If I have a diagonal matrix, is it necessarily the product of two other diagonal matrices? We know that the product of two diagonal matrices forms another diagonal matrix, since we just multiply the entries.
So my question is, does the converse necessarily hold? In other words, if I have a diagonal matrix, did it necessarily come from the product of two other diagonal matrices?
My gut feeling tells me "No" since in Linear Algebra, all sorts of 'intuition' seem to go wrong.
But I would like a confirmation and perhaps a counter example would be nice.
 A: Of course yes, and there are infinitely many such decompositions $D = AB$. For each diagonal entry $d_i$, you just decompose it into a product $d_i = a_i b_i$ arbitrarily. Letting $A = diag(a_i)$ and $B = diag(b_i)$ you have $D=AB$. 
For example, you can have $A = D$ and $B = I$. 
Another example is $A = diag(\vert d_i \vert)$ and $B = diag(sgn(d_i))$.
A: If $D$ is diagonal then $D=ID$ where $I$ is the identity matrix. :)
A: I guess we can easily arrive at a counter-example:
Let us take two $2 \times 2$  matrices and multiply them:
$$
\begin{bmatrix}
a&b\\
c&d\\
\end{bmatrix}
\times
\begin{bmatrix}
e&f\\
g&h\\
\end{bmatrix}
=
\begin{bmatrix}
ae+bg&af+bh\\
ce+dg&cf+dh\\
\end{bmatrix}
$$
Now for the result to be a diagonal matrix:
$ce=-dg $ and $ af=-bh$
So resulting in :
$$
\begin{bmatrix}
ae+bg&0\\
0&cf+dh\\
\end{bmatrix}
$$
So Let us take an example as :
Let $\large c=1,d=-\dfrac{1}{2},e=1,g=2\space$ and $\large \space a=1,b=-\dfrac{1}{4},f=2,h=8$ 
Hence we arrive at:
$$
\begin{bmatrix}
1&-\dfrac{1}{4}\\
1&-\dfrac{1}{2}\\
\end{bmatrix}
\times
\begin{bmatrix}
1&2\\
2&8\\
\end{bmatrix}
=
\begin{bmatrix}
\dfrac{1}{2}&0\\
0&-2\\
\end{bmatrix}
$$

Thus it is not always necessary to have two diagonal matrices give a diagonal matrix...

Hope this helps
