0

If $A$ is PD, then $A$ is invertible. Proof:

Let $x\in\mathbb{R}^{n}$ such that $x\neq0$. If it weren't the case that $A$ is invertible, then: $$ Ax=0\implies x^{T}Ax=0 $$ not positive definite. Hence, proof complete.

My question is, if this is enough to show that the implication is true. Should I also mention that if $x\neq0$ and $A$ is PD then $x^{T}Ax>0$ with $Ax\neq0\implies A$ is invertible?

ZeinabQom
  • 57
  • 1
    It's not true that $Ax = 0$ for an arbitrary nonzero vector when $A$ is not invertible. (That would only be true if $A = 0$.) You have to choose a nonzero $x$ in the null space of $A$. With that modification, your proof would be OK. –  Jul 05 '21 at 20:13
  • What I mentioned is that if $A$ is not invertible, then we may write $Ax=0$ for non-zero vector $x$. I am not sure if this implication is true. Because I know that if $A$ is invertible then the null space must contain the zero vector only but this is not the case since $x\neq0$ @Bungo – ZeinabQom Jul 05 '21 at 20:21
  • 1
    It is true. If $A$ is not invertible, then the linear map induced by $A$ is not injective, so it has a nontrivial null space, so there's a nonzero $x$ such that $Ax = 0$. –  Jul 05 '21 at 20:22
  • Alright, but I did not quite understand what you said in your first comment can you point out where I made this error in my proof @Bungo – ZeinabQom Jul 05 '21 at 20:23
  • 1
    If I'm interpreting correctly, you chose an arbitrary nonzero $x \in \mathbb R^n$ and said that if $A$ is not invertible, then $Ax = 0$. This is not true. If $A$ is not invertible, it doesn't mean that $Ax = 0$ for all nonzero $x$, only that there exist nonzero $x$ such that $Ax = 0$. –  Jul 05 '21 at 20:25
  • 1
    Your proof would be correct if you replaced the first sentence by "Let $x\in \mathbb{R}^n\setminus {0}$ satisfy $Ax=0$" (which exists since you assume that $A$ is not invertible). – projectilemotion Jul 05 '21 at 20:28
  • Ah so the main issue is that if I say "$A$ is not invertible, then for a non-zero $x$ we have $Ax=0$" is not always the case for non-invertible matrices? @Bungo – ZeinabQom Jul 05 '21 at 20:29
  • Alright, but I can't seem to understand how it differs from the statement I used. @projectilemotion – ZeinabQom Jul 05 '21 at 20:32
  • You didn't say "if $A$ is not invertible, then for a non-zero $x$ we have $Ax = 0$". That is ambiguous but can be interpreted as true if "for a non-zero $x$" means "there exists a non-zero $x$". But you didn't say either of those things. You started with an arbitrary non-zero $x$. There's no reason that $Ax$ has to be $0$ for that arbitrary choice of $x$. –  Jul 05 '21 at 20:32
  • The issue is that $Ax=0$ is not true for all $x\in \mathbb{R}\setminus {0}$ if $A$ is not invertible. What is true is that there exists $x\in \mathbb{R}\setminus {0}$ such that $Ax=0$ if $A$ is not invertible. – projectilemotion Jul 05 '21 at 20:33
  • 1
    Maybe an example will help. Let $\displaystyle A = \begin{pmatrix}1 & 0 \ 0 & 0\end{pmatrix}$. Then $A$ is not invertible (it has a column of zeros). But if we choose $\displaystyle x = \begin{pmatrix}1 \ 0 \end{pmatrix}$ then $\displaystyle Ax = \begin{pmatrix}1 \ 0 \end{pmatrix}$, which is not zero. However, there exist nonzero choices of $x$ such that $Ax = 0$. For example, $\displaystyle x = \begin{pmatrix}0 \ 1 \end{pmatrix}$. –  Jul 05 '21 at 20:35
  • Ah alright, now I fully understand it! thank you very much @projectilemotion – ZeinabQom Jul 05 '21 at 20:35
  • and thank you @Bungo for the explanation and the counter example as well. – ZeinabQom Jul 05 '21 at 20:35

0 Answers0