将矩阵的每一行与其转置的自身相乘

Question

The formula I have to translate to Octave/Matlab goes something like this:

\sum (v_i - m) (v_i - m)^T

I have a matrix, and I need to take each row, subtract m from it and then multiply it with its own transpose. I wrote the inner part as a function:

function w = str(v, m)
    y = v - m
    w = y * transpose(y)
end

My matrix is like this

xx = [1 2 3 4 5; 1 2 3 4 5; 1 2 3 4 5]

Now I have no idea how to apply this function to each row in a matrix and then sum them up to a new matrix. Maybe someone can help me here.

EDIT: The result is not the dot product. I'm looking for v * v^T , which has a matrix as result!

Answer 1

Probably you need this

X = bsxfun( @minus, A, m );
Y = X'* X;

Answer 2

You can subtract the mean using bsxfun

>> v_m = bsxfun( @minus, v, m );

For the sum of outer product of all vectors you can use bsxfun again

>> op = bsxfun( @times, permute( v, [3 1 2]), permute( v, [1 3 2] ) );
>> op = sum( op, 3 );

Answer 3

Suppose the matrix is A, then the solution is

total = sum(sum((A-m).*(A-m),2));

A.*A is an element wise multiplication, hence sum(A.*A,2) returns a column vector, with each element being the self dot product of each row in A .

If m is a vector then, it is slightly more complicated.

[p,~]=size(A);
    total = sum(sum((A-repmat(m,p,1)).*(A-repmat(m,p,1)),2));

Cheers.

Answer 4

In the end, I wrote this:

function w = str(v, m)
    y = v - m;
    w = y' * y;
end

y = zeros(5,5);
for i=1:12
    y = y + str(A(i,:), m);
end

Surely not the most elegant way to do this, but it seems to work.

Answer 5

There are two ways to solve this issue:

Assume A is your matrix:

sum(drag(A' * A))

will do the job. However, it is slightly more efficient with the following:

sum((A .* A)(:))