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6.D Jordan chains and Jordan bases

Given a matrix $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , and an eigenvalue $\lambda$ , in the previous section we put a lot of effort into finding a basis for each of the generalized eigenspaces of $\lambda$ . They are subspaces, and by Theorem 6.21 there is a number $r\geq 1$ such that:

\{\vec{0}\}\subsetneqq V_{\lambda}^{(1)}\subsetneqq V_{\lambda}^{(2)}% \subsetneqq\cdots\subsetneqq V_{\lambda}^{(r)}=V_{\lambda}^{(r+1)}=V_{\lambda}% ^{(r+2)}=\cdots\subset{\mathbb{C}}^{n}.

An important observation is that if we pick a vector in one of these subspace, repeatedly multiplying that vector by the matrix $A-\lambda I_{n}$ moves it along these subspaces from the right to the left, creating a “chain” of vectors. In other words:

Theorem 6.26.

If $\vec{x}\in V_{\lambda}^{(i)}$ , for some $i\geq 1$ , then

(A-\lambda I_{n})\vec{x}\in V_{\lambda}^{(i-1)}.

Proof.

The proof is because $\vec{x}\in V_{\lambda}^{(i)}$ means that $(A-\lambda I_{n})^{i}\vec{x}=\vec{0}$ , by definition of the generalized eigenspace. This implies that $(A-\lambda I_{n})^{i-1}((A-\lambda I_{n})\vec{x})=\vec{0}$ , which is what we wanted to prove.

Notice that this formula still works with $i=1$ , because we have that $V_{\lambda}^{(0)}=\{\vec{0}\}$ , the zero subspace. ∎

Exercise 6.27:

Let $A=\begin{bmatrix}-1&0\\ 1&-1\end{bmatrix}$ .

i.

Prove that $V_{-1}^{(1)}=\operatorname{span}\{(0,1)\}$ and that $V_{-1}^{(2)}={\mathbb{C}}^{2}$ .
ii.

Find a non-zero vector $\vec{x}$ which is in $V_{-1}^{(2)}$ but is not in $V_{-1}^{(1)}$ .
iii.

Prove that $(A+I_{2})\vec{x}\in V_{-1}^{(1)}$ .

[End of Exercise]

In the following definition, notice that a “Jordan chain of length 1 for $\lambda$ ” is exactly the same thing as an “eigenvector for $\lambda$ ”.

Recall that if $Y\subset X$ are sets, then $a\in X\backslash Y$ means that $a\in X$ but $a\notin Y$ .

Definition 6.28:

Given a square matrix $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ and an eigenvalue $\lambda$ , a sequence of vectors $\vec{x_{1}},\cdots,\vec{x_{k}}$ is called a Jordan chain of length $k$ for $\lambda$ if:

•

$\vec{x_{k}}\in V_{\lambda}^{(k)}\backslash V_{\lambda}^{(k-1)}$ , and
•

$\vec{x_{i}}=(A-\lambda I_{n})\vec{x_{i+1}}$ , for every $i=1,\cdots,k-1$ .

A Jordan basis (for $A$ ) is a basis of ${\mathbb{C}}^{n}$ which consists only of Jordan chains (for different eigenvalues, in general).

Any Jordan chain, $\vec{x_{1}},\cdots,\vec{x_{k}}$ , must obey $\vec{x_{1}}\neq\vec{0}$ . This is because $\vec{x_{k}}\notin V_{\lambda}^{(k-1)}$ is another way of saying $\vec{x_{1}}=(A-\lambda I_{n})^{k-1}\vec{x_{k}}\neq\vec{0}$ .

Example 6.29.

Let $A=\begin{bmatrix}5&1&0\\ 0&5&1\\ 0&0&5\end{bmatrix}$ . You may use that

$V_{5}^{(1)}=\operatorname{span}\{(1,0,0)\}$
$V_{5}^{(2)}=\operatorname{span}\{(1,0,0),(0,1,0)\}$
$V_{5}^{(3)}={\mathbb{C}}^{3}$

Find a Jordan chain of length 3.

Solution: First pick an element in $V_{5}^{(3)}\backslash V_{5}^{(2)}$ . Say $\vec{x_{3}}=(0,0,1)$ . Then define $\vec{x_{2}}=(A-5I_{3})\vec{x_{3}}=(0,1,0)$ and $\vec{x_{1}}=(A-5I_{3})\vec{x_{2}}=(1,0,0)$ . Then sequence $\vec{x_{1}},\vec{x_{2}},\vec{x_{3}}$ is a Jordan chain of length 3. In fact this sequence is a Jordan basis, since it is a basis of ${\mathbb{C}}^{3}$ and it is made up of Jordan chains (in this case, just one).

Example 6.30.

Let $A=\begin{bmatrix}3&2&1\\ 0&3&1\\ -1&-4&-1\end{bmatrix}$ as in Example 6.23. Find a Jordan basis.

(Solution:) First we find the generalized eigenspaces for each eigenvalue.

$\lambda=1$ : All of the generalized eigenspaces are 1-dimensional and are spanned by the vector $\vec{y_{1}}:=(0,1,-2)$ . In particular, this eigenvector forms a Jordan chain of length 1 for the eigenvalue $\lambda=1$ , and no longer chains are possible.

$\lambda=2$ : Based on our computation of the dimensions of the generalized eigenspaces, a Jordan chain for $\lambda=2$ has length at most 2. Let’s choose a vector in $V_{2}^{(2)}\backslash V_{2}^{(1)}$ . One such vector is $\vec{x_{2}}=(3,-1,0)$ . Then $\vec{x_{1}}:=(A-2I_{3})\vec{x_{2}}=(1,-1,1)$ . So $\vec{x_{1}},\vec{x_{2}}$ is a Jordan chain of length 2.

Now $\vec{y_{1}},\vec{x_{1}},\vec{x_{2}}$ forms a basis of ${\mathbb{C}}^{3}$ , so our search ends. In other words, we have found a Jordan basis for $A$ ; it is the union of two Jordan chains.

Let’s see how the matrix $A$ from Example 6.30 looks in the new Jordan basis

\mathcal{B}:=(\vec{y_{1}},\vec{x_{1}},\vec{x_{2}})=((0,1,-2),(1,-1,1),(3,-1,0)).

If $T(\vec{v})=A\vec{v}$ is the associated linear transformation, then we want to compute ${}_{\mathcal{B}}[T]_{\mathcal{B}}$ . Using the method from Section 4.A we compute:

$T(\vec{y_{1}})=\vec{y_{1}}+0\vec{x_{1}}+0\vec{x_{2}}$
$T(\vec{x_{1}})=0\vec{y_{1}}+2\vec{x_{1}}+0\vec{x_{2}}$
$T(\vec{x_{2}})=0\vec{y_{1}}+\vec{x_{1}}+2\vec{x_{2}}$

This calculation shows that

{}_{\mathcal{B}}[T]_{\mathcal{B}}=\begin{bmatrix}1&0&0\\ 0&2&1\\ 0&0&2\end{bmatrix}.

Alternately, one could do a much longer calculation by using the change of basis matrix from $\mathcal{B}$ to the standard basis:

P={{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}}=\begin{bmatrix}% 0&1&3\\ 1&-1&-1\\ -2&1&0\end{bmatrix}.

Then we need to compute the inverse of $P$ , and finally verify that ${}_{\mathcal{B}}[T]_{\mathcal{B}}=P^{-1}AP$ gives the same matrix as above.

The resulting matrix ${}_{\mathcal{B}}[T]_{\mathcal{B}}$ is not diagonal, but it is as close as we can get to diagonalizing. In the next section we will see that this matrix is in Jordan normal form.

Exercise 6.31:

Find a Jordan basis for $\begin{bmatrix}0&4\\ -1&-4\end{bmatrix}$ (see also Exercise 6.18).

Exercise 6.32:

Find a Jordan basis for $\begin{bmatrix}-1&0\\ 1&-1\end{bmatrix}$ (see also Exercise 6.27).

[End of Exercise]

We can’t always find a basis of eigenvectors, but in the above examples, we were able to find a basis of Jordan chains. The remarkable thing about these Jordan bases, and the reason why this method should be considered a superior extension to diagonalizing a matrix, is that they always exist:

Theorem 6.33.

For any matrix $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , there is a Jordan basis for $A$ .

In other words, there is always a basis of ${\mathbb{C}}^{n}$ consisting of Jordan chains for $A$ .

At the end of the next section is an algorithm for finding a Jordan basis.

Exercise 6.34:

Find a Jordan basis for $A=\begin{bmatrix}2&0&0\\ 0&2&0\\ 1&-1&2\end{bmatrix}$ .

[End of Exercise]