When do Eigenvalues of $Q^TAQ$ and $A$ coincide? Lets say I have some $n\times n$ matrix  $A$ and a $n\times l$ matrix $Q$, whose columns for a basis for some subspace $\mathcal{L} \subset \mathbb{R^n}$. My intuition tells me that I would need $Q$ to be a full orthogonal basis of $\mathbb{R^n}$ (i.e. $l=n$), such that all eigenvalues $\lambda_i$ of  $Q^TAQ$ and $A$ coincide.
However, I don't see where in the proof for the latter assertion the same dimensionality is needed:


*

*Be $v\in\mathbb{R}^n\setminus\{0\}$ an eigenvector of A, i.e. $Av=\lambda v$. Then $\lambda$ is an eigenvalue of $Q^TAQ$ with eigenvector $Q^Tv$:
$$
(Q^TAQ)(Q^Tv)=Q'A(QQ^T)v=Q^TAv=\lambda(Q^Tv).
$$

*Be $v\in\mathcal{L}\setminus\{0\}$ an eigenvector of (Q^TAQ), i.e. $(Q^TAQ)v=\lambda v$. Then $\lambda$ is an eigenvalue of $A$ with eigenvector $Qv$:
$$
A(Qv)=IAQv=QQ^TAQv=Q(Q^TAQv)=\lambda(Qv).
$$


Maybe you can help me out here. Thanks a lot in advance!

In case Q forms an orthonormal basis of $\mathcal{L}$, you have $Q^TQ = I$, but as long as $l<n$, $QQ^T \neq I$.  Notably, $QQ^T$ is the orthogonal projection onto the subspace spanned by the columns of $Q$.
