How do I find the bases of the Jordan Canonical Form of $C$?

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 $$4$$.

1. How do I find the geometric multiplicity of $$1$$?
2. If the geometric multiplicity of $$1$$ is less than $$4$$, how do I find the bases that give me the Jordan canonical form?
• Geo-multi = $\dim (\mathrm {Ker} \,(C - 1I)) = n - \mathrm {rank} (C - 1I)$, hence you could find $\mathrm {rank} (C-1I)$ via RREF or some other thing. – xbh Mar 14 at 5:51
• $C$ is not invertible, so shouldn’t it’s characteristic polynomial be divisible by $z$? – Joppy Mar 14 at 6:41

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 $$\overset{C}{\left[\begin{array}{rrrr} 0 & -1 & -2 & 3 \\ 0 & 0 & -2 & 3 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & -1 & 2 \end{array}\right]} = \overset{P}{\left[\begin{array}{rrrr} 1 & 0 & -1 & 0 \\ 0 & -1 & -1 & 1 \\ 0 & -1 & 1 & 0 \\ 0 & -1 & 0 & 0 \end{array}\right]} \overset{J}{\left[\begin{array}{r|rrr} 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right]} \overset{P^{-1}}{\left[\begin{array}{rrrr} 1 & 0 & 1 & -1 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 1 & 1 & -2 \end{array}\right]}$$

• "In general, knowing these multiplicities is not enough to infer 𝐽." What do we do in that case? I would greatly appreciate you if you could tell me. – ErotemeObelus Mar 17 at 3:17