Eigenvector Rotation in 2x2

What you are seeing: a 2x2 matrix $M = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$ acting on the unit circle. The blue ellipse is the image of the unit circle under $M$ (semi-axes equal to the singular values). The accent arrows are the eigenvectors of $M$ , scaled by their eigenvalues. The eigenvalue formula $\lambda_\pm = (a+d)/2 \pm \sqrt{(a+d)^2/4 - (ad - bc)}$ has real solutions when the discriminant is non-negative.

Drag the sliders for $a, b, c, d$ . With $a = d$ and $b = -c$ you get a rotation; the eigenvalues become complex and the eigenvector arrows disappear (the readout flags the complex spectrum). With $b = c$ (symmetric $M$ ) the eigenvectors are exactly orthogonal. With $b = c = 0$ the eigenvectors stay axis-aligned regardless of $a, d$ .

Figure 1. 2x2 eigendecomposition: unit circle, its image, and eigenvectors. Method: closed-form quadratic for eigenvalues; null-space row pick for eigenvectors.

a2.00

b1.00

c1.00

d3.00

WHAT TO TRY

Watch the sweeping arrow: its image (warm) usually points a different way (it turns). At the eigenvector directions the two arrows line up, the turn drops to zero in the plot below.
Set $b = -c$ (and $a = d$) for a rotation: every direction turns, the turn curve never touches zero, and there are no real eigenvectors.
Set $b = c$ (symmetric $M$) and the two eigenvector directions come out exactly at right angles.