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.
