First, we need the characteristic polynomial.
The characteristic polynomial of \(A\) is given by
\begin{equation*}
c_A(x)=(x-2)^2(x-3)^3\text{.}
\end{equation*}
We thus have two eigenvalues: \(2\text{,}\) of multiplicity \(2\text{,}\) and \(3\text{,}\) of multiplicity \(3\text{.}\) We next find the \(E_2(A)\) eigenspace.
The computer gives us
\begin{equation*}
E_2(A)=\nll(A-2I) = \spn\{\xx_1\}, \text{ where } \xx_1=\bbm -1\\0\\-1\\1\\0\ebm\text{,}
\end{equation*}
so we have only one independent eigenvector, which means that \(G_2(A)=\nll(A-2I)^2\text{.}\)
Following
Insight 5.6.5, we extend
\(\{\xx_1\}\) to a basis of
\(G_2(A)\) by solving the system
\begin{equation*}
(A-2I)\xx = \xx_1\text{.}
\end{equation*}
Using the results above from the computer (or Gaussian elimination), we find a general solution
\begin{equation*}
\xx = \bbm -t\\-1\\-t\\t\\0\ebm = t\bbm -1\\0\\-1\\1\\0\ebm + \bbm 0\\-1\\0\\0\\0\ebm\text{.}
\end{equation*}
Note that our solution is of the form \(\xx = t\xx_1+\xx_2\text{.}\) We set \(t=0\text{,}\) and get \(\xx_2 = \bbm 0\amp -1\amp 0\amp 0\amp 0\ebm^T\text{.}\)
Next, we consider the eigenvalue \(\lambda=3\text{.}\) The computer gives us the following:
Rescaling to remove fractions, we find
\begin{equation*}
E_3(A) = \nll(A-3I) = \spn\{\yy_1,\yy_2\}, \text{ where } \yy_1 = \bbm 1\\-2\\2\\2\\0\ebm, \yy_2 = \bbm -1\\2\\0\\0\\2\ebm\text{.}
\end{equation*}
Again, we’re one eigenvector short of the multiplicity, so we need to consider \(G_3(A)=\nll(A-3I)^3\text{.}\)
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 \(\yy_1\) 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 \(\yy_2\) as well!
The systems \((A-3I)\yy = \yy_1\) and \((A-3I)\yy=\yy_2\) are both inconsistent, but we can salvage the situation by replacing the eigenvector \(\yy_2\) by some linear combination \(\zz_2 = a\yy_1+b\yy_2\text{.}\) We row-reduce, and look for values of \(a\) and \(b\) 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 \(a=b\text{,}\) the system is inconsistent.
We find that \(a=b\) does the job, so we set
\begin{equation*}
\zz_2 = \yy_1+\yy_2 = \bbm 0\\0\\2\\2\\2\ebm\text{.}
\end{equation*}
Solving the system \((A-3I)\zz = \yy_1+\yy_2\text{,}\) using the code above, we find
\begin{align*}
\zz \amp = \bbm \frac12 +\frac12 s-\frac12 t\\1-s+t\\1+s\\s\\t\ebm\\
\amp = \bbm \frac12\\1\\1\\0\\0\ebm + s\bbm\frac12\\-1\\1\\1\\0\ebm+t\bbm -\frac12\\1\\0\\0\\1\ebm\\
\amp = \bbm \frac12\\1\\1\\0\\0\ebm \frac{s}{2}\yy_1+\frac{t}{2}\yy_2\text{.}
\end{align*}
We let \(\zz_3 = \bbm 1 \\ 2 \\ 2 \\ 0 \\ 0\ebm\text{,}\) and check that
\begin{equation*}
A\zz_3 = 3\zz_3+\zz_2\text{,}
\end{equation*}
as required:
This gives us the basis \(B = \{\xx_1,\xx_2,\yy_1,\zz_2,\zz_3\}\) for \(\R^5\text{,}\) and with respect to this basis, we have the Jordan canonical form
\begin{equation*}
M_B(T) = \bbm 2 \amp 1\amp 0\amp 0 \amp 0\\
0 \amp 2\amp 0\amp 0 \amp 0\\
0 \amp 0\amp 3\amp 0 \amp 0\\
0 \amp 0\amp 0\amp 3 \amp 1\\
0 \amp 0\amp 0\amp 0 \amp 3\ebm\text{.}
\end{equation*}
Now that we’ve done all the work required for
Example 5.6.6, we should confess that there was an easier way all along:
The jordan_form()
command returns a pair \(P,J\text{,}\) where \(J\) is the Jordan canonical form of \(A\text{,}\) and \(P\) is an invertible matrix such that \(P^{-1}AP=J\text{.}\) 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 \(P\text{,}\) which are given by a basis of (generalized eigenvectors).