What's the spectrum of this element

Question

I am reading that

$$ x = {1 \over 2}\left ( \begin{array}{cc}3 & 2 \\ 2 & 1 \end{array}\right )$$

is not positive since it has a negative eigenvalue.

I think that $x$ is positive because it is self-adjoint and the only two $\lambda \in \mathbb C$ for which $x - \lambda I$ is not invertible are $-3/2$ and $-1/2$ hence $\sigma (x) = \{-3/2, -1/2\}$. What am I missing?

Edit The spectrum is defined as follows:

$$ \sigma (x) = \{\lambda \in \mathbb C \mid x - \lambda I \text{ is not invertible }\}$$

I made a sign error in my previous calculation. If $\lambda = -1/2$ then the first row of $x -\lambda I$ becomes $(2,2)$ so in this case $x - \lambda I$ is not invertible therefore $-1/2 \in \sigma (x)$. Similarly for $-3/2$ and the second row.

Second edit Now that I corrected the sign error I no longer believe that the spectrum is non-negative. So my question boils down to:

I have shown that $ \{-3/2, -1/2\} \subseteq \sigma (x)$. Does $\sigma (x)$ contain any other values?

Answer 1

Note that the product of the eigenvalues is $\det(x) = -\frac 14$. Clearly, one of them is negative.

The spectrum consists of the roots of the characteristic polynomial $$ x^2 - 2x - \frac 14 $$ given by $$ x = 1 \pm \frac{\sqrt 5}{2} $$

Answer 2

Simple method: the matrix has determinant $-\frac14$. The determinant is the product of the eigenvalues so one of them must be negative and the other positive.