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Given $x>0$ and \begin{bmatrix} 1 & 1+x & 1\\ 1+x & 1 & 1+x\\ 1 & 1+x & 1 \end{bmatrix} Show that there is exactly one positive and one negative eigenvalue.

I solved this problem. This is easy to do by simply computing the characteristic polynomial, $0$ is an eigenvalue and two others of different signs. Now, I want to generalize this to $n \times n$ case ($x>0$).

Note that the $n \times n$ analogue will have rank $2$ and hence exactly two non-zero eigenvalues both of which are real because the matrix is symmetric. Now also the trace is $n$ so there is a positive eigenvalue. Now I want to show that there is a negative eigenvalue. Computing the characteristic polynomial will be clumsy now, even though it would finally look like $t^{n-2}(t^2 +at +b)$.

If both the other roots are positive then we have that the spectral radius is strictly less than $4$. Can we use this?

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    $\begingroup$ What do you mean as the generalized matrix ? Is it a "checkerboard" matrix with $a_{i,j}=1$ or $1+x$ according to the parity of $i+j$ ? $\endgroup$ Commented Jul 21, 2022 at 9:37
  • $\begingroup$ @JeanMarie yes, exactly $\endgroup$ Commented Jul 21, 2022 at 12:00

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Without $x>0$ it's not true, for instance if $x=-1$ the eigenvalues are $0, 1, 2$. Also, if $x$ is zero, then the only nonzero eigenvalue is always $n$.

If $n$ is even, then you can easily guess the non-zero eigenvectors $(1,1,\ldots, 1)$ and $(1, -1, \ldots, 1, -1)$, with eigenvalues $\frac{n}{2}(2 + x)$ and $\frac{n}{2}(-x)$. One is positive and one negative for $x>0$.

More generally (also including the odd case $n>1$), you can easily prove that for symmetric matrices $A$ and any vector $v$ it holds that $$ \|Av\| \leq \rho(A)\|v\| $$ see the last line of this paragraph. Apply this to a vector $(1,1,\ldots, 1)$ and you get that the spectral radius is at least $n + \lfloor n/2 \rfloor x > n$, so the other eigenvalue has to be negative.

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  • $\begingroup$ Sorry I wanted to put condition that $x$ is positive $\endgroup$ Commented Jul 21, 2022 at 12:02
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in the $n\times n$ case you don't even need symmetry since the matrix under consideration is positive in the Perron Theory sense. Key idea is to compare sum along diagonal vs minimum row sum.

$\text{rank}(A) = 2\implies n=\text{trace}\big(A\big)= \lambda_1 + \lambda_2 + \sum_{k=3}^n \lambda_k =\lambda_1 + \lambda_2$
(geometric multiplicity of $n-2$ is a lower bound on algebraic multiplicity for eigenvalue 0).

For a crude bound on the minimum row sum, you have
$n + x \leq \text{min row sum}$

But by Perron Theory, the Perron Root $\lambda_1$ satisfies
$\text{min row sum} \leq \lambda_1 \leq \text{max row sum}$
and these inequalities are strict unless the minimum row sum = maximum row sum. Thus
$\implies n=\text{trace}\big(A\big)=\lambda_1 + \lambda_2\gt (n+x) + \lambda_2\implies -x\gt \lambda_2$
i.e. $\lambda_2$ is negative

note: the above technique is flexible enough to tell you e.g. that if you left multiply your original $3\times 3$ matrix by a permutation matrix that swaps rows 1 and 2, then the result is a matrix of rank 2 that has 2 positive eigenvalues.

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Since your matrix is symmetric and it contains an indefinite principal $2\times2$ submatrix $\pmatrix{1&1+x\\ 1+x&1}$, it must have some positive eigenvalues and some negative eigenvalues. As the matrix has rank two, the numbers of positive or negative eigenvalues must each be equal to one.

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