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Earlier today I posted about finding eigenvalues by inspection (Eigenvalues of a matrix A by inspection) when the matrix is of the form.

$$ \begin{bmatrix} a&b&b\\ b&a&b\\ b&b&a \end{bmatrix} $$
For example:
$$ \begin{bmatrix} 3&2&2\\ 2&3&2\\ 2&2&3 \end{bmatrix} $$

And I was made aware of an "algorithm" in such cases. Is there such shortcuts for matrices of the form:

$$ \begin{bmatrix} a&b&0\\ b&c&b\\ 0&b&a \end{bmatrix} $$

For example:

$$ \begin{bmatrix} 1&1&0\\ 1&2&1\\ 0&1&1 \end{bmatrix} $$

I might be starting to push my luck with this question because this matrix is slightly more complex than the previous example, but just putting it out there.
Thanks.

Bucephalus
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    The product of a matrix and column vector is a linear combination of the columns of the matrix with coefficients given by the vector. Playing with simple linear combinations of the columns of the matrix can sometimes lead you to some of the eigenvalue/eigenvector pairs (column 1 minus column 3 jumps out at me here). Another fact that can be useful is that the eigenspaces of a real symmetric matrix are orthogonal. – amd Oct 04 '17 at 20:03

1 Answers1

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One of the eigenvalues of $$\begin{bmatrix} a&b&0\\ b&c&b\\ 0&b&a\end{bmatrix}$$ is $a$, but the other two are $$ \frac{ a + c\pm \sqrt{a^2 - 2 a c + 8 b^2 + c^2}}2 $$ which would be a surprise to me if they could be found "by inspection".

In the particular case $a=1$, $b=1$, $c=2$, you obtain $0,1,3$, so there might be a trick for that particular matrix.

Martin Argerami
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  • That's really good @MartinArgerami. I got your answer the old-fashioned way, and I did it with that expression and got the same answer. That's very clever. I would expect that that would be about as close as I can get to "by inspection" for such a matrix. – Bucephalus Oct 04 '17 at 14:52