Let $$C = \left[ {\begin{array}{cccc}
0 & -1 & -2 & 3 \\
0 & 0 & -2 & 3 \\
0 & 1 & 1 & -1 \\
0 & 0 & -1 & 2
\end{array} } \right].$$ What is the Jordan canonical form of C?
We know that the characteristic polynomial
$$\chi_C(\lambda) = (z – 1)^4$$
So the algebraic multiplicity of $1$ is $4$.
- How do I find the geometric multiplicity of $1$?
- If the geometric multiplicity of $1$ is less than $4$, how do I find the bases that give me the Jordan canonical form?
Best Answer
Let's follow this algorithm described by Stefan Friedl.
A little work shows that the characteristic polynomial of $C$ is $$ \chi_C(t) = t \cdot (t - 1)^{3} $$ which gives a table of eigenvalues $$ \begin{array}{c|c|c} \lambda & \operatorname{am}_C(\lambda) & \operatorname{gm}_C(\lambda) \\ \hline 0 & 1 & ? \\ 1 & 3 & ? \end{array} $$ Here, $\operatorname{am}_C(\lambda)$ is the algebraic multiplicity of $\lambda$ as an eigenvalue of $C$ and $\operatorname{gm}_C(\lambda)$ is the geometric multiplicity.
Our factorization of the characteristic polynomial allowed us to fill in the algebraic multiplicities in our table. The geometric multiplicities can be computed from the definition $\operatorname{gm}_C(\lambda)=\operatorname{nullity}(\lambda\cdot I-C)$. In our case, we have $$ \begin{array}{c|c|c} \lambda & \operatorname{am}_C(\lambda) & \operatorname{gm}_C(\lambda) \\ \hline 0 & 1 & 1 \\ 1 & 3 & 1 \end{array} $$ Note that $\operatorname{gm_C}(0)$ can also be quickly inferred from the inequality $1\leq\operatorname{gm}_C(0)\leq\operatorname{am}_C(0)=1$.
At this stage, we can infer the Jordan form $J$ of $C$. Recall the interpretations of the multiplicities of the eigenvalues as \begin{align*} \operatorname{am}_C(\lambda) &= \text{number of $\lambda$'s on the diagonal of $J$} \\ \operatorname{gm}_C(\lambda) &= \text{size of the largest Jordan block corresponding to $\lambda$ inside $J$} \end{align*} In general, knowing these multiplicities is not enough to infer $J$. However, in our case we can see that $$ J=\left[\begin{array}{r|rrr} 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right] $$
Now, we proceed to compute the change of basis matrix $P$. The easiest way to start is to note that $$ E_0 = \operatorname{Span}\{\langle1, 0, 0, 0\rangle\} $$ This gives our first column of $P$, so $$ P= \left[\begin{array}{rrrr} 1 & ? & ? & ? \\ 0 & ? & ? & ? \\ 0 & ? & ? & ? \\ 0 & ? & ? & ? \end{array}\right] $$ Now, to build the other three columns, we compute the numbers $$ d_k=\operatorname{nullity}((\lambda\cdot I-C)^k)-\operatorname{nullity}((\lambda\cdot I-C)^{k-1}) $$ for $1\leq k\leq\operatorname{gm}_C(\lambda)$ where $\lambda=1$. For us, these numbers turn out to be \begin{align*} d_1 &= 1 & d_2 &= 1 & d_3 &= 1 \end{align*} We now take these numbers and build a diagram of empty boxes $$ \begin{array}{c} \Box\\ \Box\\ \Box \end{array} $$ The algorithm we're following demands that we start at the bottom of this diagram and fill the boxes in row $k$ with linearly independent vectors that belong to $\operatorname{Null}((\lambda\cdot I-C)^k)$ but not $\operatorname{Null}((\lambda\cdot I-C)^{k-1})$. Once a box in the diagram is filled with a vector $\vec{v}$, the box immediately above it is filled with $(\lambda\cdot I-C)\vec{v}$.
In our situation, we have \begin{align*} (I-C)^2 &= \left[\begin{array}{rrrr} 1 & 0 & 1 & -1 \\ 0 & -1 & -1 & 2 \\ 0 & -1 & -1 & 2 \\ 0 & -1 & -1 & 2 \end{array}\right] & (I-C)^3 &= \left[\begin{array}{rrrr} 1 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{align*} We easily see that $\langle0,1,0,0\rangle\in\operatorname{Null}((I-C)^3)$ but $\langle0,1,0,0\rangle\notin\operatorname{Null}((I-C)^2)$. This allows us to fill out our diagram $$ \begin{array}{c} \fbox{$\left\langle0,\,-1,\,-1,\,-1\right\rangle$}\\ \fbox{$\left\langle-1,\,-1,\,1,\,0\right\rangle$} \\ \fbox{$\left\langle0,\,1,\,0,\,0\right\rangle$} \end{array} $$ This defines our matrix $P$ as $$ P = \left[\begin{array}{rrrr} 1 & 0 & -1 & 0 \\ 0 & -1 & -1 & 1 \\ 0 & -1 & 1 & 0 \\ 0 & -1 & 0 & 0 \end{array}\right] $$ We can verify ourselves that this is indeed correct $$ \overset{C}{\left[\begin{array}{rrrr} 0 & -1 & -2 & 3 \\ 0 & 0 & -2 & 3 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & -1 & 2 \end{array}\right]} = \overset{P}{\left[\begin{array}{rrrr} 1 & 0 & -1 & 0 \\ 0 & -1 & -1 & 1 \\ 0 & -1 & 1 & 0 \\ 0 & -1 & 0 & 0 \end{array}\right]} \overset{J}{\left[\begin{array}{r|rrr} 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right]} \overset{P^{-1}}{\left[\begin{array}{rrrr} 1 & 0 & 1 & -1 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 1 & 1 & -2 \end{array}\right]} $$