# Frobenius norm is not induced

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.

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

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

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.