How to convert this vector to a matrix? 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.
 A: 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 $1$.
A: $\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}
$
A: 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
$$\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}
\bmat 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \emat \bmat a \\ b \\ c\emat \bmat 1 & 0 & 0 \emat
+ \bmat 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \emat \bmat a \\ b \\ c \emat \bmat 0 & 1 & 0 \emat 
+ \bmat 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \emat \bmat a \\ b \\ c \emat \bmat 0 & 0 & 1 \emat 
$$
$$=\bmat a & b & 0 \\ 0 & a & b \\ 0 & b & c \emat$$
A: 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).
