let be $A$ a symmetric matrix, which has $k$ ones and $n-k$ zeros in every row and for any two rows there is exactly one column where they both have an one. The main diagonal consists of zeros (only). ($A$ is supposed to be an adjacency matrix of a $k$-regular simple graph on $n$ vertices, $k>2$.) Suppose that $n=k^2-k+1$ holds. Then
The higlighted parts are problematic for me.
If I have matrix $X$ with eigenvalue $x$ and matrix $Y$ with eigenvalue $y$, then $X+Y$ has eigenvalue $x+y$? If so, why? And what mupltiplicity does $x+y$ have?
The orange part is almost clear, perhaps. But for sure -- $A$ is similar with a diagonal matrix $B$, which has eigenvalues of $A$ on its diagonal. Then $B=C\cdot C$, where $C$ is diagonal as well and the only possible way is, that $c_{ii}=\pm\sqrt{b_{ii}}$. Hence $A$ has eigenvalue $\pm\sqrt{k-1+n}=\pm k$.
Why is it $+k$ and not $-k$ and why does it have multiplicity 1.
Thanks anybody for helps.
For the yellow: the identity matrix is special here. If $\lambda$ is an eigenvalue of $M$, then $M + \alpha I$ has $\lambda + \alpha$ as an eigenvalue.
For the orange: $A^2$ has eigenvalue $k - 1 + n$, which is to say $k^2$ (using the supposed equation), with multiplicity $1$. $A$ must have an eigenvalue $\pm \sqrt{k^2} = \pm k$, with matching multiplicity. So, you're right: they should have said $\pm k$. The multiplicity is $1$ to match that of $k^2$.