Is it true that the eigenvalues of $A + B$ are the sum of some eigenvalue of $A$ and some eigenvalue of $B$?

Question

Is it true that the eigenvalues of $A + B$ are the sum of some eigenvalue of $A$ and some eigenvalue of $B$?

I'm taking a linear algebra class, and I recently learned about eigenvalues. I think that this claim is true, but it is not a theorem that I can find anywhere. I've tried it for many examples, and I just wanted someone to confirm.

Thanks

Answer 1

Look at $$A=\begin{pmatrix} 0&0\\1&0\end{pmatrix}+B=\begin{pmatrix} 0&-1\\1&0\end{pmatrix}=\begin{pmatrix} 0&-1\\2&0\end{pmatrix}$$

The eigenvalues of $A$ are zero while $B$ are $\pm i$ and the eigenvalues of their sum matrix are $\pm i\sqrt{2}$. But $$0 \pm i \neq \pm i\sqrt{2}$$

Answer 2

Take

$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix},~B=\begin{pmatrix}1&0\\1&-1\end{pmatrix}$$

The eigenvalues are given by $\lambda_{1/2}=1$ and $\lambda'_{1/2}=\pm1$. The sum is of those is

$$C=A+B=\begin{pmatrix}2&1\\1&0\end{pmatrix}$$

The corresponding eigenvalues are $\mu_{1/2}=1\pm\sqrt 2$ which cannot be interpreted as sum of the aforementioned eigenvalues.

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I think the reason why you were not able to find this conjectured theorem within the literature is that it fails in general. As shown by finding a single counterexample we have to admit that it does not hold for all matrices, but it may be true for certain ones.

Answer 3

It is not hard to find a counterexample.

The Eigenvalues of

$$\begin{pmatrix}1&1\\0&2\end{pmatrix}$$ and \begin{pmatrix}3&0\\1&4\end{pmatrix} are found on the diagonals and are integer.

The Eigenvalues of

$$\begin{pmatrix}4&1\\1&6\end{pmatrix}$$

are the roots of $(4-\lambda)(6-\lambda)-1$, which are irrational (the discriminant is not a perfect square).

Answer 4

As others have already shown by providing counterexamples, the result you mentioned doesn't generally hold. Yet, if $A$ and $B$ commute then it is true.