually vectors themselves? My comment from before still holds for A identity $\endgroup$ – JMoravitz Sep 9 '15 at 2:58
  • $\begingroup$ Yes this exactly the case, why does the prduct make no sense? $\endgroup$ – Henry Sep 9 '15 at 2:59
  • $\begingroup$ I don't understand your question. Are you asking whether $AD=BA\operatorname{diag}(D)$ for some matrix $B$? $\endgroup$ – user1551 Sep 9 '15 at 3:43
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Suppose $M=1$ (so that $A$ is $4\times4$) and the $a_i$'s are all nonzero. Then $A$ is invertible. Do you think the equality $AD=A\operatorname{diag}(D)$ always holds?

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