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So, I find myself choosing to prove that a matrix $A \in M_{n×n}(F)$ is invertible if and only if $\operatorname{rank}(A) = n$.

My thoughts are that I should incorporate the fact that such a matrix $A$ is invertible if and only if $\det{A} \neq 0$, and use this to show the linear independence of the column space of $A$.

Trancot
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  • I mean, if there are two identical columns, then $\det{A}=0$, but does this also apply for $m>2$ identical columns? – Trancot May 09 '13 at 06:12
  • http://math.stackexchange.com/questions/21990/proof-square-matrix-has-maximal-rank-if-and-only-if-it-is-invertible – Alex Wertheim May 09 '13 at 06:14
  • @BarisaBarukh Since $\det A = 0$ for two identical columns then it certainly holds that $\det A = 0$ for more than two identical columns (just ignore all but $2$ if you will). – EuYu May 09 '13 at 06:16
  • Fortunately, the title of your question is not what you found yourself choosing to prove. Use words instead of formulas. – Martin May 09 '13 at 06:18
  • There are many, many ways to prove this. Can you tell us more about what you know so far? Determinants are probably not the most natural way to do this. – EuYu May 09 '13 at 06:34
  • Can it be shown that $\det{A}=0 \implies A$ has $m\geq 2$ identical columns? This would be helpful. Perhaps $n-$linear forms are required here instead. – Trancot May 09 '13 at 06:35
  • The determinant is $0$ if and only if the columns are linearly dependent. There doesn't have to be any identical columns. For example, consider the matrix with first column all $1$s, second column all $2$s, so on... This is a very elementary result, there's really no need to involve $n$-linear forms. – EuYu May 09 '13 at 06:36
  • @EuYu Prove it. – Trancot May 09 '13 at 06:37
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    @BarisaBarukh Prove what? – EuYu May 09 '13 at 06:38

1 Answers1

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Hint:

$\operatorname{rank}(A)=n\iff$ the columns of $A$ are linearly independent $\overset{\text{why ?}}\iff$ there are linear combinations of the columns of $A$ which give the vectors $$e_i=\begin{pmatrix}0\\\vdots\\0\\1\\0\\\vdots\\0\end{pmatrix}.$$

Marc van Leeuwen
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