Imagine a linear transformation A of stretching all vectors by a factor of 2 in the three-dimensional space. Any vector v would be transformed into 2v. Therefore, for this transformation all vectors are eigenvectors with eigenvalue of 2.
Consider a rotation of a three-dimensional space around Z-axis by an angle of 90^o. Obviously, all vectors except those along the Z-axis will change the direction and, therefore, cannot be eigenvectors. But those vectors along Z-axis (their coordinates are of the form [0,0,z]) will retain their direction and length, therefore they are eigenvectors with eigenvalue of 1.
Finally, consider a rotation by 180^o in a three-dimensional space around Z-axis. As before, all vectors long Z-axis will not change, so they are eigenvectors with eigenvalue of 1.
In addition, all vectors in the XY-plane (their coordinates are of the form [x,y,0]) will change the direction to opposite, while retaining the length. Therefore, they are also eigenvectors with eigenvalues of -1.
Any linear transformation of a vector space can be expressed as multiplication of a vector by a matrix. For instance, the first example of stretching is described as multiplication by a matrix A
| 2 | 0 | 0 |
| 0 | 2 | 0 |
| 0 | 0 | 2 |
Such a matrix, multiplied by any vector v={x,y,z} will produce A*v={2x,2y,2z}
This is obviously equals to 2*v. So, we have
A*v = 2*v,
which proves that any vector v is an eigenvector with an eigenvalue 2.
The second example (rotation by 90^o around Z-axis) can be described as multiplication by a matrix A
| 0 | -1 | 0 |
| 1 | 0 | 0 |
| 0 | 0 | 1 |
Such a matrix, multiplied by any vector v={x,y,z} will produce A*v={-y,x,z},
which can have the same direction as original vector v={x,y,z} only if x=y=0, that is if original vector is directed along the Z-axis.
