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Is it true that given a matrix $A_{m\times n}$, $A$ is regular / invertible if and only if $m=n$ and $A$ is a basis in $\mathbb{R}^n$?

Seems so to me, but I haven't seen anything in my book yet that says it directly.

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Yes, almost true. What do you mean by "$A$ is a basis"? If you mean the columns/rows of $A$ to be a basis of $\mathbb R^n$, it is fine. –  martini Apr 17 '12 at 8:19
    
Yes, that's my intent. Thanks. –  Robert S. Barnes Apr 17 '12 at 11:29

2 Answers 2

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$\mathrm{rk}(A_{m \times n}) \leq \min(m, n)$, from which follows $\mathrm{rk}(A_{m \times n}B_{n \times m}) \leq \min(m, n)$, while $\mathrm{rk}(E_{m \times m}) = m$, so, obviously, $m$ should be equal to $n$ in order for $A$ to be invertible.

Also, if the strings (or columns) of $A_{n \times n}$ do not form the basis in ${\mathbb R}^n$, it means that some of the strings / columns are linearly dependent, and thus $\mathrm{rk}(A) < n$, $|A| = 0$, so that A is not invertible.

From the other hand, if the strings (or columns) of $A_{n \times n}$ do form the basis, it means that they are linearly independent, and $|A| \ne 0$. In that case, you can find $A^{-1}$ by the Cramer's rule.

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Aren't both statements simply saying that A has full rank?

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I don't know what rank is, haven't gotten to that yet I guess. –  Robert S. Barnes Apr 17 '12 at 8:22
    
Ok, check here: en.wikipedia.org/wiki/Rank_(linear_algebra) . Look at the 'properties' section. –  Wonder Apr 17 '12 at 8:25

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