5.D Jordan chains and Jordan bases‣ Chapter 5 Jordan normal form (2024)

Given a matrix $A\in\operatorname{M}_{{n}}({\operatorname{\mathbb{C}}})$ , 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 5.22 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\operatorname{\mathbb{C}}^{n}.

Recall that $X\subsetneqq Y$ means $X$ is a subset of $Y$ but $X\neq Y$ .

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 5.26.

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.∎

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 $a\notin Y$ . For example $\{1,2,3\}\backslash\{2\}$ is the set $\{1,3\}$ .

Definition 5.27:

Given a square matrix $A\in\operatorname{M}_{{n}}({\operatorname{\mathbb{C}}})$ 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:

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$\vec{x_{k}}\in V_{\lambda}^{(k)}\backslash V_{\lambda}^{(k-1)}$ , and
See Also
Jordan's Math Work—Free Games and Resources — Mashup Math 5.G Exercises‣ Chapter 5 Jordan normal form
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$\vec{x_{i}}=(A-\lambda I_{n})\vec{x_{i+1}}$ , for every $i=1,\cdots,k-1$ .

A Jordan basis (for $A$ ) is a basis of $\operatorname{\mathbb{C}}^{n}$ 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 $\vec{x_{1}}=(A-\lambda I_{n})^{k-1}\vec{x_{k}}\neq\vec{0}$ .

Exercise 5.28:

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

i.
Prove that $V_{-1}^{(1)}=\operatorname{span}\{(0,1)\}$ and that $V_{-1}^{(2)}=\operatorname{\mathbb{C}}^{2}$ .
ii.
Find a non-zero vector $\vec{x}$ which is in $V_{-1}^{(2)}$ but is not in $V_{-1}^{(1)}$ .
iii.
Prove that $(A+I_{2})\vec{x}\in V_{-1}^{(1)}$ .

Solution.[5.28] (i) $A=\left[\begin{smallmatrix}-1&0\\1&-1\end{smallmatrix}\right]$ . We find the generalized eigenspaces as follows:

	$\displaystyle V_{-1}^{(1)}=\ker(A+I_{2})=\ker\begin{bmatrix}0&0\\1&0\end{bmatrix}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{%bmatrix}0\\1\end{bmatrix}\},$
	$\displaystyle V_{-1}^{(2)}=\ker\begin{bmatrix}0&0\\1&0\end{bmatrix}^{2}=\ker\begin{bmatrix}0&0\\0&0\end{bmatrix}=\operatorname{\mathbb{C}}^{2}.$

(ii) Since $\operatorname{\mathbb{C}}^{2}=V_{-1}^{(2)}$ , we just need any vector which is not in $V_{-1}^{(1)}$ . Let’s take $\vec{x}=(1,0)$ ; this is not in a scalar multiple of $(0,1)$ , and therefore it is not in $V_{-1}^{(1)}$ .

(iii) We make the following calculation with the vector chosen above:

(A+I_{2})\vec{x}=\begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}\in V_{-1}^{(1)}.

This is all we were asked to check.

Example 5.29.

Let $A=\begin{bmatrix}5&1&0\\0&5&1\\0&0&5\end{bmatrix}$ . You may use that $V_{5}^{(1)}=\operatorname{span}\{(1,0,0)\}$ , $V_{5}^{(2)}=\operatorname{span}\{(1,0,0),(0,1,0)\}$ , and $V_{5}^{(3)}=\operatorname{\mathbb{C}}^{3}$ .

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 $\operatorname{\mathbb{C}}^{3}$ and it is made up of Jordan chains (in this case, just one).

Example 5.30.

Let $A=\begin{bmatrix}3&2&1\\0&3&1\\-1&-4&-1\end{bmatrix}$ as in Example 5.24. 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 we use Corollary 1.59, and check that the determinant of the matrix is $-1$ , and hence $\vec{y_{1}},\vec{x_{1}},\vec{x_{2}}$ forms a basis of $\operatorname{\mathbb{C}}^{3}$ , 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 5.30 looks in the new Jordan basis

\operatorname{\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 map, then we want to compute ${}_{\operatorname{\mathcal{B}}}[T]_{\operatorname{\mathcal{B}}}$ . Using the method from Section 3.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

{}_{\operatorname{\mathcal{B}}}[T]_{\operatorname{\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 $\operatorname{\mathcal{B}}$ to the standard basis:

P={{}_{\operatorname{\mathcal{C}}}[}\operatorname{Id}\nolimits]_{\operatorname%{\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 ${}_{\operatorname{\mathcal{B}}}[T]_{\operatorname{\mathcal{B}}}=P^{-1}AP$ gives the same matrix as above.

The resulting matrix ${}_{\operatorname{\mathcal{B}}}[T]_{\operatorname{\mathcal{B}}}$ 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 5.31:

Find a Jordan basis for $\begin{bmatrix}0&4\\-1&-4\end{bmatrix}$ (see also Exercise 5.19).

Solution.[5.31] $A=\left[\begin{smallmatrix}0&4\\-1&-4\end{smallmatrix}\right]$ . Recall from Exercise 5.19 that $-2$ is the only eigenvalue, $V_{-2}^{(1)}=\operatorname{span}_{\operatorname{\mathbb{C}}}\{\begin{bmatrix}2%\\-1\end{bmatrix}\}$ , and $V_{-2}^{(2)}=\operatorname{\mathbb{C}}^{2}$ . So, to find a Jordan basis, we need an element in $V_{-2}^{(2)}\backslash V_{-2}^{(1)}$ , for example, the vector $\vec{x_{2}}=(1,0)$ . Then, following Definition 5.27 we will make a Jordan chain of length 2 by defining $\vec{x_{2}}=\begin{bmatrix}1\\0\end{bmatrix}$ , and $\vec{x_{1}}:=(A-\lambda I_{2})\vec{x_{2}}=\begin{bmatrix}2&4\\-1&-2\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}2\\-1\end{bmatrix}$ . These two vectors form a basis of $\operatorname{\mathbb{C}}^{2}$ , and therefore the sequence $\vec{x_{1}},\vec{x_{2}}$ forms a Jordan basis.

Notice that many other choices of the vector $\vec{x_{2}}$ are possible, which would potentially lead to a different vector $\vec{x_{1}}$ ; but result would still be a Jordan basis.

Exercise 5.32:

Find a Jordan basis for $\begin{bmatrix}-1&0\\1&-1\end{bmatrix}$ (see also Exercise 5.28).

Solution.[5.32] $A=\left[\begin{smallmatrix}-1&0\\1&-1\end{smallmatrix}\right]$ . Recall from Exercise 5.28 that there is only one eigenvalue, $-1$ , and that $V_{-1}^{(1)}=\operatorname{span}\{(0,1)\}$ and $V_{-1}^{(2)}=\operatorname{\mathbb{C}}^{2}$ . So to make a Jordan basis, we just need to choose a vectors as in Definition 5.27:

$\vec{x_{2}}:=\begin{bmatrix}1\\0\end{bmatrix}$ , $\vec{x_{1}}:=(A+I_{2})\vec{x_{2}}=\begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$ .

There two vectors are linearly independent, and therefore form a basis of $\operatorname{\mathbb{C}}^{2}$ . Since they form a Jordan chain, the sequence $\vec{x_{1}},\vec{x_{2}}$ forms a Jordan basis, as required.

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 (see Theorem 5.41).

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

Exercise 5.33:

Find a Jordan basis for $A=\begin{bmatrix}2&0&0\\0&2&0\\1&-1&2\end{bmatrix}$ . State the length of each Jordan chain in your basis.