Matrix subtraction errors using repmat

I am trying to subtract a 1x1 matrix with a 10000x1. In order to have proper matrix subtraction my formula was

A=unitwt*z; % (a 1x1 matrix)
B=a;        % 10000x1 matrix
AB=repmat((A,10000,1)-B).*(a 10x1 matrix)

However when doing this, I keep getting the error:

noncomformant arguments (op1 is 1x1, op2 is 10000x1).

Is there anything I am missing?

Answer 1

[OCDER]

Why do you need to use repmat? a 1x1 matrix is a scalar, hence you can just do A-B without repmat.

A = 2; %this is a 1x1 matrix
B = rand(10)  %this is a 10x10 matrix

%these will work:
A + B  
A - B
A * B
B / A

The real question is, what do you expect a 10000x1 matrix times by a 10x1 matrix to yield?

AB =            (A-B) .* C
     (10000x1 matrix) .* (10x1 matrix) = ?????

Or did you mean

AB =            (A-B) .* C.'
     (10000x1 matrix) .* (1x10 matrix) = 10000x10 matrix?

Answer 2

[dpb]

1. You can subtract/add/multiply/divide any sized array with a scalar without repmat so you're "fixing" a problem that isn't a problem to begin with ( :) ), and
2. You've got parentheses-nesting problem in the way you wrote the arguments to repmat if you are going to do it...

AB=(repmat(A,size(B))-B).*a_10x1_matrix;

will get as far as the subtraction, but then the matrix multiply will be non-conforming as you'll have [10000x1]*[10x1] and the inner dimensions must match. You could write the transpose there for the second if the result is intended to be [10000x10].

But, as pointed out at the top, just write

AB=(A-B).*a_10x1_matrix.';

or otherwise solve the latter problem on the end result size.

ADDENDUM

I just noticed both OCDER and I missed the 'dot' in the .* element multiply. There's an issue there in commensurate sizes as well; the problem in answering the Q? is that the intended result is ill-defined such that the intent is not known.

Explain what the output result is intended to be; perhaps illustrate the desired result with a set of small sample inputs (making up such an example might just let you see the answer to the problem on your own, even).