# Direct sum of the eigenspaces equals V?

1. Aug 6, 2013

### Bipolarity

I am following Friedberg's text and having some trouble understanding some of the theorems regarding diagonalizability. The proofs seem to skip some steps, so I guess I need to work through them a bit more slowly.

Given a linear operator $T:V → V$, with eigenspaces $\{ E_{ \lambda_{1}},E_{ \lambda_{2}},...E_{ \lambda_{k}} \}$, is it true that $E_{ \lambda_{1}} \bigoplus E_{ \lambda_{2}} \bigoplus E_{ \lambda_{2}} ... \bigoplus E_{ \lambda_{k}} = V$ ?

What if T is diagonalizable? This is just a thought which may perhaps help me to understand diagonalizability a bit better. I simply need to know whether this is right or not. Please no explanations, as I will prove it myself. Thanks!

BiP

2. Aug 6, 2013

### mathman

Assuming all eigenvalues are distinct (V is k dimensional), it is correct.

Things get slightly more complicated if an eigenvalue has multiplicity. The eigenvectors for such an eigenvalue define a multidimensional space (dimension = multiplicity of eigenvalue).

3. Aug 7, 2013

### Bipolarity

Suppose the eigenvalues are distinct, but dim(V) > k so that the eigenvalues all have multiplicity. Then what?

BiP

4. Aug 7, 2013

### HallsofIvy

Staff Emeritus
If the n by n matrix, V, is diagonalizable then we have n independent eigenvectors. If a given eigenvalue has multiplicity i then we can find i independent vectors that span its "eigenspace". The whole point of "diagonal" is that there exist a basis such that all vectors in the basis are eigenvectors.