Direct sum of the eigenspaces equals V?

[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

 

[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).

 

[Bipolarity]

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

BiP

 

[HallsofIvy]

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.

 

[blue_raver22]

olarBear,

Yes, it is true that the direct sum of the eigenspaces of a linear operator $T:V \to V$ equals $V$. This is known as the diagonalizability theorem, and it states that if $T$ has $n$ distinct eigenvalues $\lambda_1, \lambda_2, ..., \lambda_n$, then $V$ can be decomposed as the direct sum of the corresponding eigenspaces, i.e. $V = E_{\lambda_1} \oplus E_{\lambda_2} \oplus ... \oplus E_{\lambda_n}$.

Furthermore, if $T$ is diagonalizable, then it means that there exists a basis of $V$ consisting of eigenvectors of $T$. This is a very useful property, as it allows us to easily compute the action of $T$ on any vector in $V$, by simply multiplying it by the corresponding eigenvalue. This simplifies many calculations and makes working with $T$ much easier.

I understand that the proofs in Friedberg's text may seem to skip some steps, but it is important to remember that the proofs are meant to be guides and not necessarily exhaustive. It is always helpful to take the time to work through the proofs and understand each step in detail. This will not only help you understand the theorems better, but also improve your overall understanding of linear algebra.

I hope this helps you in your understanding of diagonalizability. Keep up the good work in your studies!