Rank of a matrix.

Question

Let a non-zero column matrix $A_{m\times 1}$ be multiplied with a non-zero row matrix $B_{1\times n}$ to get a matrix $X_{m\times n}$ . Then how to find rank of $X$?

score 5 · Answer 1 · edited Apr 04 '13 at 10:58

5

Hint: To find the rank of a matrix, one usually row (or column) reduces it. In your case, all the rows (or columns) are multiples of one row (column). What can you conclude about the reduced matrix?

edited Apr 04 '13 at 10:58

SBF

36,041

answered Apr 04 '13 at 10:55

Dennis Gulko

15,640

Can't get it. I have just read basic meaning of rank and nothing more than that about rank. – ABC Apr 04 '13 at 11:11
@exploringnet: in case any two rows are linearly dependent (they are both multiples of $B$), can it happen that the rank is $2$? – SBF Apr 04 '13 at 11:30
@Ilya Yes , but then answer is $min(m,n)$ yes ? . I don't have answer with me. – ABC Apr 04 '13 at 11:35
@exploringnet: well, the rank of $\Bbb R^{m\times n}$ matrix is at most $\min(m,n)$. In your case, clearly, the matrix is very degenerate. – SBF Apr 04 '13 at 11:39
@exploringnet: the rank of a matrix is the dimension of the space its columns or rows span. Since all the rows of the product are multiples of the row matrix, what is the dimension of the space they span? Since all the columns of the product are multiples of the column matrix, what is the dimension of the space they span? – robjohn Apr 04 '13 at 11:46

score 3 · Answer 2 · answered Apr 04 '13 at 12:19

Hint: You always have

$$rk(AB)\leq min\{rk(A),rk(B)\}$$

It would be helpful to convince yourself of this fact. @DennisGulko's answer gives you an idea how.

Now can the rank of $AB$ be zero? That would mean that the matrix has only zero-entries...

Edit: I assume the entries of the matrix are elements of a field, i.e there are no zero-divisors.

score 0 · Answer 3 · answered Mar 08 '17 at 10:19

Let me discuss a shortcut for finding the rank of a matrix .

Rank of a matrix is always equal to the number of independent equations . The number of equations are equal to the number of rows and the variables in one equation are equal to number of columns .

Suppose there is a 3X3 matrix with elements as : row 1 : 1 2 3 row 2 : 3 4 2 row 3 : 4 5 6

So there will be three equations as x + 2y + 3z = 0 -1 3x + 4y + 2z = 0 -2 4x + 5y + 2z = 0 -3

Any of the above equation cannot be obtained by adding or subtracting two equations or multiplying or dividing a single equation by a constant . So there are three independent equations . So rank of above matrix is 3 .

Consider another matrix of order 3X3 with elements as : row 1 : 10 11 12 row 2 : 1 2 7 row 3 : 11 13 19

equations : 10x + 11y + 12z =0 - 4
x + 2y + 7z =0 - 5 11x + 13z + 19z =0 - 6

equation 6 can be obtained by adding equations 4 and 5 . So there are only two independent equations . So rank of this matrix is 2 .

This method can be applied to matrix of any order .

Rank of a matrix.

3 Answers3