Okay, what have you done? You are told that one eigenvalue (not eigenvector) has algebraic multiplicity 2 so that the characteristic polynomial has only two distinct roots- and that means they must both be real numbers. Shouldn't be hard to find.
A =
[ 3 2 4 ]
[ 2 0 2 ]
[ 4 2 3 ]
show that A is diagonalizable even though one eigenvector has algebraic multiplicity 2. Do this by brute force computation. write A = Q/\Q^T where Q's columns are orthogonal unit vectors of A.
Okay, what have you done? You are told that one eigenvalue (not eigenvector) has algebraic multiplicity 2 so that the characteristic polynomial has only two distinct roots- and that means they must both be real numbers. Shouldn't be hard to find.
You are told exactly what to do! "write A = Q/\Q^T where Q's columns are orthogonal unit vectors of A".
I take it that "/\" is the diagonal matrix with the eigenvalues on the diagonal.
You got, as eigenvectors corrresponding to eigenvalue -1, <-1, 2, 0> and <-1, 0, 1> (I would prefer you had posted them here rather than sending them as private messages so that others can profit by them). Then are not however "orthogonal unit vectors". You need to find another vector that is a linear combination of <-1, 2, 0> and <-1 0, 1> but orthogonal to <-1, 2 0>. Once you have done that, divide each by its length to get "unit" vectors and put them in as the first two columns of Q. You got <2, 1, 2> as the eigenvector corresponding to eigenvalue 8 and that is aready orthogonal to the first two vectors (it can be shown that eigenvectors corresponding to different eigenvalues are orthogonal). Divide that by its length to get another unit vector and use that at the third column of Q.