# Thread: Any rank one matrix is diagonalizable?

1. ## Any rank one matrix is diagonalizable?

How to show that any rank one matrix is diagonalizable?

2. ## 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.

3. ## 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?

,
,

,

# is evrry matrix of rank 1 is diagonslizable

Click on a term to search for related topics.