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My Linear algebra is a bit rusty, I want to transform this vector:

$\begin{bmatrix}a \\ b \\ c \end{bmatrix}$

To this matrix:

$\begin{bmatrix} a & b & 0 \\ 0 & a & b \\ 0 & b & c \end{bmatrix}$

How do I do it with matrix multiplications (can have multiple steps)? I've tried to create a diagonal matrix from the vector but had no success onward, can someone show me the light? Thanks.

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The desired matrix has determinant $a(ac-b^2)$, so it is, in general, a regular matrix. It can't be obtained multipliyng matrices if one of them has rank $1$.

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  • $\begingroup$ Can you explain the contradiction part in a bit more detail? Thanks! $\endgroup$ – Rocky Li Jun 18 at 19:04
  • $\begingroup$ No, but technically it doesn't have to be a single multiplication. The question really is: Is there an limited amount of matrix multiplication that can get me there, or is it impossible? $\endgroup$ – Rocky Li Jun 18 at 19:10
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$\mathrm{reshape} \left( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} \right) = \begin{bmatrix} a & b & 0 \\ 0 & a & b \\ 0 & b & c \end{bmatrix} $

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  • $\begingroup$ This multiplication is not possible. $\endgroup$ – ajotatxe Jun 18 at 19:18
  • $\begingroup$ Thanks! This is really @jgon's addition wrapped in a single way. the best possible solution around even when I asked of the impossible. $\endgroup$ – Rocky Li Jun 18 at 19:36
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I'm not sure what you mean by allowing matrix multiplications with multiple steps.

If you allow matrix addition, this is possible. Otherwise, with only matrix multiplication this is impossible, as ajotatxe says.

With matrix addition, we can write $$\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}} \bmat 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \emat \bmat a \\ b \\ c\emat \bmat 1 & 0 & 0 \emat + \bmat 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \emat \bmat a \\ b \\ c \emat \bmat 0 & 1 & 0 \emat + \bmat 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \emat \bmat a \\ b \\ c \emat \bmat 0 & 0 & 1 \emat $$ $$=\bmat a & b & 0 \\ 0 & a & b \\ 0 & b & c \emat$$

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  • $\begingroup$ Thanks for this pragmatic answer. $\endgroup$ – Rocky Li Jun 18 at 19:48
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The only way to bring the vector into a $3 \times 3$ matrix is to multiply it on the right by a horizontal vector, as @ajotatxe already suggested. $$ {\bf P} = \left[ {\matrix{ a \cr b \cr c \cr } } \right]\left[ {\matrix{ x & y & z \cr } } \right] = \left[ {\matrix{ {ax} & {ay} & {az} \cr {bx} & {by} & {bz} \cr {cx} & {cy} & {cz} \cr } } \right] $$

But $\bf P$ has a null determinant, while that of your objective matrix ($\bf A$), in general, it is not null.

Now, there is no matrix, to use and multiply $\bf P$ on the left or on the right, that might render the resulting determinant not null.

So sorry, the answer is negative (of course in terms of "normal" multiplication with matrices / vectors).

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