Faster way to find the rank of the following matrix?

Question

I was wondering if there was a faster way to find the rank of the following matrix other than performing elementary row operations. $$ \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10 \\ 11 & 12 & 13 & 14 & 15 \\ 16 & 17 & 18 & 19 & 20 \\ 21 & 22 & 23 & 24 & 25 \end{bmatrix} $$ For context, I am preparing for the math GRE and would like to know how to maximize my time efficiency. Hopefully this specific example will help walk me through a "trick" or possible general procedure. Thanks.

See this answer by Robert Israel – Jyrki Lahtonen Oct 12 '15 at 18:17 — Jyrki Lahtonen, Oct 12 '15 at 18:17

score 3 · Answer 1 · answered Oct 12 '15 at 18:02

3

Try subtracting the first row from the second, third, fourth and fifth rows. The resulting four bottom rows will all be multiples of the same vector (that is different from the first row), thus the matrix has rank 2.

answered Oct 12 '15 at 18:02

aras

5,757

Seems very obvious now that you pointed that out. Thanks for the quick response. – Poko Oct 12 '15 at 18:08

score 3 · Answer 2 · answered Oct 12 '15 at 18:03

3

The row space is generated by $(1,2,3,4,5)$ and $(5,5,5,5,5)$, hence the rank is clearly $2$.

answered Oct 12 '15 at 18:03

Jack D'Aurizio

361,689

I understand now, thanks for the help. – Poko Oct 12 '15 at 18:08

score 3 · Accepted Answer · answered Oct 12 '15 at 18:13

3

It might be "more obvious" to notice that the difference between $a_{i,j+1}$ and $a_{i,j}$ is always one. Thus, one is lead to subtract the fourth column from the fifth, then the third from the fourth, etc. The result is a matrix that is clearly of rank two.

answered Oct 12 '15 at 18:13

levap

67,610

This is very helpful. Thank you. – Poko Oct 12 '15 at 23:39

Faster way to find the rank of the following matrix?

3 Answers3