matrix p-norm

A class of matrix norms, denoted $\parallel \cdot {\parallel }_p$, is defined as

$${\parallel A\parallel }_p=\underset{x\ne 0}{sup}\frac{{\parallel Ax\parallel }_p}{{\parallel x\parallel }_p}\mathit{\hspace{1em}\hspace{1em}}x\in {ℝ}^n,A\in {ℝ}^{m\times n}.$$

The matrix $p$-norms are defined in terms of the vector $p$-norms (http://planetmath.org/VectorPNorm).

An alternate definition is

$${\parallel A\parallel }_p=\underset{{\parallel x\parallel }_p=1}{\max }{\parallel Ax\parallel }_p.$$

As with vector $p$-norms, the most important are the 1, 2, and $\mathrm{\infty }$ norms.The 1 and $\mathrm{\infty }$ norms are very easy to calculate for an arbitrary matrix:

It directly follows from this that ${\parallel A\parallel }_1={\parallel A^T\parallel }_{\mathrm{\infty }}$.

The calculation of the $2$-norm is more complicated. However, it can be shown thatthe 2-norm of $A$ is the square root of the largest eigenvalue of $A^{T}A$.There are also various inequalities that allow one to make estimates on the valueof ${\parallel A\parallel }_2$:

$$\frac{1}{\sqrt{n}}{\parallel A\parallel }_{\mathrm{\infty }}\le {\parallel A\parallel }_2\le \sqrt{m}{\parallel A\parallel }_{\mathrm{\infty }}.$$

$$\frac{1}{\sqrt{m}}{\parallel A\parallel }_1\le {\parallel A\parallel }_2\le \sqrt{n}{\parallel A\parallel }_1.$$

$${\parallel A\parallel }_2^2\le {\parallel A\parallel }_{\mathrm{\infty }}\cdot {\parallel A\parallel }_1.$$

$${\parallel A\parallel }_2\le {\parallel A\parallel }_F\le \sqrt{n}{\parallel A\parallel }_2.$$

(${\parallel A\parallel }_F$ is the Frobenius matrix norm)