Why is 4-by-2 matrix compatible with 4-by-1 vector?

x = [1 2;3 4;5 6; 7 8].*[1;2;3;4] 
% Firstly the first column elements are multiplied with the elements of second matrix, with each element.
% Implies
% 1*1 is the first element in the first column
% 3*2 is the second element in the first column
% 5*3 is the thrid element in the first column
% 7*4 is the fourth element in the first column
% Now, the same applies for second column
% 2*1 is the first element in the second column
% 4*2 is the second element in the second column
% 6*3 is the third element in the second column
% 8*4 is the fourth element in the second column

It is an element-wise multiplication. If there is a normal matrix multiplication (*), then it must satify the matrix multiplication rule as stated by you.

If you transpose the matrix, it doesn't directly work with element wise operation but can perform matrix multiplication. Since, (2 x 4) matrix multiplied by (4 x 1), the column size of first matrix is same as row matrix. thus, matrix multiplication is possible.

But, for element wise multiplication, the number of rows must be the same, if the first element is a matrix with number of rows and columns greater than 1. If the first element is a column vector and second element is a row vector, it must be similar to matrix multiplication.

In simple terms, the element-wise operation takes place element by element and to perform it, each column element must be multiplied with the same column element of second matrix.

Hope this helps.

Regards,

Sriram

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Answer 2

John D'Errico 2020-7-29

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编辑：John D'Errico 2020-7-29

You are using .* to do the multiplication. NOT * which is the linear algebra operator.

MATLAB extends the vector implicitly into a 4x2 array, and THEN does an element-wise multiply.

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Answer 3

Walter Roberson 2020-7-29

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This is because of "implicit expansion".

It used to be the case (in R2016a and earlier) that when you did binary operations such as + or .* between two arrays, that the sizes had to match in every dimension. As of R2016b, the rules changed, and operations became similar to as-if bsxfun() had been used (but implemented differently internally.)

For implicit expansion (and bsxfun), the rule is no longer that the sizes must match exactly: the rule instead is that the sizes must match except that a dimension in one of them may be 1 when the other one is not 1, and that when that happens, the one that is 1 will be internally copied as many times as needed to match the size of the other operand.

So with a 4 x 2 matrix A being operated on together with a 4 x 1 matrix B, then the operation would be carried out "as if" you had done A .* repmat(B, [1 size(A,2)]), so for your calculation the result would be the same as if you had used

x = [1 2;3 4;5 6; 7 8].*[[1;2;3;4], [1;2;3;4]]