$A$ is Hermitian and $(Ax,x)=0 \Rightarrow x=0$.

Question

Let $A$ be $n\times n$ Hermitian matrix, and $(\cdot ,\cdot)$ be Hermitian inner product.

Define the condition (i) as :

(i) $(Ax,x)=0 \Rightarrow x=0$.

Prove that if $A$ satisfies (i), then $A$ is regular (invertible) matrix.

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I have my proof but I'm not sure this is correct.

Proof

For showing $A$ is regular, it suffices to show $Ax=0 \Rightarrow x=0$ (The equation $Ax=0$ has only trivial solution.).

So, suppose $Ax=0$.

Then, $(Ax,x)=(0,0)=0.$

From (i), I get $x=0.$

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Is this proof correct ?

I'm not sure this is correct because I didn't use the Hermitianness of $A.$

Answer 1

No, it isn't correct, but only up to a detail which, however, isn't minor.

From $Ax=0$ you get $$ \langle Ax,x\rangle=\langle 0,x\rangle=0 $$ In your text there's $\langle 0,0\rangle$ in the middle step, which is wrong.

Now that this is fixed, we derive $x=0$ from condition (i). So we have proved that $Ax=0$ implies $x=0$ and this proves that $A$ is invertible.

Answer 2

I think the idea seems right, maybe we write like this: The counter-statement of the condition is $$x\neq 0 \implies \langle Ax,x\rangle \neq 0.$$ Assume $A$ is not invertible, then $\exists$ $x'\neq 0$ such that $Ax'=0$. Then $\langle Ax',x'\rangle =0$, a contradiction. I don't think we need the hermitian condition here.