If we have a n x n matrix with 1 on the diagonal entries apart from the ith column which has a -1. As well as this ith row can have any real number in each entry. Other than this the matrix is 0 everywhere.
Show this matrix is diagonalisable.
The Attempt at a Solution
I know the eigenvalues of the matrix are -1 and 1 (with multiplicity n-1). I can find the eigenvector corresponding to the eigenvalue -1.
However I can't find eigenvectors corresponding to the eigenvalues 1. I assume that if I could do this then they would be n-1 linearly independent eigenvectors which I think would be enough to show that the matrix is diagonalisable.