Hello!Recently I have been asked the following problem at the interview for the internship position of quant analyst With a hint I solved it, but.. Find the eigenvalues and eigenvectors of the following n x n matrix:all entries on both of the diagonals are equal to 1;all the other entries of the matrix are equal to 0Good luck!

if n is evenall eigenvalues are 0 and eigenvectors are [0, 0 ,..0]if n is oddn eigenvalues are 1-(nth root of unity) 's

QuoteOriginally posted by: amit7ulif n is evenall eigenvalues are 0 and eigenvectors are [0, 0 ,..0]if n is oddn eigenvalues are 1-(nth root of unity) 'sI guess it would be fair to give a hint I had been given too:Split the initial matrix to the sum of two, and calculate the powers of the non-identity component

QuoteOriginally posted by: amit7ulif n is evenall eigenvalues are 0 and eigenvectors are [0, 0 ,..0]if n is oddn eigenvalues are 1-(nth root of unity) 'sIt can't be so. the n=2 matrix is (1, 1 ; 1, 1) whose eigenvalues are 0 and 2.Your solution is true for every n: The eigenvalues are 1-(nth root of the cyclotomic polynomial).I think that, even if they give u the hint that Barvinok mentioned, this is a pretty tough question for an interview...

cyclotomic polynomial ??

QuoteOriginally posted by: amit7ulcyclotomic polynomial ??Yeah... the x^n-1 polynomial... cyclotomic is a fancy word for 'cutting the circle'...

One can directly calculate the det of the eigenvalue matrix by row reductionWhen n is even, det[A] = [(1-x)^2-1]^(n/2)wehn n is odd, det[A] = (1-x) [(1-x)^2-1]^[(n-1)/2]

..minimal polynomial of the matrix..and something about its roots..?

if n is even n/2 eigenvalues are 0 and eigenvectors are [1, 0 , 0,..,0, 0, -1], [0, 1, 0,.., 0, -1, 0], [0, 0, 1,.., -1, 0, 0], ...n/2 eigenvalues are 2 and eigenvectors are [1, 0 , 0,..,0, 0, 1], [0, 1, 0,.., 0, 1, 0], [0, 0, 1,.., 1, 0, 0], ... if n is odd then (n-1)/2 eigenvalues are 0, (n-1)/2 eigenvalues are 2 (eigenvectors like in even case). And last eigenvalue is 1 and eigenvector is [0, 0 , ..0, 1, 0..., 0, 0]

QuoteOriginally posted by: Tapacbif n is even n/2 eigenvalues are 0 and eigenvectors are [1, 0 , 0,..,0, 0, -1], [0, 1, 0,.., 0, -1, 0], [0, 0, 1,.., -1, 0, 0], ...n/2 eigenvalues are 2 and eigenvectors are [1, 0 , 0,..,0, 0, 1], [0, 1, 0,.., 0, 1, 0], [0, 0, 1,.., 1, 0, 0], ... if n is odd then (n-1)/2 eigenvalues are 0, (n-1)/2 eigenvalues are 2 (eigenvectors like in even case). And last eigenvalue is 1 and eigenvector is [0, 0 , ..0, 1, 0..., 0, 0]Good for you

QuoteOriginally posted by: BarvinokHello!Recently I have been asked the following problem at the interview for the internship position of quant analyst With a hint I solved it, but.. Find the eigenvalues and eigenvectors of the following n x n matrix:all entries on both of the diagonals are equal to 1;all the other entries of the matrix are equal to 0Good luck!Why not to just proceed by definition of eigenvector?Assume we have eigenvector (a_1,..., a_n) and corresponding eigenvalue \lambda, thenBy multiplying the first row of the matrix by the eigenvector you get: a_1+a_nBut by definition of eigenvector it should equal to \lambda a_1. Thusa_1+a_n = \lambda a_1a_1+a_n = \lambda a_n Similarlya_2+a_{n-1}= \lambda a_2a_2+a_{n-1}= \lambda a_{n-1}and so on.So there are only 2 possibilities for eigenvalues.1) \lambda = 2. Then a_1 = a_n, a_2 = a_{n-1} etc.2) \lambda = 0. Then a_1 = -a_n, a_2=-a_{n-1} etc.We can also notice that "half" of the matrix is linearly dependent on the other "half", so "half" of the eigenvalues must be 0, then all of the other "half" clearly must be 2.We have to pay a little attention on what we mean by "half" depending on whether n is even or odd.

QuoteOriginally posted by: Tapacbif n is even n/2 eigenvalues are 0 and eigenvectors are [1, 0 , 0,..,0, 0, -1], [0, 1, 0,.., 0, -1, 0], [0, 0, 1,.., -1, 0, 0], ...n/2 eigenvalues are 2 and eigenvectors are [1, 0 , 0,..,0, 0, 1], [0, 1, 0,.., 0, 1, 0], [0, 0, 1,.., 1, 0, 0], ... if n is odd then (n-1)/2 eigenvalues are 0, (n-1)/2 eigenvalues are 2 (eigenvectors like in even case). And last eigenvalue is 1 and eigenvector is [0, 0 , ..0, 1, 0..., 0, 0]I got basically the same results:if n is even n/2 eigenvalues are 0 and eigenvectors are [-1, 0 , 0,..,0, 0, 1], [0, -1, 0,.., 0, 1, 0], [0, 0, -1,.., 1, 0, 0], ...n/2 eigenvalues are 2 and eigenvectors are [1, 0 , 0,..,0, 0, 1], [0, 1, 0,.., 0, 1, 0], [0, 0, 1,.., 1, 0, 0], ...if n is odd(n-1)/2 eigenvalues are 0 and eigenvectors are [-1, 0 , 0,..,0, 0, 1], [0, -1, 0,.., 0, 1, 0], [0, 0, -1,.., 1, 0, 0], ...(n-1)/2 eigenvalues are 2 and eigenvectors are [1, 0 , 0,..,0, 0, 1], [0, 1, 0,.., 0, 1, 0], [0, 0, 1,.., 1, 0, 0], ...The last eigenvalue is 1 and eigenvector is [0, 0 , ..0, 1, 0..., 0, 0]

Hi!I obtain the same answer than tapacb and Mcarreira, but I do not make any use of your hint...in fact it is a bit confusing for me...do you need to split the matrix??... What advantage you get from it (without using that hint the problem is solved quite easily...just apply the same elemental operations to half the rows of the matrix) PS: when you say both the diagonals you mean and "x shaped matrix", arent you?