eigenvalues of "diagonal matrix $+$ rank one matrix"

Question

I am confused about the following claim:

Let $v\in \mathbb{R}^n$ such that

$$v = \begin{cases} 1 & i \text{-th entry}\\ -1 & j \text{-th entry} \\ 0 & \text{otherwise} \end{cases}$$

Let $\Lambda\in \mathbb{R}^{n\times n}$ be a diagonal matrix.

We know that $vv^T$ is rank one, so only one nonzero eigenvalue (actually it is $2$ in my case.)

Will $\Lambda + vv^T$ only increase one eigenvalue of $\Lambda$?

I do not believe this is correct and it can be checked by Matlab.

Is there any extra conditions or related theorem for changing only one eigenvalue of a diagonal matrix by adding a specific matrix (such as rank one)?

Answer 1

As I state in my comment, Weyl's inequalities tell you everything you can say using only the length of $v$.

We can clearly see through an example that, in general, more than one eigenvalue will change. For example: the matrix $$ \Lambda = \pmatrix{1&0\\0&0} $$ has eigenvalues $1$ and $0$. However, $$ \Lambda + vv^T = \pmatrix{2 & -1\\-1 & 1} $$ has eigenvalues $\frac{3 \pm \sqrt{5}}{2}$, which means that both eigenvalues have changed.

It is notable, however, that for your friend's choice of $v$, only the $i$th and $j$th eigenvalues from the diagonal might change. If $v$ is an eigenvector of $\Lambda$ and $\Lambda$ is symmetric, then $\Lambda + vv^T$ will have the same eigenvectors as $\Lambda$ and only the eigenvalue corresponding to $v$ will change.