Prometheus Posted January 7, 2015 Posted January 7, 2015 I found the eigenvalues of a 2x2 matrix with all elements equal to one to be 2 and 0. The eigenvalues of a 3x3 matrix with all elements equal to one to be 3,0,0. A 4x4 matrix to be 4,0,0,0 I stopped at a 5x5 matrix. I was just wondering (as i'm rubbish generalising things) whether this results holds for any size matrix. Also, are there any links were this is explored, i couldn't find any.
Bignose Posted January 7, 2015 Posted January 7, 2015 Yes, this is not surprising, since a matrix of all 1s clearly does not have linearly independent rows or columns. That's where all the 0s come from, the non independence. The last single value represents the one row or column vector that is independent.
Prometheus Posted January 10, 2015 Author Posted January 10, 2015 Yes, i see now where all the zeros are coming from, thanks. The only way i can get the last single eigenvaluev is to use the fact that the sum of the eigenvalues is equal to the trace of the matrix and since all the other eigenvalues are zero, the last eigenvalue must be n. I was looking at the eigenvectors too. Its obvious that one eigenvector will be a vector of ones, corresponding to the eigenvalue n. For the other eigenvalues i just choose any vectors that are mutually orthogonal since the matrix is real symmetric. Is that reasoning sound?
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