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?

  • $\begingroup$ What you wrote seems rather straightforward. Can you use determinants and Cauchy's theorem in the proof? $\endgroup$ – user23211 Jan 20 '12 at 14:08
  • $\begingroup$ 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. $\endgroup$ – sony jimbo Jan 20 '12 at 14:09
  • $\begingroup$ Which canonical form are you talking about? $\endgroup$ – GeoffDS Jan 20 '12 at 15:56
  • $\begingroup$ @sonyjimbo When you said "we didn't learn that yet", do you mean Cauchy's theorem? $\endgroup$ – GeoffDS Jan 20 '12 at 16:26
  • $\begingroup$ @ymar What is Cauchy's theorem? $\endgroup$ – GeoffDS 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.

| cite | improve this answer | |
  • $\begingroup$ By the way, it is not clear the OP wants an answer without determinants. I imagine he is saying he doesn't know Cauchy's theorem. I am a PhD student and I don't know what is meant by Cauchy's theorem. If you search Wikipedia for Cauchy's theorem, there is a disambiguation page and it does not contain anything involving linear algebra. I looked in Horn and Johnson and it does not have Cauchy's theorem in the index, same with Hoffman and Kunze, same with Lay. When the OP said "we didn't learn that yet", it's probably in regards to Cauchy's theorem. $\endgroup$ – GeoffDS Jan 20 '12 at 16:24
  • $\begingroup$ Didn't learn determinants yet as well. What do you mean by LI? $\endgroup$ – sony jimbo Jan 20 '12 at 22:33
  • $\begingroup$ linearly independent $\endgroup$ – GeoffDS Jan 21 '12 at 4:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.