On the sign of a minimum eigenvalue I have a square matrix with dimension 2 which is symmetric, and all elements are positive.
I am interested in the eigenvalue of such a matrix, particularly the sign of the minimum eigenvalue.
I repeated the calculation of eigenvalues for matrices carrying the above characteristics with different values, and found that there may be case where I get a negative sign for the minimum eigenvalue, e.g., for the below matrix.
          [,1]      [,2]
[1,] 0.5828461 0.8575778
[2,] 0.8575778 0.8914404

How can I intuitively explain the case of the minimum eigenvalue as negative, as people at first glance would expect to get all positive eigenvalues, since all elements of the underlying matrix are strictly positive. Particularly, if there could be any other conditions for elements of such matrix that would result in the minimum eigenvalue as negative.
 A: Well, the algebraic way to look at this is to recall that if the eigenvalues are $\lambda_1$ and $\lambda_2$, then $\lambda_1 \lambda_2 = \det(M)$. And your matrix has a negative determinant, so we know that we have one negative and one positive eigenvalue. And in this simple 2-dimensional case, having a negative determinant is equivalent to your positive-entry matrix having a negative eigenvalue - if you'll let me ignore the degenerate $\det = 0$ case.
[They aren't equivalent for a matrix with arbitrary values, as one could have two negative eigenvalues, and thus a positive determinant. But we can use the other very useful fact that $\lambda_1 + \lambda_2 = trace(M)$, and because all our entries are positive our trace is positive, so the possibility of two negative eigenvalues is ruled out.]
But here's a more geometric, hand-wavey/intuitive way to look at it. Remember, the columns of your matrix are the images of the vectors $(1,0)$ and $(0,1)$. Now, let's think of this transformation taking place in two steps: first, this particular matrix takes $\hat x $ to $(0.58, 0.85)$, i.e. it takes the $x$-axis to the line $y = 1.47x$, which is rather more vertical than horizontal. Next, it takes $\hat y$ to $(0.85, 0.89)$, i.e. it takes the $y$-axis roughly to the line $y=x$. And to do this, it has to take the $y$-axis below* the image of the $x$-axis. When I see a negative eigenvalue, I expect there to be some sort of flipping of direction, and this "interchanging" of the $x$- and $y$- axes is what causes that here.
And we can even tie this geometric view back to the algebraic view! For your positive-entry matrix
$$
\det\left(
\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}
\right) < 0
$$
is equivalent (with a bit of algebra) to
$$ \frac{d}{b} \lt \frac{c}{a}$$
which says that the slope of the image of the $y$-axis is smaller than the slope of the image of the $x$-axis, i.e. that our transformation moves the $y$-axis below where it moves the $x$-axis.

*I said "below" because I'm really only thinking about what happens in the first quadrant. Pedantically, in the third quadrant the $y$-axis get moved above the image of the $x$-axis, but I hope this description is good enough for an intuitive answer.
