My textbook says:

Frobenius norm defined on $\mathbb{R}^{m,n}$ by a formula: $$\|A\|_F=\sqrt{\sum_{i=1}^n\sum_{j=1}^m|a_{i,j}|^2}$$ when $n>1, \ m>1$ is not induced by any vector's $p$-norm.

But there is no proof. I searched over the Internet and didn't find one. Is it very hard to prove? I would be very grateful for help.


2 Answers 2


Any induced norm of the identity matrix is $1.$

  • $\begingroup$ Correct! In addition: $||I||_F = \sqrt{\min\{m,n\}}$ hence $||I||_F > 1$ for $n>1, m>1$. $\endgroup$
    – TheWaveLad
    Dec 1, 2013 at 19:54
  • $\begingroup$ brilliant! thank you very much $\endgroup$
    – xan
    Dec 1, 2013 at 20:10

Will Jagy's answer solved the problem to a certain extent, but a new question arose. If we normalize the F-norm (like $\frac{1}{\sqrt{n}}\|\cdot\|_F$ for $ \mathbb{C}^{n\times n}$), will it be an induced norm?

I found a theorem about this question. The original paper is https://haddad.gatech.edu/journal/00478340.pdf. The theorem is as following.

Let $\|\cdot\|$ denote a unitarily invariant matrix norm on $\mathbb{C}^{m \times n}$. Then there exist vector norms $\|\cdot\|^{\prime}$ and $\|\cdot\|^{\prime \prime}$ on $\mathbb{C}^n$ and $\mathbb{C}^m$, respectively, such that $\|\cdot\|$ is the matrix norm induced by $\|\cdot\|^{\prime}$ and $\|\cdot\|^{\prime \prime}$ if and only if there exists $k>0$ such that $\|A\|=k \sigma_{\max }(A)$ for all $A \in \mathbb{C}^{m \times n}$. Furthermore, if $\|A\|=k \sigma_{\max }(A)$ for all $A \in \mathbb{C}^{m \times n}$ then $k=\left\|E_{11}\right\|$.

We will proof this in three steps:

  1. Frobenius norm and matrix 2 norm are unitarily invariant matrix norm, which means $\|QAP\|=\|A\|$ for unitaty matrices $Q\in\mathbb{R}^{m\times m}$, $P\in\mathbb{R}^{n\times n}$.
  2. Let $\|\cdot\|$ denote the matrix norm on $\mathbb{C}^{m \times n}$ induced by vector norms $\|\cdot\|^{\prime}$ and $\|\cdot\|^{\prime \prime}$ on $\mathbb{C}^n$ and $\mathbb{C}^m$ and let $y \in \mathbb{C}^m, x \in \mathbb{C}^n$. Then $\left\|y x^*\right\|=\|y\|^{\prime \prime}\|x\|_D^{\prime}$, where $\|\cdot\|_D$ denote the dual norm.
  3. Apply SVD to $yx^*$, $\|y\|^{\prime \prime}\|x\|_D^{\prime}=\left\|y x^*\right\|=\sigma_{max}(yx^*)\|E(1,1)\|=c\|y\|_2\|x\|_2$. So we have $\|\cdot\|''=k_1\|\cdot\|_2,\ \ \|\cdot\|'_D=\|\cdot\|_2$

First, Frobenius norm is unitarily invariant norm. To see it, for $A\in\mathbb{C}^{m\times n}$ and unitaty matrices $Q\in\mathbb{C}^{m\times m}$, $P\in\mathbb{C}^{n\times n}$ $$ \|QAP\|_F^2=tr((QAP)^*QAP)=tr(P^*A^*Q^*QAP)=tr(A^* A)=\|A\|_F^2 $$ Matrix 2-norm is also unitatily invariant, as $\sigma_{max}(QAP)=\sigma_{max}(A)$

Second, let $\|\cdot\|$ denote the matrix norm on $\mathbb{C}^{m \times n}$ induced by vector norms $\|\cdot\|^{\prime}$ and $\|\cdot\|^{\prime \prime}$ on $\mathbb{C}^n$ and $\mathbb{C}^m$. Let $y \in \mathbb{C}^m, x \in \mathbb{C}^n$. Then $$\left\|y x^*\right\|=\max_{\|z\|'=1}\|yx^*z\|''=\|y\|''\max_{\|z\|'=1}|x^*z|=\|y\|^{\prime \prime}\|x\|_D^{\prime}$$ The dual norm of $\|\cdot\|'$ is defined as: $$ \|x\|'_D=\max_{\|z\|'=1}|x^*z| $$ For example, the dual norm of vector 2-norm is still 2-norm; the dual norm of infity norm is 1-norm.

Third, by SVD, for a unitarily invariant matrix norm, we have $$ \|yx^*\|=\|\sigma_{max}(yx^*)E(1,1)\|=\sigma_{max}(yx^*)\|E(1,1)\|=c\|y\|_2\|x\|_2, $$ where $c$ is a constant. Combining with former statement, we have $$ c\|x\|_2\|y\|_2=\|y\|^{\prime \prime}\|x\|_D^{\prime} $$ for all $x \in \mathbb{C}^m$ and $y \in \mathbb{C}^n$. Fixing $x$, it implies that there exists $k_1>0$ such that $\|y\|^{\prime \prime}=k_1\|y\|_2$. Similarly, fixing $y$, it implies that there exists $k_2>0$ such that $\|x\|^{\prime} =k_2\|x\|_2$. Hence, it follows that $\|A\|=\left(k_1 / k_2\right) \sigma_{\max }(A)$ for all $A \in \mathbb{C}^{m \times n}$ as required.


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