Find the eigenvalues and their multiplicities of a special matrix

Question

Find the eigenvalues (with multiplicities) of the matrix $M=M_{a,b}\in Mat_n(\mathbb R)$ that has $a$'s on the main diagonal and $b$'s elsewhere.

I tried to adapt the great method suggested by @Lord Shark the Unknown in this answer.

For simplicity first assume $a < b$. Then $M=B-(b-a)I$, where $B$ is the matrix with $b$'s everywhere. We have $$\det(tI-M)=\det(tI-B+(b-a)I)=\det([t+b-a]I-B).$$ Thus it suffices to find the eigenvalues with multiplicities of $B$. The product of eigenvalues is $0$, the sum is $nb$. But the only thing I can conclude from this is that there is the eigenvalue $0$ of unknown multiplicity. How to find the other eigenvalues and their multiplicities?

Answer 1

Think about the possible eigenvectors.

You can have an eigenvector with all the entries are $1$, giving an eigenvalue of $a+(n-1)b$. (with multiplicity of $1$)

You can have eigenvectors with one entry $1$ , one entry $-1$ and all other entries are zero, this gives an eigenvalue of $a-b$ and this eigenspace has multiplicity $n-1$.

Answer 2

Note that $M= be e^T + (a-b)I$.

The $(a-b)I$ just shifts the eigenvalues of $b ee^T$ by $a-b$.

Note that $be e^T$ is symmetric so it has a full set of eigenvectors.

Note that $be e^T e = nb e$ and if $v \in \{e\}^\bot$ we have $be^T v = 0$.

Hence $be e^T$ has eigenvalues $0$ (multiplicity 1) and $nb$ (multiplicity $n-1$).

Hence $M$ has eigenvalues $a-b$ (multiplicity 1) and $a+(n-1))b$ (multiplicity $n-1$).