timsea81 said:
Is the following statement correct?
To find the rank of a matrix, reduce the matrix using elementary row operations to row-echelon form. Count the number of not-all-zero rows and not-all-zero columns. The rank is smaller of those 2 numbers.
yes, but...
due to the nature of row-reduction, there will always be the same number or fewer non-zero rows than non-zero columns. every non-zero row has a "leading 1", which then also means the column containing that 1 (often called "pivot columns") is also non-zero.
but it may be that between two successive non-zero rows, the leading 1 in the lower row is more than 1 place to the right of the leading 1 in the upper row. since the entries in the upper row in the columns between the leading 1's are not constrained to be 0 (being neither above, nor below a leading 1), it can happen that they are, in fact, non-zero, leading to more non-zero columns than rows.
that's why it's called ROW reduction, because the rank of the original (and reduced row-echelon form) matrix is equal to the number of non-zero rows of the rref form.
one can also perform column reduction operations, leading to the number of non-zero columns being minimal. in this procedure, one obtains (perhaps) more non-zero rows than columns.
either way, the column rank of a column-reduced matrix, or the row rank of a row-reduced matrix, will give you the same number, which is simply called the rank of the original matrix. row rank is sometimes called "the dimension of the solution space", and column rank "the dimension of the image space", but in any case, one thing is clear: the rank of a matrix tells you essentially "how many (row, or column) vectors really matter".
