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Let $a$, $b$, $c$ be positive reals such that $b^2 + c^2 < a < 1$. Consider the $3\times 3$ matrix $M = \begin{bmatrix} 1 &b &c \\b & a &0 \\c & 0 & 1 \end{bmatrix}$. Show that all the eigenvalues of $M$ are positive real numbers.

I wanna prove this without using quadratic forms if possible.

Attempt: Since $M$ is a real symmetric matrix all its eigenvalues are real, Let $\lambda_1,\lambda_2,\lambda_3$ be the eigenvalues of $M$, so $\det(M)= \lambda_1 \lambda_2 \lambda_3$ but $\det(M)= a - (b^2 + c^2) + c^2(1-a) > 0$, from here it's clear that at least one of the eigenvalue is positive, but how do I conclude that the other eigenvalues are also positive as well? from $\mathrm{trace}(M)$ I can't conclude that other eigenvalues are positive.

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    $\begingroup$ How about the characteristic polynomial? $\endgroup$ – user403337 Oct 19 '20 at 21:06
  • $\begingroup$ Okay, lemme try $\endgroup$ – lucas Oct 19 '20 at 21:07
  • $\begingroup$ $\chi_M (x) = x^3 -(a+2)x^2 +(1+2a-b^2-c^2)x - (a-b^2-c^2 a) = 0$ from here all I get is the sum of eigen values taken two at the time is positive $\endgroup$ – lucas Oct 19 '20 at 21:18
  • $\begingroup$ Shouldn't it be $2x^3$? $\endgroup$ – user403337 Oct 19 '20 at 21:26
  • $\begingroup$ characteristic polynomial is monic 😁 $\endgroup$ – lucas Oct 19 '20 at 21:33
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1.) the eigenvalues are real since $M$ is real symmetric thus you have $\lambda_1\geq \lambda_2 \geq \lambda_3$

2.) the eigenvalues of $M$ (Cauchy) interlace with those of
$M' = \begin{bmatrix} a & 0\\0 & 1 \end{bmatrix}$
which means
$\lambda_3 \leq a \leq \lambda_2 \leq 1 \leq \lambda_1$
Thus $\lambda_3$ is the only possible non-positive eigenvalue

3.) you already computed
$\lambda_1\cdot \lambda_2 \cdot \lambda_3 = \det(M)= a - (b^2 + c^2) + c^2(1-a) > 0\implies \lambda_3 \gt 0$

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  • $\begingroup$ Wow! Cauchy's eigenvalue interlacing theorem is so cool! didn't know about this before. Thanks $\endgroup$ – lucas Oct 19 '20 at 23:52
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It suffices to show that matrix $M$ is positive definite, which is guaranteed by if all the leading principal minors are positive. Indeed \begin{align*} & M\begin{pmatrix} 1 \\ 1 \end{pmatrix} = 1 > 0, \\ & M\begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} = a - b^2 > c^2 > 0, \\ & \det(M) = a - b^2 - ac^2 > c^2 - ac^2 = c^2(1 - a) > 0. \end{align*} This completes the proof.

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  • $\begingroup$ What are the matrices after $M$? I have never seen that notation. $\endgroup$ – LinAlg Oct 19 '20 at 22:53
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    $\begingroup$ @LinAlg The matrix defines the rows (first row) and columns (second row) from $M$. Some used $M_{1; 1}, M_{1, 2; 1, 2}$ instead. I prefer to this notation as the $M_{1; 1}$ notation may refer to the submatrix instead of determinants. $\endgroup$ – Zhanxiong Oct 19 '20 at 23:40

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