# eigendecomposition of large dimensional matrix from eigendecomposition of projected matrix

Let $$A$$ be an $$n\times n$$ diagonalizable matrix. Consider an $$n \times r$$ matrix $$U$$ with orthonormal columns, i.e. $$U^TU = I$$ with $$n > r$$.

Let us construct the projected matrix $$\tilde{A} = U^T A U \in \mathbb{R}^{r \times r}.$$ If I know the eigenvalues and eigenvectors of $$\tilde{A}$$, how can I find the eigenvalues and eigenvectors of $$A$$?

I read a page in Wikipedia that say if $$y$$ is an eigenvector of $$\tilde{A}$$, then $$Uy$$ is an eigenvector of $$A$$ with the same eigenvalue $$\lambda$$. However, I do not see why $$AUy = \lambda Uy$$. If this is true, then certainly by multiplying both sides by $$U^T$$, we get $$\tilde{A}y = \lambda y$$.

But I want to show the other direction, i.e. if $$\tilde{A}y = \lambda y$$ then $$AUy = \lambda Uy$$.

Suppose $$n=2m$$ and consider the example of
$$U:=\begin{bmatrix}\mathbf I_r \\ \mathbf 0\end{bmatrix}$$
$$A := \begin{bmatrix}\mathbf 0 &B \\ B^T &\mathbf 0\end{bmatrix}$$
for invertible $$B$$.
Then $$\tilde{A}= U^TAU = \mathbf 0$$
so every non-zero vector of $$\tilde{A}$$ is an eigenvector with eigenvalue 0,
but the choice of $$B\in GL_m\big(\mathbb R\big)$$ was arbitrary and in general there is no way that every non-zero vector is an eigenvector of $$A$$.
You may want to specialize to real symmetric $$A$$ and look into Cauchy Eigenvalue Interlacing.