Consider the n x n matrix M4 that contains 1, 2, 3, ..., n^2

[KidKat1]

Consider the n x n matrix M4 that contains all integers 1,2,3,...,n^2 as it's entries, written in sequence, column by column.

b) Determine the rank of Mn for any artitary n greater than or equal to 2

I sent up a matrix interms of n, however, when I try to put it in RREF I get really lost. Any help with be a big help.

Thanks

 

[pka]

Are you constructing the matrix correctly?
Can you follow this tranformation?
\(\displaystyle \left[ {\begin{array}{llcr}
1 & 5 & 9 & {13} \\
2 & 6 & {10} & {14} \\
3 & 7 & {11} & {15} \\
4 & 8 & {12} & {16} \\
\end{array}} \right] \to \left[ {\begin{array}{lllc}
1 & 5 & 9 & {13} \\
1 & 1 & 1 & 1 \\
2 & 2 & 2 & 2 \\
3 & 3 & 3 & 3 \\
\end{array}} \right].\)

If so do you see the rank?
Can you generalize?

 

[KidKat1]

I didn't contruct it that same way... for a M4 is reduced it to by row (1,5,9,13) (0,1,2,3) (0,1,2,3) and (0,1,2,3) which gave me a rank of 2. However, I'm not sure where I can generalize because is everything from the second row down always going to be indentical and how do you prove that? [/tex][/code]

 

[pka]

Becuse the problem says "written in sequence, column by column", I think that it should be written as I did.
Try a 5x5 and a 6x6. See if the rank is 2.

 

[KidKat1]

I got that the rank was 2, but I don't know how you reduced it to get the 1's 2's and 3's in the rows. But, can generalize that the rank is always going to be 2 because the all the rows after the first can be reduced to the same thing? So therefore they can be subtracted to get a row of zeros, or many rows of zeros?