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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?

user557
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  • The rank of $B$ is $1,$ and the vector of all ones is an eigenvector. – saulspatz Jul 16 '18 at 23:57
  • @saulspatz So this means that there is an eigenvalue $n$. But what about multiplicities? What does the statement about the rank tell us? – user557 Jul 17 '18 at 00:02
  • If the rank is $1,$ the kernel has dimension $n-1,$ and the kernel is the eigenspace corresponding to the eigenvalue $0$. – saulspatz Jul 17 '18 at 00:13
  • Related: https://math.stackexchange.com/questions/904926/determinant-of-a-rank-1-update-of-a-scalar-matrix-or-characteristic-polynomia – user557 Jul 17 '18 at 00:59

2 Answers2

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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 $n-1$.

Donald Splutterwit
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  • I used your idea on $B$. Every vector from the list $e_1-e_n,\ e_2-e_n,\dots, e_{n-1}-e_n$ is an eigenvector of $B$ with eigenvalue $0$. Thus $B$ has $n-1$ eigenvalues $0$ and $1$ eigenvalue $n$. Then the characteristic polynomial of $B$ is $x^{n-1}(x-n)$ and thus the characteristic polynomial of $M$ is $(x+b-1)^{n-1}(x-n+b-a)$, so the eigenvalues are $1-b$ (of multiplicity $n-1$) and $n-b+a$ (of multiplicity $1$). Does this look correct? – user557 Jul 17 '18 at 00:17
  • I think the characteristic equation is $(x-(a-b))^{n-1} (x-(a+(n-1)b))$. – Donald Splutterwit Jul 17 '18 at 00:24
  • I spotted a few mistakes in my previous comment. Without them I get $(x-(a-b))^{n-1}(x-(nb+a-b))$, which agrees with your answer. The eigenvalue corresponding to $e_1+\dots+e_n$ is $bn$, not $n$ as I thought. – user557 Jul 17 '18 at 00:29
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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 $n-1$).

copper.hat
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