Inner products on $M_n(F)$ inducing the same norm on all matrices of rank one

Question

This question is related to exercise 11 on page 310 of Hoffman and Kunze's Linear Algebra book. Let $F:=\mathbb R/\mathbb C$. Let $\langle \ , \ \rangle_1,\langle \ , \ \rangle_2$ be two inner products on $M_n(F)$ with induced norms $\lVert \ \rVert_1,\lVert \ \rVert_2$, respectively. If $\lVert A\rVert_1=\lVert A\rVert_2$ for all $n\times n$ matrices $A$ of rank one, does it follow that $\langle \ , \ \rangle_1=\langle \ , \ \rangle_2$?

Answer 1

Let $\varphi$ be a fixed nonzero linear functional on $F^n$. For example $$\varphi (x_1,x_2,\ldots,x_n)=x_1$$ For any $0\neq y\in F^n$ the operator $A_y:F^n\to F^n$ defined by $$A_yx=\varphi(x)y$$ has rank one and $$\|A_y\|_j=\max_{\|x\|_j=1}\|A_yx\|_j=\|y\|_j\max_{\|x\|_j=1}|\varphi(x)|,\qquad j=1,2$$ As $\|A_y\|_1=\|A_y\|_2$ we get $$\|y\|_2=c\|y\|_1,\quad c={\displaystyle \max_{\|x\|_1=1}|\varphi(x)|\over \displaystyle\max_{\|x\|_2=1}|\varphi(x)|}$$ By the polar identity we obtain $$\langle y_1,y_2\rangle_2=c^2\langle y_1,y_2\rangle_1$$ Summarizing the answer is yes, inner products coincide up to a positive multiplicative constant.

Remark The positive constant $c$ cannot be replaced by $1,$ as the norm of an operator does not change if we replace the original norm by its positive multiple.

Answer 2

The answer is negative when $F=\mathbb R$ but positive when $F=\mathbb C$. Let us write the matrix representation of an inner product with respect to the standard basis of $M_n(F)$ as a block matrix $A=(A_{ij})_{i,j\in\{1,2,\ldots,n\}}\in M_{n^2}(F)$, where $A_{ij}=A_{ji}^\ast\in M_n(F)$. The inner product is then given by $$ \langle X,Y\rangle_A=\operatorname{vec}(X)^\ast A\operatorname{vec}(Y). $$ Therefore, when $X=uv^T$ is a rank-one matrix, we have $$ \begin{align} \langle X,X\rangle_A &=\operatorname{vec}(X)^\ast A\operatorname{vec}(X)\\ &=\pmatrix{\overline{v}_1u^\ast&\cdots&\overline{v}_2u^\ast} A\pmatrix{v_1u\\ \vdots\\ v_nu}\\ &=\sum_{i,j}\overline{v}_iv_ju^\ast A_{ij}u\\ &=v^\ast\pmatrix{u^\ast A_{11}u&\cdots&u^\ast A_{1n}u\\ \vdots&&\vdots\\ u^\ast A_{n1}u&\cdots&u^\ast A_{nn}u}v.\tag{1} \end{align}\\ $$ Now, when $F=\mathbb R$, let $A_{ii}=I_n$ and $A_{ij}=A_{ji}^T=K$ for all $i<j$ where $K$ is a skew-symmetric matrix. When $K$ is sufficiently small, $A$ is positive definite. Since $K$ is skew-symmetric, $u^TKu$ is identically zero. Thus $(1)$ gives $$ \langle uv^T,uv^T\rangle_A=\|v\|_2^2\|u\|_2^2. $$ That is, the norm $\|X\|_A=\langle X,X\rangle_A^{1/2}$ does not depend on $K$ whenever $X$ is rank-one. Yet, as the inner product is uniquely determined by its matrix representation, by choosing a different $K$ we obtain a different $\langle \cdot,\cdot\rangle_A$. Hence the answer to your question is negative.

When $F=\mathbb C$, if $A$ and $B$ are two positive definite matrices in $M_{n^2}(\mathbb C)$ and $\|uv^\ast\|_A=\|uv^\ast\|_B$ for all $u,v\in\mathbb C^n$, then $(1)$ implies that $A_{ij}=B_{ij}$ for all $i$ and $j$. Hence $\langle \cdot,\cdot\rangle_A=\langle \cdot,\cdot\rangle_B$.

The decisive difference between the real case and the complex case is that a complex matrix $M$ such that $x^\ast Mx=0$ for all complex vectors $x$ is necessarily zero, while a real matrix $M$ such that $x^TMx=0$ for all real vectors $x$ is only skew-symmetric.