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 April 9th, 2010, 02:59 PM #1 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 Computing the rank of this matrix quickly? How do I quickly find the rank of this matrix? A = $\left( \begin{array}{ccc} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10 \\ 11 & 12 & 13 & 14 & 15 \\ 16 & 17 & 18 & 19 & 20 \\ 21 & 22 & 23 & 24 & 25 \end{array} \right)$ element $a_{ij}= j + 5(i-1)$ and subtracting from each of the rows i > 1 a multiple of row 1 that gives 0 in the first column for each row after i >1 we have that $a'_{ij} = a_{ij} - a_{i1}a_{1j} = j + 5(i-1) - (1 + 5(i-1))*j = 5(i-1)(1-j)$ for i > 2. This gives similar matrix A' A' = $\left( \begin{array}{ccc} 1 & 2 & 3 & 4 & 5 \\ 0 & -5 & -10 & -15 & -20 \\ 0 & -10 & -20 & -30 & -40 \\ 0 & -15 & -30 & -45 & -60 \\ 0 & -20 & -40 & -60 & -80 \end{array} \right)$ Divide each of these elements by -5(i-1) = 5(1- i) (which you can do because you're dividing each element of the same row by the same number) and you get that (for each row after row 1) the elements are just j-1 , row 2,3,4, and 5 become [0 1 2 3 4]. So the last 3 rows are redundant, and thus the rank is 2. That is A' becomes similar matrix A'': A'' = $\left( \begin{array}{ccc} 1 & 2 & 3 & 4 & 5 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 & 4 \end{array} \right)$
 April 10th, 2010, 11:39 PM #2 Member   Joined: Nov 2009 From: France Posts: 98 Thanks: 0 Re: Computing the rank of this matrix quickly? You can take the line $i$ and substract the line $i-1$ for $i$ from $2$ to $5$. You will obtain : $A' = \begin{pmatrix} 1&2&3&4&5\\ 5&5&5&5&5\\ 5&5&5&5&5\\ 5&5&5&5&5\\ 5&5&5&5&5 \end{pmatrix}$
 April 12th, 2010, 02:39 PM #3 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: Computing the rank of this matrix quickly? Oh, nicely done.
 April 16th, 2010, 12:30 PM #4 Newbie   Joined: Apr 2010 Posts: 11 Thanks: 0 Re: Computing the rank of this matrix quickly? smart

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