# Inverible matrix and canonical form relation

Suppose $A$ is invertible matrix, we want to prove that in the canonical form that matches $A$ there is no zero rows.

So I proved the following that if the canonical form that matches there is a row of zeros then $A$ isn't invertible matrix then in something along that, let $\hat{A}$ be the canonical form of A. Since $\hat{A}$ have at least one row of zeros, there is less then $n$ rows with leading '1', so there is at least one free variable. this is the solution dimension of the homogeneous equations. Hence there are infinite solutions.

Suppose we have $T_{A}\colon\mathbb{F}^{n}\to\mathbb{F}^{n}$, linear transformation defined by multiplying by $A$, then $\ker T_{A}=\operatorname{null}\left(A\right)$

So $T_{A}$ isn't invertible, then $A$ isn't invertible, cause $A$ is invertible if and only if $T_{A}$ is invertible.

Those are just some guidlines I thought of to prove it,also forgive me if the terms are translated bad.

But I'm interested to know, if proving it straightforward $A$ invertible $\Rightarrow$ canonical form have no zero rows in it. Could be easier to prove? Or how would you do it?

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What you wrote seems rather straightforward. Can you use determinants and Cauchy's theorem in the proof? –  user23211 Jan 20 '12 at 14:08
we didn't learn that yet. But what I mean is that instead of proving an equivalent claim, proving the claim itself seems interesting to me. –  sony jimbo Jan 20 '12 at 14:09
Which canonical form are you talking about? –  Graphth Jan 20 '12 at 15:56
@sonyjimbo When you said "we didn't learn that yet", do you mean Cauchy's theorem? –  Graphth Jan 20 '12 at 16:26
@ymar What is Cauchy's theorem? –  Graphth Jan 20 '12 at 16:26

Without using determinants you can say that a matrix $A$ is invertible if and only if the rows (columns) form a basis, in particular they are LI. Then a row can't be the zero vector.