# Properties of norm induced by positive definite matrix

Suppose that $$A$$ is a $$n \times n$$ positive definite matrix. Consider the norm $$\|x\|_A = \sqrt{x^T A x}$$ where $$x \in \mathbb{R}^n$$.

Consequently, define $$\|B\|_A$$ for a $$n \times n$$ matrix $$B$$ to be $$\|B\|_A = \sup_{\|x\|_A = 1} \|Bx\|_A$$.

Is it true that this norm is submultiplicative, i.e. is $$\|BC\|_A \le \|B\|_A \|C\|_A$$ where $$C$$ is another $$n \times n$$ matrix?

Also, is there a simple formula to compute $$\|B\|_A$$? For example, $$\|B\|_2 = \sigma_{max}(B)$$ for the spectral norm. Does a nice formula exist for the induced matrix norm?

For your first question, yes, it is submultiplicative. Note that for any matrix $$M$$ and vector $$x$$, $$\|Mx\|_A\leq \|M\|_A\|x\|_A$$ by definition and homogeneity, so that $$$$\|BC\|_A=\sup_{\|x\|_A=1} \|BCx\|_A\leq \|B\|_A \sup_{\|x\|_A=1} \|Cx\|_A\leq \|B\|_A\|C\|_A.$$$$
For the second question, I think so. Observe that by scaling, $$$$\|B\|^2_A=\sup_{x\neq 0} \frac{\|Bx\|^2_A}{\|x\|_A^2}=\sup_{x\neq 0} \frac{x^TB^TABx}{x^TAx}=\sup_{y\neq 0} \frac{y^TA^{-1/2}B^TABA^{-1/2}y}{y^Ty}=\lambda_{\max}(A^{-1/2}B^TABA^{-1/2}),$$$$ where we use the substitution $$x=A^{-1/2}y$$, and then the usual variational characterization of eigenvalues of symmetric matrices.