# Show that all the eigenvalues of $M$ are positive real numbers.

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.

• How about the characteristic polynomial? – user403337 Oct 19 '20 at 21:06
• Okay, lemme try – lucas Oct 19 '20 at 21:07
• $\chi_M (x) = x^3 -(a+2)x^2 +(1+2a-b^2-c^2)x - (a-b^2-c^2 a) = 0$ from here all I get is the sum of eigen values taken two at the time is positive – lucas Oct 19 '20 at 21:18
• Shouldn't it be $2x^3$? – user403337 Oct 19 '20 at 21:26
• characteristic polynomial is monic 😁 – lucas Oct 19 '20 at 21:33

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

$$\lambda_1\cdot \lambda_2 \cdot \lambda_3 = \det(M)= a - (b^2 + c^2) + c^2(1-a) > 0\implies \lambda_3 \gt 0$$
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.
• What are the matrices after $M$? I have never seen that notation. – LinAlg Oct 19 '20 at 22:53
• @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. – Zhanxiong Oct 19 '20 at 23:40