My apologies if this is a silly question. I feel like I'm misunderstanding something basic and just need a quick clarification.
I have an eigenvalue problem stated in terms of the inner product. More specifically, I have an operator $O: X \to Y$, where $X$ and $Y$ are Hilbert spaces. I would like to find the largest $n$ eigenvalues $\lambda_i \in \mathbb{R}^+$ and corresponding $n$ eigenfunctions $\varphi_i \in X$ that satisfy $\langle O \varphi_i, Ov \rangle_Y = \lambda_i \langle \varphi_i, v \rangle_X$ for all $v \in X$. The $O \varphi_i$ for $i = 1,...n$ will span the space that I want to find.
I need to do this using the inner product matrices of $X$ and $Y$ (e.g for $X$ this means some matrix $I_X$ such that $\langle u, v \rangle_X = u^T I_X v$). My attempt so far is as follows
I think I can represent $\langle O \varphi_i, Ov \rangle_Y = \lambda_i \langle \varphi_i, v \rangle_X$ as
$M_{sol}^T I_Y M_{sol} = \lambda_i M_{input}^T I_X M_{input}$
where $M_{input}$ is a matrix whose columns are the basis functions in $X$ (basically the identity matrix) and $M_{sol}$ is a matrix whose $i^{th}$ column is a vector in $Y$ representing the value of the operator applied to the $i^{th}$ basis function, ie. $O(M_{input}(:,i))$.
So would this mean that I could find my eigenvalues by finding the eigenvalues of $M_{sol}^T I_Y M_{sol}$?
I feel weird about this because I feel I should be using $M_{input}^T I_X M_{input}$ somewhere in the calculations but I don't see how to connect it.
Any input is greatly appreciated!
I think I now partially understand your question. Let me write it down in my words, and we'll see if that helps you.
It's possible that I have made a mistake somewhere, I didn't know the problem before.
Let $X$ and $Y$ be real Hilbert spaces and $O:X\rightarrow Y$ be a (continuous) linear operator. It should work similarly with complex Hilbert spaces, but you tagged real analysis, so maybe that's all you need.
You're looking for $\varphi\in X$ and $\lambda\in\mathbb{R}^+$ such that $$\langle O \varphi, Ov \rangle_Y = \lambda \langle \varphi, v \rangle_X$$ for all $v\in X$, and you're wondering whether this can be regarded as finding the eigenvalues and eigenvectors of some matrix or operator.
High-level version
Let us introduce the adjoint operator $O^*:Y\rightarrow X$ like here. Then our condition becomes $$\left\langle O^*O\varphi,v\right\rangle_X=\lambda\langle\varphi,v\rangle_X$$ for all $v\in V$, or in other words $$\left\langle O^*O\varphi,\cdot\right\rangle_X=\langle\lambda\varphi,\cdot\rangle_X.$$ Since $x\mapsto\langle x,\cdot\rangle_X$ is an isomorphism of $H$ onto the space $\mathcal{L}(X;\mathbb{R})$ of continuous linear functionals on $H$, this is equivalent to $$O^*O\varphi=\lambda\varphi.$$ So the problem is finding the eigenvalues and -vectors of the operator $O^*O$.
Low-level version with columns and matrices
Let us assume that the underlying vector spaces of $X$ and $Y$ are $\mathbb{R}^n$ and $\mathbb{R}^m$ for some $n,m\in\mathbb{N}$, regard the elements of $X$ and $Y$ as columns. Also, define $I_X$ and $I_Y$ as in the question. Also, we can directly interpret $O$ as a matrix. This is probably the same as your $M_{sol}$, while $M_{input}$ is the unit matrix (not only basically).
Then the condition becomes $$(O\varphi)^TI_Y(Ov)=\lambda\varphi^TI_Xv$$ for all $v\in X$, which is equivalent to $$\varphi^TO^TI_YO=\lambda\varphi^TI_X,$$ and therefore, by transposing on both sides, to $$O^TI_YO\varphi=\lambda I_X\varphi,$$ i.e. $$I_X^{-1}O^TI_YO\varphi=\lambda\varphi.$$
So the problem is finding the eigenvalues and -vectors of the matrix $I_X^{-1}O^TI_YO$. This seems to be what you found out, but with the factor $I_X^{-1}$.