Let $AB$ be a diagonal matrix. Is A invertible? Let $AB \in K^{n\times n}$, $AB$ is a diagonal matrix and the elements on the diagonal are non-zero.
Is $A$ invertible?
Since $AB$ is a diagonal matrix and the elements on the diagonal are non-zero, $AB$ is invertible. Also neither $A$ nor $B$ can possibly contain zero-vectors or $AB$ would too which it doesn't. However not containing zero-vectors is not sufficient for a matrix to be invertible, right?
Please be aware that I don't know much about linear algebra yet, so a thorough explanation would be much appreciated.
 A: $$AB = \Lambda$$
Since $\Lambda$ is a diagonal matrix with non-zero elements it's invertible
$$\left (
AB
\right)^{-1} = \Lambda^{-1}$$
If a product is invertible, then multiplicands are also invertible
$$\left(
AB
\right)^{-1} = B^{-1}A^{-1} = \Lambda^{-1}
$$
therefore $A^{-1} = B\Lambda^{-1}$
A: Use the determinant:
$$AB=\begin{pmatrix} a_1&0&...&0\\...&...&...&...\\0&0&...&a_n\end{pmatrix}\,\,,\,a_i\neq 0\Longrightarrow \det A\cdot\det B=\det AB=\prod_{i=1}^na_i\neq 0\Longrightarrow$$
$$\Longrightarrow \det A\neq 0\Longrightarrow \,\,A\;\;\text{ is regular and thus invertible}$$
A: Yes, both $A$ and $B$ are invertible provided they are square in the first place. We could however say that $A$ and $B$ both have full-rank since
$$\text{rank(AB)} \leq \min  \{\text{rank(A), rank(B)}\}$$ Since $AB$ is diagonal with non-zero entries, $\text{rank(AB)} = n$. If $A \in \mathbb{R}^{n \times m}, B \in \mathbb{R}^{m \times n}$, then the previous result means that $m \geq n$. If $m=n$, we could also conclude that $A$ and $B$ are invertible.
