Diagonalization of an almost diagonal matrix

[TheHup]

Homework Statement

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.

Homework Equations

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.

 

[mighty2000]

Dear forum post author,

Thank you for your question. It is correct that the eigenvalues of this matrix are -1 and 1 with multiplicity n-1. To find the eigenvectors corresponding to the eigenvalue 1, you can use the fact that the matrix is symmetric. This means that the eigenvectors corresponding to different eigenvalues are orthogonal to each other. So, you can find the eigenvectors corresponding to the eigenvalue 1 by finding the null space of the matrix (A-I), where I is the n x n identity matrix. This will give you n-1 linearly independent eigenvectors, which is enough to show that the matrix is diagonalisable.

I hope this helps. Good luck with your research!