0

I know it's impossible to transpose a matrix through matrix multiplication, but is it possible to convert a $2 \times 2$ matrix, e.g.

\begin{bmatrix}a&b\\c&d\end{bmatrix}

into a $4 \times 4$ matrix with $a, b, c, d$ on the diagonal, through left/right matrix multiplication?

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

In other words, are there some conformable $L$ and $R$ such that $LAR$ equals this matrix? I initially considered these:

$$ L= \begin{bmatrix} 1&0\\1&0\\0&1\\0&1 \end{bmatrix} $$

and

$$ R= \begin{bmatrix} 1&0&1&0\\ 0&1&0&1 \end{bmatrix} $$

but those that doesn't work, and I haven't had any luck juggling 1s and 0s to get this to work, and I'm curious.

Simon Fraser
  • 2,528
Michael A
  • 125

1 Answers1

3

It is not possible to find such $L$ and $R$ that $$ L \begin{bmatrix}a&b\\c&d\end{bmatrix} R = \begin{bmatrix}a&0&0&0\\0&b&0&0\\0&0&c&0\\0&0&0&d\end{bmatrix}$$ for all $a,b,c,d$.

If all $a,b,c,d$ are non-zero then LHS has rank at most $2$ but RHS has rank $4$.

Matsmir
  • 2,545