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I know the title isn't very clear. The question is the following. Let $\|\cdot\|$ be a vector norm on $\mathbb{C}^n$. Define the induced norm on $\mathbb{C}^{n\times n}$ as: $$\|A\|_{op}=\max\limits_{x\in\mathbb{C}^n\smallsetminus\{0\}}\dfrac{\|Ax\|}{\|x\|}.$$ How do I prove that:

  1. If $\|\cdot\|_{2}$ is the Euclidean norm, then: $$\|A\|_{2}=\rho^{\frac12}(A^{*}A),$$ with $\rho(A)$ the spectral radius of $A$ and $A^{*}$ the conjugate transpose of $A$?
  2. If $\|\cdot\|_{\infty}$ is the max norm ($\|x\|_{\infty}=\max|x_i|$, $x_i$ being the components of $x$) then: $$\|A\|_{\infty}=\max_{i=1,\dots,n}\left(\sum\limits_{j=1}^n|a_{ij}|\right)?$$
  3. And that if $\|\cdot\|_{1}$ is the sum norm ($\|x\|_{1}=\sum_{i=1}^n|x_i|$, $x_i$ being, again, the components of $x$), then: $$\|A\|_{1}=\max\limits_{j=1,\dots,n}\left(\sum\limits_{i=1}^n|a_{ij}|\right)?$$
Rodrigo de Azevedo
  • 1
MickG
  • 8,645

1 Answers1

7

  1. Note that $\left\|Ax\right\|_{2}^{2} = \left< Ax, Ax \right> = \left<x, A^{*}Ax\right>$. Since $A^{*}A$ is Hermitian and positive-semidefinite, then it has an orthonormal basis of eigenvectors $\{u_{j}\}_{j=1}^{n}$ with associated eigenvalues $\mu_{j}^{2} \in \mathbb{R}_{\ge 0}$. Then for any $x\in\mathbb{C}^{n}$, we can write $x = \sum_{j} \alpha_{j}u_{j}$ so that \begin{align} \left\|Ax\right\|_{2}^{2} &= \left<\sum_{j} \alpha_{j}u_{j}, \sum_{k} \mu_{k}^{2}\alpha_{k}u_{k} \right> = \sum_{k}\mu_{k}^{2}|\alpha_{k}|^{2} \le\mu_{\max}^{2}\sum_{k}|\alpha_{k}|^{2}\\ & = \rho(A^{*}A)\left\|x\right\|^{2} \end{align} Hence $\left\|A\right\|_{2} \le \sqrt{\rho(A^{*}A)}$. Choosing $x = u_{k}$ such that $\mu_{k} = \mu_{\max}$ shows that $\left\|A\right\|_{2} \ge \sqrt{\rho(A^{*}A)}$.

  2. Beginning with the definition: \begin{align} \left\|Ax\right\|_{\infty} &= \max_{i \in \{1,2,\ldots n\}} \left| \sum_{j} a_{ij}x_{j} \right|\\ &\le\max_{i \in \{1,2,\ldots n\}} \sum_{j} \left|a_{ij}\right|\left|x_{j} \right|\\ &\le\max_{i \in \{1,2,\ldots n\}} \sum_{j} \left|a_{ij}\right|\left|x_{\max} \right|\\ &=\max_{i \in \{1,2,\ldots n\}} \sum_{j} \left|a_{ij}\right| \left\|x\right\|_{\infty} \end{align} So $\left\|A\right\|_{\infty} \le \max_{i \in \{1,2,\ldots n\}} \sum_{j} \left|a_{ij}\right|$. Choose $k$ such that \begin{align} \sum_{j} \left|a_{kj}\right| = \max_{i \in \{1,2,\ldots n\}} \sum_{j} \left|a_{ij}\right| \end{align} Then choose $x$ such that $x_{j} = \overline{a_{kj}}/|a_{kj}|$ if $a_{kj} \ne 0$ and $x_{j} = 1$ if $a_{kj} = 0$. Then $\left\|x\right\|_{\infty} = 1$ and \begin{align} \left\|Ax\right\|_{\infty} &\ge \left| \sum_{j} a_{kj}x_{j} \right|\\ & = \sum_{j} |a_{kj}| \\ & = \max_{i \in \{1,2,\ldots n\}} \sum_{j} \left|a_{ij}\right| \end{align}

  3. Again starting with the definition: \begin{align} \left\|Ax\right\|_{1} &=\sum_{j} \left|\sum_{k} a_{jk}x_{k} \right| \\ &\le \sum_{k} |x_{k}|\sum_{j} |a_{jk}| \\ &\le \max_{k'} \sum_{j} |a_{jk'}| \sum_{k} |x_{k}| \\ &\le \max_{k} \sum_{j} |a_{jk}| \left\|x\right\|_{1} \end{align}

So $\left\|A \right\|_{1} \le \max_{k} \sum_{j} |a_{jk}|$.

Now choose $\ell$ such that \begin{align} \sum_{j} |a_{j\ell}| = \max_{k} \sum_{j} |a_{jk}|. \end{align} Then choose $x$ such that $x_{\ell} = 1$ and $x_{k} = 0$ for $k\ne \ell$. Then \begin{align} \left\|Ax\right\|_{1} &= \sum_{j}\left|\sum_{k} a_{jk}x_{k} \right| \\ &= \sum_{j} |a_{j\ell}| \\ &= \max_{k} \sum_{j} |a_{jk}| \end{align} Since $\left\|x\right\|_{1} = 1$, this completes the proof.

user14717
  • 4,902
  • Assuming $\mu_k^2\alpha_k\mu_k$ is a typo and the last $\mu$ was supposed to be a $u$, let me explicit that the first inequality divided by $|x|_2^2$ gives, passing to the $\max$ on the left, that $|A|_2^2\leq\rho(A^HA)$, and the square root applied to both sides (licit) gives the first inline inequality. The vector choice gives us $\mu_k^2|x|_2=|\mu_k^2x|_2=|A^HAx|_2\leq|A^H|_2|A|_2|x|_2\implies\mu_k^2\leq|A|_2|A^H|_2$, and then? – MickG Aug 18 '14 at 11:36
  • But then I have elsewhere a proof that every induced norm (operator norm) is not less than the spectral radius, so we are done here. – MickG Aug 18 '14 at 11:40
  • Point 2, again we divide by $|x|_\infty$ on both sides and pass to the max on the left. And I bet point 3 does the same. – MickG Aug 18 '14 at 11:42
  • To finish and explicit point 2: $$|A|{\infty}|x|{\infty}\geq|Ax|{\infty}\geq\left|\sum\limits{j=1}^na_{kj}\right|=$$ $$=\left|\sum\limits_{j:a_{kj}\neq0}a_{kj}\cdot\dfrac{\overline{a_{kj}}}{|a_{kj}|}\right|=\left|\sum\limits_{j:a_{kj}\neq0}\dfrac{|a_{kj}|^2}{|a_{kj}|}\right|=$$ $$=\left|\sum\limits_{j:a_{kj}\neq0}|a_{kj}|\right|=\sum\limits_{j:a_{kj}\neq0}|a_{kj}|+0=\sum\limits_{j:a_{kj}\neq0}|a_{kj}|+\sum\limits_{j:a_{kj}=0}|a_{kj}|=$$ $$=\sum\limits_{j=1}^n|a_{kj}|=\max\limits_{i=1,\dots,n}\left(\sum\limits_{j=1}^n|a_{kj}|\right),$$ MathJax won't put the $n$ up in the last sum, … – MickG Aug 18 '14 at 12:23
  • … and since $|x|{\infty}$ is one by choice of $x$ we get $|A|{\infty}$ is not smaller than the final maximum. – MickG Aug 18 '14 at 12:26
  • And point 3 is much the same as point 2, with the same implicit passages. – MickG Aug 18 '14 at 12:40
  • There was a typo. It is fixed now. – user14717 Aug 18 '14 at 20:04