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Suppose $T \colon \mathbb{R}^n \to \mathbb{R}^n$ is a linear operator. In following, we will consider $T$ as a matrix. The operator norm induced by $2$-norm on $\mathbb R^n$ is given by $\|T\|_{op,2} = \max_{\|x\|_2 = 1} \|Tx\|_2$. Let $T$ act by matrix multiplication on the vector space $\mathcal{M}(n \times m, \mathbb{R})$, i.e., the space of $n \times m$ matrices. What is operator norm $\|T\|_{op,F}$ induced by Frobenius norm $\| \cdot \|_F$ on $\mathcal {M}(n \times m)$?

Here is my thought: For a given matrix $A \in \mathcal{M}(n \times m, \mathbb{R})$, let $A = (a_1, \dots, a_m)$ where $a_j \in \mathbb R^n$ \begin{align*} \| TA\|_F^2 &= \|(Ta_1, \dots, Ta_m)\|_F^2 \\ &= \| Ta_1\|_2^2 + \dots \|Ta_m\|_2^2 \le \|T\|_{op,2}^2 \|a_1\|_2^2 + \dots + \|T\|_{op,2}^2 \|a_m\|_2^2 \\&= \|T\|_{op,2}^2 \|A\|_F^2. \end{align*} It seems like the operator norm $\|T\|_{op,F}$ should be upper bounded by $\|T\|_{op,2}$. Are they indeed equal? Thanks.

EDIT: @erfink points to this link. Indeed, I am asking a quite different question from this. I am not asking the Frobenius norm of $T$ but the operator norm induced by Frobenius norm of $\mathcal{M}(n \times m)$ when $T$ acts on this space by matrix multiplication. $T$ itself even is not an element of space but an elemnt of $\mathcal{M} (n \times n)$.

Rodrigo de Azevedo
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user1101010
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1 Answers1

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Yes, $\max_{\|A\|_F=1}\|TA\|_F=\|T\|_2$, because $$ \|TA\|_F^2 =\sum_j \|T\mathbf a_j\|_2^2 \le\sum_j \|T\|_2^2\|\mathbf a_j\|_2^2 =\|T\|_2^2\|A\|_F^2 $$ and equality occurs when $A=[\mathbf x|0|\cdots|0]$, where $\mathbf x$ is a unit singular vector corresponding to the largest singular value of $T$.

Alternatively, if you vectorise $A$ and $TA$, the mapping $A\mapsto TA$ is equivalent to $\operatorname{vec}(A)\mapsto(I\otimes T)\operatorname{vec}(A)$. Hence $$ \max_{\|A\|_F=1}\|TA\|_F = \max_{\|\operatorname{vec}(A)\,\|_2=1}\|(I\otimes T)\operatorname{vec}(A)\|_2=\|I\otimes T\|_2=\|T\|_2. $$

user1551
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