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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.

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    $A(B(AB)^{-1})=I$, done.2012-11-18

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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.

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$AB = \Lambda$ 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 $A^{-1} = B\Lambda^{-1}$

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    @Kaster , your proof assumes what must be proved, i.e. that $\,A\,$ is invertible. To write $\,(AB)^{-1}=B^{-1}A^{-1}\,$ is meaningless *unless* we know a priori that both $\,A,B\,$ are invertible.2012-11-18
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Use the determinant:

$AB=\begin{pmatrix} a_1&0&...&0\\...&...&...&...\\0&0&...&a_n\end{pmatrix}\,\,,\,a_i\neq 0\Longrightarrow \det A\cdot\det B=\det AB=\prod_{i=1}^na_i\neq 0\Longrightarrow$

$\Longrightarrow \det A\neq 0\Longrightarrow \,\,A\;\;\text{ is regular and thus invertible}$

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    Indeed so, @Christian2012-11-18