Effects of increasing a matrix's values on the eigenvalue decomposition

Question

Let $A \in \mathbb{R}^{n \times n}$ be a real, symmetric positive semi-definite matrix, and let $A = UVU^T$ be $A$'s eigendecomposition.
Suppose that the matrix $A'$ was obtained by $A$ by making some values of $A$ larger, in a way that $A'$ is still symmetric and positive semi-definite matrix.
Can we say something about $A'$ eigendecomposition in terms of $A$'s eigendecomposition? In particular, does $A'$ eigenvalues necessarily larger?

Answer 1

A very simple example for some n x n matrix

consider

$B(\tau) = \tau \cdot2\mathbf{11}^T + (1-\tau) \cdot I$
for $\tau \in [0,1]$
at
$B(0)=I $
$B(1)=\mathbf{11}^T $

the dominant eigenvalue (perron root) $\lambda_1$ icreases with $\tau$ and all other eigenvalues decrease as $\tau$ grows. Yet all matrix components increase as $\tau$ increases.

Answer 2

You are asking if the eigenvalues of $(A+\Delta)$ are larger than the values of $A$. If the entries of the symmetric $\Delta$ are positive, this might not be true. We may have $\det(A+\Delta)< \det A$, just increase some off-diagonal elements of a symmetric $2\times 2$ matrix. However, if $\Delta$ itself is positive semi-definite, then Yes.