# The Length of a Matrix- Double Sum

1. Apr 9, 2013

### Astrum

I'm really confused about the double sum given by my text book. Here's what it says:

If A is an nxm matrix, its length is the square root of the sum of its squares of all its entries.

$$\left|A\right|^{2}=\sum^{n}_{i=1}\sum^{m}_{j=1}a_{i,j}^{2}$$

The double sum is what has me caught up. How do you visualize this, and how do you compute it?

2. Apr 9, 2013

### SteamKing

Staff Emeritus
Start working from left to right.

1. For the first sigma, set i = 1.
2. Go to the second sigma. Here, let j run from 1 to m.
3. The partial sums will be: a(1,1)^2 + a(1,2)^2 + ... + a(1,m-1)^2 + a(1,m)^2 Notice, i stays constant
4. Set i = 2
5. Go to Step 2. Repeat Step 3, adding the partial sum for i = 2 to the partial sum for i = 1
6. Repeat Steps 2 and 3 until i = n, adding the partial sums together.

Hope this helps.

3. Apr 9, 2013

### HallsofIvy

Staff Emeritus
Note that, with i= column number, j= row number $\sum_i(\sum_j a^2_{ij})$ means that for i= 1 you sum over all numbers in the first column, for i= 2, you sum over all numbers in the second column, etc.

That is, the double sum is the just the sum of squares of all the numbers in the matrix.

4. Apr 10, 2013

### Astrum

I'm still confused.

Let's try a concrete example. $$A = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$$

How would you wright this out? like this:

$$\sum^{2}_{i=1}\sum^{2}_{i=1}a^{2}_{i,j} = (\sum^{2}_{j=1}a^{2}_{j})+(\sum^{2}_{j=1}a^{2}_{j})$$

But where does it go from here? What happened to the i?

5. Apr 10, 2013

### slider142

It is not the sum of two sums, it is the sum of a sum. That is:
$$\sum_{i=1}^{n} \sum_{j=1}^m a_{i,j}^2 = \sum_{i=1}^{n} \left(\sum_{j=1}^m a_{i,j}^2\right)$$
Write the sum over j out inside the parentheses first, then sum the result over i.
In the case you raised,
$$\sum_{i=1}^2 \left(\sum_{j=1}^2 a_{i,j}^2\right) = \sum_{i=1}^2 (a_{i,1}^2 + a_{i, 2}^2)$$

6. Apr 10, 2013

### HallsofIvy

Staff Emeritus
Both indices, i and j, label the terms of the matrix. When you broke this into two sums, your first sum corresponds to i= 1 and the second to i= 2.
With this particular A, that would be $$(1^2+ 2^2)+ (0^2+ 1^2)= 5+ 1= 6$$.
Swapping i and j, that is swapping row and columns, we would get $$(1^2+ 0^2)+ (2^2+ 1^2)= 1+ 5= 6$$.

Either way, the sum is just the sum of squares or all members orf the matrix: $$1^2+ 2^2+ 0^2+ 1^2= 1+ 4+ 0+ 1= 6$$

However, I would not call this the "length" of a matrix, I would call it the "norm".