Using Multiple Eigenvectors for the Same Eigenvalue

[Fermat1]

I am trying to prove the spectral decomposition theorem for normal compact operators. Now, my book says the space H is the closure of the direct sum of $F_{t}$ where we the $F_{t}$ are eigenspaces and we sum over all eigenvalues $t$.
My question concerns what happens when there are 2 linearly independent eigenvectors associated to one eigenvalue. The above says we can write any $x$ in H as

$x=a_{1}e_{1}+a_{2}e_{2}+...$ where the $a_{i}$ are scalars and the $e_{i}$ are the eigenvectors, one from each eigenspace. So this tells me that 2 eigenvectors associated to the same eigenvalue cannot be used in the same linear combination. BUt I know that's not right.

 

[jvicens]

Thank you for your question regarding the spectral decomposition theorem for normal compact operators. I can understand your confusion about the statement in your book that eigenvectors associated to the same eigenvalue cannot be used in the same linear combination.

First, let me clarify that the spectral decomposition theorem states that any normal compact operator on a Hilbert space can be decomposed into a sum of projections onto its eigenspaces. This means that for each eigenvalue, there exists an eigenspace associated with it, and any vector in the Hilbert space can be written as a linear combination of eigenvectors from different eigenspaces.

Now, going back to your question about using two eigenvectors associated to the same eigenvalue in the same linear combination - this is actually possible. The key here is to remember that eigenvectors associated to the same eigenvalue are not necessarily linearly independent. In fact, if there are two linearly independent eigenvectors associated to the same eigenvalue, then they can be used in the same linear combination to represent any vector in the Hilbert space.

To see this, let's take a simple example. Consider a 2x2 matrix with eigenvalue $\lambda$ and two linearly independent eigenvectors $e_1$ and $e_2$. Then, any vector $x$ in the Hilbert space can be written as $x = ae_1 + be_2$, where $a$ and $b$ are scalars. Since both $e_1$ and $e_2$ are eigenvectors associated to the same eigenvalue $\lambda$, we can write this as $x = (a+b)e_1 = (a+b)e_2$, which shows that we can indeed use both eigenvectors in the same linear combination.

In summary, the statement in your book is correct, but it is important to remember that eigenvectors associated to the same eigenvalue may not be linearly independent and can therefore be used in the same linear combination. I hope this helps to clarify your understanding of the spectral decomposition theorem. Good luck with your research!