Any rank one matrix is diagonalizable?

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• Nov 21st 2012, 02:31 AM
wesleybrown
Any rank one matrix is diagonalizable?
How to show that any rank one matrix is diagonalizable?

I've no idea how to start with.
• Nov 21st 2012, 08:25 AM
HallsofIvy
Re: Any rank one matrix is diagonalizable?
An n by n matrix is "diagonalizable" if and only if there exist a basis for $R^n$ consisting entirely of eigenvectors of the matrix. If the matrix has "rank 1" then it has one non-zero eigenvalue. An eigenvector corresponding to that eigenvalue will span a one dimensional subspace. The matrix, applied to any vector in the n-1 dimensional orthogonal subspace, will give 0 so any basis for that subspace will be a basis consisting of eigenvectors corresponding to eigenvalue 0. Adding the first eigenvector to that will give a basis for the entire space consisting of eigenvectors.
• Nov 21st 2012, 09:09 AM
wesleybrown
Re: Any rank one matrix is diagonalizable?
Thank you,
"If the matrix has "rank 1" then it has one non-zero eigenvalue."
This is exactly the point where I stuck, can you explain more on this sentence?