First, we need the characteristic polynomial.
Next, we consider the eigenvalue The computer gives us the following:
In the next cell, note that we doubled the eigenvectors in E3
to avoid fractions. To follow the solution in our example, we append 2*E3[0]
, and reduce the resulting matrix. You should find that using the eigenvector corresponding to E3[0]
leads to an inconsistent system. Once you confirm this, replace E3[0]
with E3[1]
and re-run the cell to see that we get an inconsistent system using as well!
The systems and are both inconsistent, but we can salvage the situation by replacing the eigenvector by some linear combination We row-reduce, and look for values of and that give a consistent system.
The rref
command takes things a bit farther than we’d like, so we use the command echelon_form()
instead. Note that if the system is inconsistent.
The jordan_form()
command returns a pair where is the Jordan canonical form of and is an invertible matrix such that You might find that the computer’s answer is not quite the same as ours. This is because the Jordan canonical form is only unique up to permutation of the Jordan blocks. Changing the order of the blocks amounts to changing the order of the columns of which are given by a basis of (generalized eigenvectors).