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How to show that, if $A$ element of $\mathbb{C}_{p\times q}$, then there exist a left invertible matrix $B$ and a right invertible matrix $C$ such that $A=BC$? use singular value decomposition

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Hint: what is the singular value decomposition of $A$? Do you see any "somewhat invertible" matrices which you can use to find $B$ and $C$? –  user108903 Jan 21 '13 at 9:11

1 Answer 1

Think of linear maps rather than matrices. So $A\colon \mathbb{C}^q\to\mathbb{C}^p$ is a linear map. Let $r$ be the rank of $A$, and pick an $r$-dimensional subspace $V\subseteq\mathbb{C}^q$ whose intersection with the null space of $A$ is trivial. Let $C'\colon\mathbb{C}^q\to V$ be the projection which vanishes on the null space of $A$, and let $D'\colon V\to\mathbb{C}^p$ be the restriction of $A$ to $V$. Finally, let $D\colon V\to\mathbb{C}^r$ be an isomorphism, and let $C=DC'$, $B=B'D^{-1}$.

I leave the details up to you. (Hint: $C'$ is surjective, $B'$ is injective.)

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I want to do it with Singular value decomposition but I didnt understood this solution. –  Iya Jan 21 '13 at 8:14

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