- Thread starter
- #1
- Feb 5, 2012
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Hi everyone, 
Here's a question I was stuck on. Hope you people can help me out.
Now a \(n\times n\) matrix can be diagonalized if it has \(n\) distinct eigenvalues. So I don't see how the given condition (all root vectors are eigenvectors) can ensure that there are \(n\) distinct eigenvalues.
Here's a question I was stuck on. Hope you people can help me out.
The definition of root vectors is given >>here<<.Prove that \(f:\,V\rightarrow V\) over \(\mathbb{C}\) is diagonalizable iff all root vectors are eigenvectors.
Now a \(n\times n\) matrix can be diagonalized if it has \(n\) distinct eigenvalues. So I don't see how the given condition (all root vectors are eigenvectors) can ensure that there are \(n\) distinct eigenvalues.