0
$\begingroup$

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 $AUy = \lambda Uy$.

$\endgroup$

1 Answer 1

1
$\begingroup$

This can't be right.
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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .