If $V$ is a finite dimensional normed vector space and I identify row and column vectors then I can define the action of a matrix both from the left and from the right. I can define the left and right operator norms of matrices by $\vert \vert \cdot \vert \vert_{\tilde \infty}$ and $\vert \vert \cdot \vert \vert_{\infty}$.

Is it the case for any matrix $T$ and any norm on $V$ that $\vert \vert T \vert \vert_\infty = \vert \vert T \vert \vert_{\tilde \infty}$?

Generalization: Suppose that I have a normed space $V$ with norm $\vert \vert \cdot \vert \vert$ and the bounded operators on $V$ with the operator norm $\vert \vert \cdot \vert \vert_\infty$. Suppose that I can define left multiplication of $v T$ also an an element of $V$ for example because I have a fixed basis $\{e_i\}$ with respect to which I can do "a transpose" such that if $Tv = \sum_i (\sum_j T_{i,j} v_i)e_i $ then $vT =T^{T}v$ what conditions do we need on $V$ and on the norm for the left and right norms to be equal?


1 Answer 1


This is false in general if your norm is non-Euclidean. Suppose that $w\in \mathbb{R^d}$ and $A\in\mathbb{R}^{d\times d}$, then we have that $$A^Tw=(w^TA)^T. $$ This identity shows that the transpose acting on column vectors corresponds to the original matrix acting on row vectors from the right. It therefore suffices to find an operator norm and a matrix $A$ such that $A$ and $A^T$ have different operator norms.

In order to have an explicit example, consider the matrix $$A=\begin{pmatrix} 1&1\\0 &0 \end{pmatrix}.$$ Let $\|\cdot \|_{\ell^{\infty}\to\ell^{\infty}}$ be the operator norm induced from the $\ell^{\infty}$-norm on $\mathbb{R}^2$, then one can easily check that $\|A\|_{\ell^{\infty}\to\ell^{\infty}}=2$ (consider the normalized vector $v$ defined via $v^T=\begin{pmatrix}1&1 \end{pmatrix}$ ). Its transpose is given by $$A^T=\begin{pmatrix} 1&0\\1&0 \end{pmatrix} $$ and one can easily check that $\|A^T\|_{\ell^{\infty}\to\ell^{\infty}}=1$.


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