How many negative eigenvalues can a $3 \times 3$ symmetric matrix have? Suppose that I have a symmetric matrix of the form
\begin{equation} 
M= \begin{pmatrix}
1 & B & C \\
B & 1 & E \\ 
C & E & 1\\ 
\end{pmatrix},
\end{equation}
where $B, C, E$, all belong to $[-1, 1]$. What is the most number of negative eigenvalues that this matrix can have? The trace of the matrix is $3$, so it cannot have more than $2$ negative eigenvalues.
 A: It is impossible to have 2 negative eigenvalues under these conditions.  By Gershgorin's bound we have all eigenvalues $\lambda\in[-1,3]$, so it is impossible to have 2 negative and a positive eigenvalues that sum to 3.
But one negative eigenvalue is possible, such as
$$
\begin{pmatrix}
1&0&\cos\theta\\
0&1&\sin\phi\\
\cos\theta&\sin\phi&1\\
\end{pmatrix}
$$
with determinant $\frac12(\cos 2\theta-\cos 2\phi)$ which you can easily make negative.
A: You can also do this directly with Cauchy Eigenvalue Interlacing.  Consider
$\begin{equation} 
M'= \begin{bmatrix}
1 & B  \\
B & 1\\
\end{bmatrix},
\end{equation}$
$\text{trace}\big(M'\big) = 2$
$\text{det}\big(M'\big)= 1 - B^2 \geq 0$
hence $M$ has eigenvalues $\sigma_1 \geq \sigma_2 \geq 0$
And the eigenvalues of $M$ and $M'$ interlace so we have
$\lambda_1 \geq \sigma_1 \geq \lambda_2 \geq \sigma_2 \geq \lambda_3\implies\lambda_1 \geq \lambda_2 \geq \sigma_2\geq 0$
So there is at most 1 negative eigenvalue in $M$
A: Here's a strange application of the formula for the roots of a cubic equation.  For a degree-3 polynomial in depressed form
$$
p(x) = x^3 + px +q,
$$
there is a formula for the roots, given by
$$
r_k = 2\sqrt{-\frac{p}{3}}\cos\left(
\frac{1}{3}\arccos\left(
\frac{3q}{2p}\sqrt{-\frac{3}{p}}
\right)
-\frac{2\pi k}{3}
\right),~~~~~k=0,1,2.
$$
We'll apply this to the characteristic polynomial of the matrix, given by
$$p(x) = (x - 1)^3 + (-b^2-c^2-e^2)(x - 1) - 2 e b c.$$
Then, the expression for the roots is
$$
r_k = 1 + 2\sqrt{\frac{b^2+e^2+c^2}{3}}\cos\left(
\frac{1}{3}\arccos\left(
{\frac{3\sqrt{3}ebc}{(b^2+e^2+c^2)^{3/2}}}
\right)
-\frac{2\pi k}{3}
\right),~~~~~k=0,1,2.
$$
The smallest this expression can be is when the $\cos$ evaluates to $-1$, and since the off-diagonal elements of the matrix are restricted to be between $-1$ and 1, the largest that the factor multiplying the $\cos$ can be is $2$, i.e.,
$$
2\sqrt{\frac{b^2+e^2+c^2}{3}}\leq2\sqrt{\frac{1^2+1^2+1^2}{3}}=2.
$$
Thus, the smallest a root can be is $-1$. Similarly the largest root is at most 3, and so, as mentioned in another answer, this means that there can be at most one negative eigenvalue (since the trace is 3).
From this formula, we can see that if all the off-diagonals are equal to $-1$, then the $k=2$ root is $r_2=-1$, with the other two roots both being $2$.
