The Length of a Matrix- Double Sum

[Astrum]

I'm really confused about the double sum given by my textbook. 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?

 

[SteamKing]

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.

 

[HallsofIvy]

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.

 

[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?

 

[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,
$$\sum_{i=1}^2 \left(\sum_{j=1}^2 a_{i,j}^2\right) = \sum_{i=1}^2 (a_{i,1}^2 + a_{i, 2}^2)$$

 

[HallsofIvy]

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?

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".