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I was looking at the properties of invertible matrix, and I came across this statement.

If $A$ is $m$ by $n$ matrix and the $rank$ of $A$ is equal to $m$, then $A$ has a right inverse.

Can any give me a proof for this statement?

Mark
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    A map has a right inverse if and only if it is surjective. The rank of the matrix being equal to the number of rows means the induced linear map is surjective. – Daniel Fischer Sep 27 '13 at 21:09
  • @DanielFischer I'm not sure I understand your terms. Could you explain more explicitly? – Mark Sep 27 '13 at 21:19
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    Which terms are you not sure you understand? Do you know that a matrix induces a linear map via multiplication of the matrix with a (column) vector? – Daniel Fischer Sep 27 '13 at 21:23
  • @DanielFischer I was sure about the mapping. – Mark Sep 27 '13 at 21:59

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A right inverse means that there exists some $B$ such that $AB = I$, in this case the identity is on $\mathbb{R}^{m \times m}$.

The fun fact here is that $\ker A^T = \ker (A A^T)$. Since ${\cal R} A = \mathbb{R}^m$, we have $\ker A^T = \{0\}$, that is, $A A^T$ is invertible.

To find $B$ above, we try to solve $Ax = y$. Since ${\cal R} A = {\cal R} A A^T$, we can start by looking for a solution to $A (A^T z) = y$, and substitute back to get $x$ later.

Since $A A^T$ is invertible, we have $z = (A A^T) ^{-1} y$, and so $x = A^T (A A^T) ^{-1} y$, which suggests setting $B = A^T (A A^T) ^{-1}$.

It is easy to verify that $B = A^T (A A^T) ^{-1}$ is a right inverse of $A$.

Note that this is not, in general, unique. For example, suppose $C$ is such that ${\cal R} C \subset \ker A$, then we have $AC = 0$, and so $B+\lambda C$ is also a right inverse for any $\lambda$.

copper.hat
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