MHB Matrix multiplication simplified to Vector multiplication

[CyanBC]

Hello, I'm not sure where to put this. I have spent the last week (14+ hour days) editing some code I have for selecting representative spectra for a remote sensing masters thesis I'm working on. The program is very-very slow, and I've been trying to speed it up as much as possible by NOT performing any conversions that are unnecessary. Which leads me to the problem I've been struggling with for the last 10 hours.

I'll use a simple example:

I have a vector (V1)
[0,1,2,3]
Which I reform into a matrix (M1)
[0,1
2,3]

and perform matrix multiplication on itself, M1*M1
which returns a matrix
[2,3
6,11]
from which I take the sum total of all items in the matrix. So the desired answer is = 22

Is there any way I can do this directly with the original vector (V1), without having to convert the original vector to a matrix? I know conversion is the easiest way - but not the most computationally efficient. And I'm nor so good at the maths.

 

[Opalg]

Hello, I'm not sure where to put this. I have spent the last week (14+ hour days) editing some code I have for selecting representative spectra for a remote sensing masters thesis I'm working on. The program is very-very slow, and I've been trying to speed it up as much as possible by NOT performing any conversions that are unnecessary. Which leads me to the problem I've been struggling with for the last 10 hours.

I'll use a simple example:

I have a vector (V1)
[0,1,2,3]
Which I reform into a matrix (M1)
[0,1
2,3]

and perform matrix multiplication on itself, M1*M1
which returns a matrix
[2,3
6,11]
from which I take the sum total of all items in the matrix. So the desired answer is = 22

Is there any way I can do this directly with the original vector (V1), without having to convert the original vector to a matrix? I know conversion is the easiest way - but not the most computationally efficient. And I'm nor so good at the maths.

Hi Cyan and welcome to MHB! Suppose you do that same sequence of calculations algebraically, starting with a vector $[a,b,c,d].$ Then the matrix is $\begin{bmatrix} a&b \\c&d \end{bmatrix}.$ When you square it you get $\begin{bmatrix} a^2 + bc&b(a+d) \\c(a+d)&bc +d^2 \end{bmatrix}.$ The sum of the elements is $\Sigma = a^2 + 2bc + d^2 + (b+c)(a+d).$ With a little bit of algebraic manipulation you can write that as $\Sigma = (a+d)(a+b+c+d) - 2(ad-bc).$

Presumably you can write a little subroutine to input $[a,b,c,d]$ and get out $\Sigma.$ That ought to be a bit faster than going via a matrix computation.

 

[CyanBC]

Thanks for illuminating that for me. Now that I see it, I think the original method may be more efficient.