# why is frobenius norm of a matrix greater than or equal to the 2 norm?

How can you prove that: $$\|A\|_2 \le \|A\|_F$$ I cannot use: $$\|A\|_2^2 = \lambda_{max}(A^TA)$$ It makes sense that the 2-norm would be less than or equal to the frobenius norm but I dont know how to prove it. I do know:

$$\|A\|_2 = \max_{\|x\|_2 = 1} {\|Ax\|_2}$$

and I know I can define the frobenius norm to be:

$$\|A\|_F^2 = \sum_{j=1}^n {\|Ae_j\|_2^2}$$ but I dont see how this could help. I dont know how else to compare the two norms though.

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Let $\Vert A\|_2 = \|Av\|$ for some $v$ with $\|v\| = 1$. Define an orthogonal matrix $U$ for which $v$ is the first column. Compute $\Vert AU\Vert_F$ and show that it is equal to $\Vert A\Vert_F$, then compare it to $\Vert Av\Vert$ by a direct computation. – Hans Engler Dec 7 '12 at 4:36
@HansEngler I dont exactly understand your answer. How would I compute $||AU||_2$ and if it is equal to $||A||_2$ then how does that prove anything about $||A||_F$? – user972276 Dec 7 '12 at 4:42
@HansEngler I dont understand how to show $||AU||_F = ||A||_F$. I understand that the first column of AU is just Av and so I can show that $||A||_2 \le ||A||_F$. – user972276 Dec 7 '12 at 4:50
$\Vert A\Vert_F^2 = trace(AA^T)$. Now compute the same thing for $AU$ and use the fact that $U$ is orthogonal. – Hans Engler Dec 7 '12 at 12:14

Write $x=\sum_{j=1}^nc_je_j$, for coefficients $c_1,\ldots,c_n$. Suppose that $\|x\|_2=1$, i.e. $\sum_j |c_j|^2=1$. Then $$\|Ax\|_2^2=\|\sum_j c_j\,Ae_j\|_2^2\leq\left(\sum_j|c_j|\,\|Ae_j\|_2\right)^{2}\\ \leq\left(\sum_j|c_j|^2\right)\sum_j\|Ae_j\|_2^2=\sum_j\|Ae_j\|_2^2=\|A\|_F^2,$$ where the triangle inequality is used in the first $\leq$ and Cauchy-Schwarz in the second.

As $x$ was arbitrary, we get $\|A\|_2\leq\|A\|_F$.

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is there a way to do it without using Cauchy-Schwartz? – user972276 Dec 7 '12 at 4:54
Not that I can think of. In any case I simplified the argument by using the triangle inequality (basically, the previous version had a built-in proof of the triangle inequality, which by the way I don't think you can prove without CS). – Martin Argerami Dec 7 '12 at 10:58

In fact, the proof from $\left\| \mathbf{A}\right\|_2 =\max_{\left\| \mathbf{x}\right\|_2=1} \left\| \mathbf{Ax} \right\|_2$ to $\left\| \mathbf{A}\right\|_2 = \sqrt{\lambda_{\max}(\mathbf{A}^H \mathbf{A})}$ is straight forward. We can first simply prove when $\mathbf{P}$ is Hermitian $$\lambda_{\max} = \max_{\| \mathbf{x} \|_2=1} \mathbf{x}^H \mathbf{Px}.$$ That's because when $\mathbf{P}$ is Hermitian, there exists one and only one unitary matrix $\mathbf{U}$ that can diagonalize $\mathbf{P}$ as $\mathbf{U}^H \mathbf{PU}=\mathbf{D}$ (so $\mathbf{P}=\mathbf{UDU}^H$), where $\mathbf{D}$ is a diagonal matrix with eigenvalues of $\mathbf{P}$ on the diagonal, and the columns of $\mathbf{U}$ are the corresponding eigenvectors. Let $\mathbf{y}=\mathbf{U}^H \mathbf{x}$ and substitute $\mathbf{x} = \mathbf{Uy}$ to the optimization problem, we obtain

$$\max_{\| \mathbf{x} \|_2=1} \mathbf{x}^H \mathbf{Px} = \max_{\| \mathbf{y} \|_2=1} \mathbf{y}^H \mathbf{Dy} = \max_{\| \mathbf{y} \|_2=1} \sum_{i=1}^n \lambda_i |y_i|^2 \le \lambda_{\max} \max_{\| \mathbf{y} \|_2=1} \sum_{i=1}^n |y_i|^2 = \lambda_{\max}$$

Thus, just by choosing $\mathbf{x}$ as the corresponding eigenvector to the eigenvalue $\lambda_{\max}$, $\max_{\| \mathbf{x} \|_2=1} \mathbf{x}^H \mathbf{Px} = \lambda_{\max}$. This proves $\left\| \mathbf{A}\right\|_2 = \sqrt{\lambda_{\max}(\mathbf{A}^H \mathbf{A})}$.

And then, because the $n\times n$ matrix $\mathbf{A}^H \mathbf{A}$ is positive semidefinite, all of its eigenvalues are not less than zero. Assume $\text{rank}~\mathbf{A}^H \mathbf{A}=r$, we can put the eigenvalues into a decrease order:

$$\lambda_1 \geq \lambda_2 \geq \lambda_r > \lambda_{r+1} = \cdots = \lambda_n = 0.$$

Because for all $\mathbf{X}\in \mathbb{C}^{n\times n}$, $$\text{trace}~\mathbf{X} = \sum\limits_{i=1}^{n} \lambda_i,$$ where $\lambda_i$, $i=1,2,\ldots,n$ are eigenvalues of $\mathbf{X}$; and besides, it's easy to verify $$\left\| \mathbf{A}\right\|_F = \sqrt{\text{trace}~ \mathbf{A}^H \mathbf{A}}.$$

Thus, through $$\sqrt{\lambda_1} \leq \sqrt{\sum_{i=1}^{n} \lambda_i} \leq \sqrt{r \cdot \lambda_1}$$ we have $$\left\| \mathbf{A}\right\|_2 \leq \left\| \mathbf{A}\right\|_F \leq \sqrt{r} \left\| \mathbf{A}\right\|_2$$

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The Frobenius norm is sub-multiplicative, therefore $||Ax||_F \leq ||A||_F ||x||_F$, which gives:

$$\forall x \neq 0 \quad \frac{ ||Ax||_2 } {||x||_2} = \frac{ ||Ax||_F } {||x||_F} \leq ||A||_F$$

So you have an upper bound for the quotient, and since the supremum (here maximum) is by definition smaller than any other upper bound, you immediately get:

$$\underset{x \neq 0}{\max}{\frac{||Ax||}{||x||}} =||A||_2 \leq ||A||_F$$

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