I know that every inner product induces a norm (that induces a metric), and a norm is induced by an inner product if it satisfies the parallelogram law. I know that $x^Tx$ induces $\begin{Vmatrix} x \end{Vmatrix}_2^2$ by definition of norm $\begin{Vmatrix} x \end{Vmatrix}:=\sqrt{x^Tx}$, and at the same time $\begin{Vmatrix} x \end{Vmatrix}_2^2$ is the norm induced by inner product $x^Tx$. I also know that $\begin{Vmatrix} x \end{Vmatrix}_1^2$ and $\begin{Vmatrix} x \end{Vmatrix}_{\infty}^2$ are not induced by inner products because it's not true that $\forall x,y\in \mathbb{R}^n$
$$\begin{Vmatrix} x+y \end{Vmatrix}_{1,\infty}^2+\begin{Vmatrix} x-y \end{Vmatrix}_{1,\infty}^2= 2(\begin{Vmatrix} x \end{Vmatrix}_{1,\infty}^2+\begin{Vmatrix} y \end{Vmatrix}_{1,\infty}^2).$$
Then, starting from $x^TAx$ with $A=LL^T$ the Cholesky decomposition of $A$, is there a way to get $\begin{Vmatrix} x \end{Vmatrix}_{1}^2$ and $\begin{Vmatrix} x \end{Vmatrix}_{\infty}^2$? Can I still write $\begin{Vmatrix} x \end{Vmatrix}_{1}^2=x^Tx$ or $\begin{Vmatrix} x \end{Vmatrix}_{\infty}^2=x^Tx$ even if these norms are not induced by $x^Tx$?
