Positive-definiteness propertys doubts of a inner product ($[x,x]=0\,\,\Leftrightarrow\,\, x=0\,\,\,$ & $\,\,\,x\neq 0\,\,\Rightarrow\,\, [x,x]>0$).

Question

I need show this:

Let $A$ an invertible matrix of $n\times n$ with entries in $\mathbb{R}$. Show that if $\langle , \rangle$ is the usual inner product in $\mathbb{R}^n$ then $[x,y]=\langle Ax,Ay \rangle$ is a inner product in $\mathbb{R}^n$.

I showed the symmetry and linearity.

I have doubts with the following two properties:

(1) $[x,x]=0$ if and only if $x=0.$

(2) If $x\neq 0$ then $[x,x]>0.$

For (1) since $A$ is invertible, $A$ is injective, therefore if $Ax=0$ then $x=0.$ So

\begin{align*} [x,x]=0\ \rangle&\Longleftrightarrow\langle Ax,Ax \rangle=0\\ &\Longleftrightarrow Ax=0\,\,\,\,\,\,\,\,\,(\,\,\langle, \rangle \,\,\,\, \rm{is\,\, inner\,\, product)}\\ &\Longleftrightarrow x=0\,\,\,\,\,\,\,\,\,(A \,\, \rm{is\,\, invertible)} \end{align*}

For (2), let $x\neq 0$, since $\langle, \rangle$ is a inner product then $\langle Ax, Ax\rangle>0$ so $[x,x]>0$.

My answers are correct? Thanks for your help.

Answer 1

Yes, all correct.

More generally, for any injective linear map $f:V\to W$, if $W$ has an inner product, it induces an inner product on $V$ in the same manner.