# Matrix Norms

Before we continue on analyzing the error in solving a linear system of equations of the form $Ax = b$, we will need to define what exactly a matrix norm is.

Definition: A Matrix Norm on the set of all $n \times n$, $M_{nn}$, whose entries are real numbers is a function that maps each matrix $A \in M_{nn}$ to a number $\| A \| \in \mathbb{R}$ that has the following properties:a) $\| A \| ≥ 0$ for all $A \in M_{nn}$ and $\| A \| = 0$ if and only if $A$ is the $n \times n$ zero matrix.b) $\| \alpha A \| = \mid \alpha \mid \| A \|$ for all $\alpha \in \mathbb{R}$ and $A \in M_{nn}$.c) $\| A + B \| ≤ \| A \| + \| B \|$ for all $A, B \in M_{nn}$.d) $\| AB \| ≤ \| A \| \| B \|$ for all $A, B \in M_{nn}$. |

There are many different matrix norms that can be defined on $M_{nn}$. The following norms will be of most importance to us.

(1)Note that $\| A \|_{1}$ is equal to the maximum of the sum of the absolute values of the entries for the columns of $A$.

(2)Note that $\| A \|_{\infty}$ is equal to the maximum of the sum of the absolute values of the entries for the rows of $A$.

As an example, we will now show that the norm $\| A \|_{\infty}$ is indeed a matrix norm.

Note though for each $i, j = 1, 2, ..., n$ we have that $\mid a_{ij} \mid ≥ 0$ and so each sum, $\sum_{j=1}^{n} \mid a_{ij} \mid = \mid a_{i1} \mid + \mid a_{i2} \mid + ... + \mid a_{in} \mid ≥ 0$, and thus $\| A \|_{\infty} = \max_{1≤i≤n} \sum_{j=1}^{n} \mid a_{ij} \mid ≥ 0$.

Now suppose that $\| A \|_{\infty} 0$. This implies that each $a_{ij} = 0$ for $i, j = 1, 2, ..., n$, and so $A$ is the $n \times n$ zero matrix. Conversely, suppose that $A$ is the $n \times n$ zero matrix. Then all entries in $A$ are zero, and the sum of the absolute values of the rows of $A$ will all be zero, so $\| A \|_{\infty} = 0$. Therefore property 1 holds.

Now let $\alpha \in \mathbb{R}$. Then:

(3)Therefore property 2 holds. Let $A, B \in M_{nn}$. Now for verifying property 3, we can apply the triangle inequality for the absolute value of two real numbers and we have that:

(4)Thus property 3 holds. Showing property 4 holds can be done in a similar manner, but we will omit showing this here and leave it as an exercise to the reader.