Is there a simple way to map this 4x1 vector to this 4x2 matrix?

Question

Is there a simple way to map a vector like $\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix}$ to $\begin{bmatrix}a&0\\b&0\\0&c\\0&d\end{bmatrix}$? I tried to do it via matrix multiplication but I couldn't figure it out. Maybe the Kronecker product? It seems like a simple question, so I wonder if anyone has run into something like this before. Thank you.

-DBL

Answer 1

We can do it via matrix arithmetic via: $$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \\ \end{bmatrix} \begin{bmatrix} 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} a & 0 \\ b & 0 \\ 0 & c \\ 0 & d \\ \end{bmatrix}. $$

Answer 2

$\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix} \to \begin{bmatrix}a&0\\b&0\\0&c\\0&d\end{bmatrix} $ itself is a well defined map from $M(4,1 )$ to $M(4,2)$. If you really want something written in mathematical language including kroneckers then do it like-

$\phi(\begin{bmatrix}a\\b\\c\\d\\\end{bmatrix})= \begin{bmatrix}a\\b\\c\\d\\\end{bmatrix} .A $ wher $A=\begin{bmatrix}\delta_{a,b}\ \delta_{c,d}\\\end{bmatrix}$ is a $1\times 2 $ matrix such that $i.\delta_{k,l}=i$ if $i=k\ \text{or}\ l$ other wise $0$ where $i,k,l \in$ {$a,b,c,d$}