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How do I find the eigenvalues & eigenvectors of the following matrix?

\begin{bmatrix}a&b&b\\b&a&b\\b&b&a\end{bmatrix}

I feel like there is an easy trick to it. I have tried using trace and determinant but no luck so far.

J. Doe
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    If you subtract $a-b$ in the diagonal, the matrix ends up with $b$ everywhere. Therefore $a-b$ is an eigenvalue. Since the resulting matrix has rank $1$ the eigenspace has dimension $2$. Therefore, this eigenvalue has multiplicity at least $2$. Then take the trace $3a$ and subtract $2(a-b)$, twice the eigenvalue. That gives you the third eigenvalue $a+2b$. – conditionalMethod Nov 16 '19 at 09:54
  • The title does not match the body. – Rodrigo de Azevedo Nov 16 '19 at 10:38
  • @RodrigodeAzevedo Thank you for pointing that out! (was very tired when I wrote the post) I just fixed it. – J. Doe Nov 16 '19 at 22:05
  • @conditionalMethod How does "the matrix ends up with b everywhere" tell us that a-b is an eigenvalue? – J. Doe Nov 17 '19 at 23:57
  • If all rows are the same, then in particular all rows are a linear combination of a single row. The rank is then at most $1$ (exactly one if $b\neq0$). – conditionalMethod Nov 17 '19 at 23:59

2 Answers2

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We have that $(1,1,1)$ is an eigenvector with eigenvalue $a+2b$ we also know that the sum of eigenvalues is equal to Tr$(A)=3a$ and their product is equal to $\det(A)=a^3+2b^3-3ab^2=(a+2b)(a-b)^2$.

Therefore the eigenvalues are: $a+2b$ and $a-b$ and we can easily find the corresponding eigenvectors $(1,1,1),(1,-1,0),(0,1,-1)$.

Also note that $A$ is a circulant matrix.

user
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Compute the eigenvalues via characteristic polynomial. For $x \in \Bbb C$, $$ \det (A - xI) = \begin{vmatrix} a - x & b & b \\ b & a-x & b \\ b & b & a - x \end{vmatrix} = \begin{vmatrix} a + 2b - x & a+ 2b - x & a + 2b - x \\ b & a-x & b \\ b & b & a - x \end{vmatrix} = (a + 2b - x) \begin{vmatrix} 1 & 1 & 1\\ b & a-x & b \\ b & b & a - x \end{vmatrix} = (a + 2b - x) \begin{vmatrix} 1 & 1 & 1\\ 0 & a-b-x & 0 \\ 0 & 0 & a-b-x \end{vmatrix} = (a+2b - x)(a-b - x)^2, $$ hence the eigenvalues are $a+2b, a-b, a-b$. Now solve linear equations $Av = (a+2b)v, A w = (a-b)w$. Easy to see that $$ v = [1\ \ 1\ \ 1] ^\mathsf {T}, w_1 = [1\ \ -1\ \ 0] ^\mathsf {T}, w_2= [1\ \ 0\ \ -1] ^\mathsf {T}. $$

xbh
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