## obtaining a wrong matrix size

### Josep (view profile)

on 22 Aug 2014
Latest activity Commented on by Star Strider

on 22 Aug 2014

### Star Strider (view profile)

I would like to know why I am obtaining as a result in this code, a matrix (15x1) and not a number. I am computing a test statistics equation that involves matrices products, so I am multiplying this matrix sizes:

((1x15)*(15x15)*(15x1))X((1x15)*(15x15)*(15x1))^2

I guess I am supposed to achieve a number, from this matrices products, but as I have said, MATLAB returns me as a solution of this equation a 15x1 matrix. Otherwise, at the end I have put the command size(T), to know the matrix size of T, and it returns me that is a 1x1 matrix (a number). Does anybody know why this two things doesn't match?

Following, here is my MATLAB code attached if you want to take a look at it

Image Analyst

### Image Analyst (view profile)

on 22 Aug 2014

The attached code also requires .xlsx workbooks, which were not attached. Attach the workbooks if you want us to try the code.

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### Star Strider (view profile)

on 22 Aug 2014

It is difficult to read your code and I can’t run it. I ran a simulation of the matrix operation you describe, and got a scalar result as expected.

I suggest you check the sizes of the matrices in your equation. One of your vectors may actually be a matrix.

Star Strider

### Star Strider (view profile)

on 22 Aug 2014

It makes the code simpler and faster to divide scalars using ‘/’ than inv.

I avoid inv where possible because the ‘\’ operator is a much more efficient and computationally stable method of multiplying by inv, although sometimes it is unavoidable. Here, inv on the first term (a scalar) is the the same as 1/(Fk(:,j)'*inv(V)*Fk(:,j)), so it makes sense to compute it as a scalar.

Josep

### Josep (view profile)

on 22 Aug 2014

Nice!

Thanks Star Strider one more time

Star Strider

### Star Strider (view profile)

on 22 Aug 2014

My pleasure!

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