These examples are carefully chosen to be rotation matrices. Note that if I gave you a random 3-by-3 matrix, it probably wouldn't be a rotation matrix, and isn't guaranteed to have any fixed vectors at all.
We'll analyse some examples of 3-by-3 rotation matrices, and then see to figure out the axis and angle of rotation for a general 3-by-3 rotation matrix.
Let A be the 3-by-3 matrix cos theta, minus sine theta, 0; sine theta, cos theta, 0; 0, 0, 1. This is an example of a 3-by-3 rotation matrix. The top-left 2-by-2 block rotates the x y-plane, and the 1 in the bottom-right tells us that the z-axis is fixed. More precisely: cos theta, minus sine theta, 0; sine theta, cos theta, 0; 0, 0, 1 times the vector x, y, z equals x cos theta minus y sine theta; x sine theta plus y cos theta; z so the vectors x, y, 0 in the x y-plane get rotated by the 2-by-2 rotation matrix cos theta, minus sine theta; sine theta, cos theta, and the vector 0, 0, 1 which points along the z-axis is fixed.
A key point here is that the axis of rotation is fixed, i.e. any vector u pointing in the z-direction satisfies u = A u.
Let B be the 3-by-3 matrix 0, 0, 1; 0, 1, 0; minus 1, 0, 0. This is another example of a 3-by-3 rotation matrix. What is the axis of rotation? We need to find a vector u equals x; y; z such that u = B u. In other words, x, y, z equals 0, 0, 1; 0, 1, 0; minus 1, 0, 0 times x, y, z, which equals z, y, minus x. This is three equations (one for each component): x equals z, y equals y and minus x equals z The equation y = y is trivially satisfied. The other two equations imply x = z = 0. Therefore u equals 0, y, 0, and u must point along the y-axis.
By comparing with Example 1, the angle of rotation can be found by:
taking a vector v which lives in the plane orthogonal to the axis,
applying B to get B v,
computing the angle between v and B v using dot products.
For example, we could take v equals 1, 0 ,0 (as v dot u equals 0 so v is orthogonal to the axis). Then B v equals 0, 0, minus 1, so v dot B v equals 0. If theta is the angle between v and B v then this implies cos theta = 0, so theta is plus or minus 90 degrees degrees.
In fact, we can understand exactly what this rotation is doing by drawing the images of the basis vectors e_1, e_2, e_3 under B (i.e. the three columns of B). The vector e_2 is fixed, e_3 goes to e_1 and e_1 goes to minus e_3, so this rotates by 90 degrees about the y-axis, sending the positive z-axis to the positive x-axis.
Take C to be the 3-by-3 matrix 0, 0, 1; 1, 0, 0; 0, 1, 0. This is another 3-by-3 rotation matrix; we'll find the axis and angle of rotation.
To find the axis u equals x, y, z, we need to solve u = C u: x, y, z equals 0, 0, 1; 1, 0, 0; 0, 1, 0 times x, y, z, which equals z, x, y The first two equations imply x = y = z, so the third is redundant, and any vector of the form u equals x, x, x is fixed. In other words, the axis points in the 1, 1, 1-direction.
To find the angle, pick a vector v orthogonal to u. For example, v equals 1, minus 1, 0 satisfies u dot v = 0 so is orthogonal to u. We compute v dot C v equals 1, minus 1, 0 dot 0, 1, minus 1, which equals minus 1. We also know that if theta is the angle between v and C v then minus 1 equals v dot C v equals norm v norm C v cos theta. Since norm v equals norm C v equals root 2, we get cos theta equals minus a half. This tells us that theta is 2 pi over 3 (or any of the other values that have cos theta equals minus a half). Let's draw a picture to convince ourselves it's really 2 pi over 3 (120 degrees).
The axis of rotation points out of the screen: