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?


1 Answer 1


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 \begin{equation} \|BC\|_A=\sup_{\|x\|_A=1} \|BCx\|_A\leq \|B\|_A \sup_{\|x\|_A=1} \|Cx\|_A\leq \|B\|_A\|C\|_A. \end{equation}

For the second question, I think so. Observe that by scaling, \begin{equation} \|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}), \end{equation} where we use the substitution $x=A^{-1/2}y$, and then the usual variational characterization of eigenvalues of symmetric matrices.


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