How to find rank of a matrix. easy ways?

Answer 1

The rank of a matrix is the maximum number of independent rows (or, the maximum number of independent columns).

1 2 0 3
1 7 3 0
0 0 4 8
2 4 0 6

Start with row 1: See if row 2, 3 or 4 are multiples. If so, cross that out.
Go to row 2: See if row 3 or 4 are multiples...
Finally compare row 3 and 4.

This is for 4 rows, but you can extend this to n rows.

Do the same for the columns.

The rank would be 3 in the example above because we crossed out a row, leaving 3. The last row is a multiple of the first.

Answer 2

it depends on what you mean by 'easy'. if you mean while standing on one foot - there is probably no 'easy' way. any way of doing this requires using software - if you want to preserve your sanity.

you don't say explicitly if the matrix (call it A) is square or not - but even if not, it can be converted to a square matrix of the same rank, by working with A'A or AA'. so i'll suppose A is square, for the most part.

excel can be 'gamed' to get rankA - but you need to work with its limitations. it doesn't seem to do eigenvalues (see below) - but it will compute determinants (the function is called MDETERM). so one can do the following:

compute the determinants of the successive principal minors of the matrix, A say. the k^th principal minor of A is the k x k matrix in the upper left corner of A. (think of it as the intersection of the first k rows with the first k columns.)

if the r^th principal minor of A has a nonzero determinant but the (r+1)^st has a zero determinant, the rank of A is r. if all principal minors of A have nonzero determinants, A is of full rank (say n).

of course one can do this process backwards - starting with det A. if nonzero - you are done: the rank of A is its order, n. ( A is n x n.) if detA = zero, remove the last row and column and find the determinant of the remaining (n-1) x (n-1) principal minor. if not zero, the rank of A is n-1.

keep eliminating the last row and column of the principal minors of A until you get one with a nonzero determinant. the order of that minor is the rank of A.

another way to get rankA is to use the gram-schmidt successive orthogonalization method on the columns of A. this can be done with excel - but can be tedious for large matrices. it is better to use software that carries out the computations automatically (see below).

the method produces a new matrix B say, whose k^th column is a linear combination of the first k columns of A. (the columns of B are all orthogonal.) if A is of rank r, all columns of B after the r^th will contain only zeros. so you count the number of non-zero columns of B to get rankA.

most linear algebra texts discuss the gram-schmidt procedure and it is not conceptually difficult to carry it out in excel - just tedious if n is large.

gram-schmidt also works if A is not square. everything said above about it remains true.

another way - which can't be done with excel, apparently - is to find the eigenvalues of A. it is better to work with the matrix A'A for this. A'A is symmetric and nonnegative definite - so its eigenvalues are all real and nonnegative. the number of nonzero eigenvalues of A'A is rankA'A = rankA.

getting eigenvalues involves finding the roots of a polynomial of degree n and it is best done with software.

the site below seems to have mathematica programs you can download that will do eigenvalues and gram-schmidt. of course you need access to mathematica.

maple and matlab can also be used to do these computations.

you can also use R, which can be downloaded for free (url below) - but learning to use it is a bit of a task - altho a very worthwhile one if you plan to make a career in math or stats. R is a very powerful program that can do many things. it is primarily for statistical analyses - but it does eigenvalues of matrices among other things.

Answer 3

Reduce the matrix to echelon form then count the pivit columns.

That's the only way I know.

Answer 4

count the the number of non zero rows of a matrix in its reduced ECHELON form.

Answer 5

-06 0
0.2 -0.8