Eigenvectors and matrix inner product

  • #1
SeM

Main Question or Discussion Point

Hi, I am trying to prove that the eigevalues, elements, eigenfunctions or/and eigenvectors of a matrix A form a Hilbert space. Can one apply the inner product formula :

\begin{equation}
\int x(t)\overline y(t) dt
\end{equation}

on the x and y coordinates of the eigenvectors [x_1,y_1] and [x_2,y_2], x_1 and y_1 and x_2 and y_2:


\begin{equation}
\int x_1\overline y_1 dt
\end{equation}

\begin{equation}
\int x_2\overline y_2 dt
\end{equation}


and by that prove that the inner product of the matrix vectors is complete and therefore forms a Hilbert space L² [a,b] ?

The reason I am asking about this is because I am looking for a way to prove that the matrix elements, its vectors and its solution is/are Hilbert space, L^2[a,b]

The solution is in general form:

\begin{equation}
\psi = \alpha v_1 e^{\lambda_1t}+\alpha v_2 e^{\lambda_2t}
\end{equation}

where v_1 and v_2 are the eigenvectors of the matrix. In Kreyszig Functional Analysis, p 132, he says "In example 2.2-7 the functions were assumed to be real-valued. In certain cases, that restriction can be removed, to consider complex valued functions. These function form a vector space, which becomes an inner product space if we define:

\begin{equation}
\int x(t)\overline y(t) dt
\end{equation}

This gives also the complex norm.

\begin{equation}
\int \big(|x(t)|^2dt \big)^{1/2}
\end{equation}

And then Kreyszig ends with "The completion of the metric space corresponding to the inner product for the complex matrix (which I just gave above) is the real space L^2[a,b] .

I would like to prove that "my" matrix satisfies this condition too. So , because the general solution given above does not explicitly show that it satisfies the inner product, can I use the eigenvectors in the inner product formula to conclude that the matrix eigenvectors form a Hilbert space?
 
Last edited by a moderator:

Answers and Replies

  • #2
13,214
10,111
I think you are confusing various things here. Given a matrix ##A## and an eigenvalue ##\lambda##, such that ##A.\vec{v}=\vec{v}## then ##\vec{v}## spans of course a vector space. For different eigenvalues, you can use the direct product of those eigenspaces. These will have an inner product if the original vector space has. There is nothing to show here.
 
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