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6.C Generalized eigenspaces

Especially starting in this section, and subsequently, we will usually consider matrices and vectors spaces with the field of the complex numbers, $F={\mathbb{C}}$ . The reason for doing so is the Fundamental Theorem of Algebra, which has two key consequences:

•

Every monic polynomial of degree $\geq 1$ is the product of linear factors $(x-a)$ ,
•

Every complex matrix has at least 1 eigenvalue.

We have seen several examples of matrices in $\operatorname{M}_{{n}}({{\mathbb{C}}})$ which are not diagonalizable; in other words, for which ${\mathbb{C}}^{n}$ does not have a basis consisting of eigenvectors. The following could be seen as a strategy to deal with this obstacle. Instead of considering only spaces of eigenvectors, we will consider a generalization of eigenvectors, as follows.

Definition 6.17:

Let $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ be a square complex matrix, and let $\lambda\in{\mathbb{C}}$ be an eigenvalue of $A$ . Then the generalized eigenspace of index $i$ is:

V_{\lambda}^{(i)}:=\{\vec{x}\in{\mathbb{C}}^{n}\;|\;(A-\lambda I_{n})^{i}\vec{% x}=\vec{0}\}=\ker\left((A-\lambda I_{n})^{i}\right).

Non-zero elements of $V_{\lambda}^{(i)}$ are called generalized eigenvectors of $A$ .

For $i=0$ , we have $V_{\lambda}^{(0)}=\ker I_{n}=\{\vec{0}\}$ .

For $i=1$ , we have $V_{\lambda}^{(1)}=\ker(A-\lambda I_{n})=V_{\lambda},$ which is the usual eigenspace for $\lambda$ .

Exercise 6.18:

Let $A=\begin{bmatrix}0&4\\ -1&-4\end{bmatrix}$ . It’s only eigenvalue is $\lambda=-2$ . Prove that:

V_{-2}^{(1)}=\operatorname{span}\{(2,-1)\}

and

V_{-2}^{(2)}={\mathbb{C}}^{2}.

Exercise 6.19:

If $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , and $\lambda$ is an eigenvalue, prove that $V_{\lambda}^{(i)}\subset V_{\lambda}^{(i+1)}$ .

[End of Exercise]

Example 6.20.

Let’s find the generalized eigenspaces with respect to the eigenvalue $\lambda=5$ , for the following matrices in $\operatorname{M}_{{3}}({{\mathbb{C}}})$ :

$A=\begin{bmatrix}5&1&0\\ 0&5&0\\ 0&0&3\end{bmatrix}$
$B=\begin{bmatrix}5&1&0\\ 0&5&1\\ 0&0&5\end{bmatrix}$

(Solution:) For $A$ , let’s compute the powers of the matrix $A-5I_{3}$ .

$A-5I_{3}=\begin{bmatrix}0&1&0\\ 0&0&0\\ 0&0&-2\end{bmatrix}$
$(A-5I_{3})^{2}=\begin{bmatrix}0&1&0\\ 0&0&0\\ 0&0&-2\end{bmatrix}\begin{bmatrix}0&1&0\\ 0&0&0\\ 0&0&-2\end{bmatrix}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&4\end{bmatrix}$

So, for any $i\geq 3$ , multiplying by more copies will still give a matrix with zeros everywhere except the lower right entry. The generalized eigenspaces are the kernels of these matrices. So

$V_{5}^{(1)}=\ker(A-5I_{3})=\operatorname{span}_{{\mathbb{C}}}\{(1,0,0)\}$
$V_{5}^{(2)}=\ker(A-5I_{3})^{2}=\ker\left(\operatorname{diag}(0,0,4)\right)=% \operatorname{span}_{{\mathbb{C}}}\{(1,0,0),(0,1,0)\}$
$V_{5}^{(i)}=\operatorname{span}_{{\mathbb{C}}}\{(1,0,0),(0,1,0)\}$ for all $i\geq 3$

Similarly for the matrix $B$ , we compute the powers of the matrix $B-5I_{3}$ :

$B-5I_{3}=\begin{bmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{bmatrix}$ .
$(B-5I_{3})^{2}=\begin{bmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{bmatrix}\begin{bmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{bmatrix}=\begin{bmatrix}0&0&1\\ 0&0&0\\ 0&0&0\end{bmatrix}$
$(B-5I_{3})^{3}=\begin{bmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{bmatrix}\begin{bmatrix}0&0&1\\ 0&0&0\\ 0&0&0\end{bmatrix}=\vec{0}$

So, for any $i\geq 4$ , multiplying by more copies will still give the zero matrix, so $(B-5I_{3})^{i}=\vec{0}$ . Now we find the kernels of the above matrices:

$V_{5}^{(1)}=\ker(B-5I_{3})=\operatorname{span}_{{\mathbb{C}}}\{(1,0,0)\}$
$V_{5}^{(2)}=\ker(B-5I_{3})^{2}=\operatorname{span}_{{\mathbb{C}}}\{(1,0,0),(0,% 1,0)\}$
$V_{5}^{(3)}=\ker\vec{0}={\mathbb{C}}^{3}$
$V_{5}^{(i)}={\mathbb{C}}^{3}$ for all $i\geq 4$ .

So, the dimensions of the generalized eigenspaces of $A$ , for the eigenvalue $\lambda=5$ are $1,2,2,2,\cdots$ , while the dimensions of the generalized eigenspaces for $B$ for the eigenvalue $\lambda=5$ are $1,2,3,3,\cdots$ .

The pattern in the previous example is that the generalized eigenspace dimension increases until a certain point, after which the dimensions stabilize. This is an example of a general phenomenon, stated in the following theorem.

Theorem 6.21.

Let $B\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ be any matrix, and let $r\geq 1$ .

\ker B^{r}=\ker B^{r+1}

implies

\ker B^{r}=\ker B^{r+1}=\ker B^{r+2}=\ker B^{r+3}=\cdots

Proof.

We already know that $\ker B^{r+1}\subset\ker B^{r+2}$ , we just need the reverse containment. Assume $\vec{x}\in\ker B^{r+2}$ , in other words, $B^{r+2}\vec{x}=\vec{0}$ . This implies $B^{r+1}(B\vec{x})=\vec{0}$ , which means $B\vec{x}\in\ker B^{r+1}$ . Using the assumption of the theorem, $B\vec{x}\in\ker B^{r}$ , in other words, $B^{r}(B\vec{x})=\vec{0}$ . This is the same as saying $\vec{x}\in\ker B^{r+1}$ . So we have proved $\ker B^{r+1}=\ker B^{r+2}$ . Using the same argument for the next exponent, and the next, etc, we have proved the theorem. This argument could also be worded using the language of induction. ∎

Corollary 6.22.

Let $\lambda$ be an eigenvalue of $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ . If $V_{\lambda}^{(r)}=V_{\lambda}^{(r+1)}$ , then all generalized eigenspaces $V_{\lambda}^{(r+i)}$ for $i\geq 0$ are equal to each other.

Proof.

This is just Theorem 6.21 applied to the case when $B=A-\lambda I_{n}$ . ∎

Example 6.23.

Let $A=\begin{bmatrix}3&2&1\\ 0&3&1\\ -1&-4&-1\end{bmatrix}$ . For each eigenvalue of $A$ , find a basis for each generalized eigenspaces of $A$ .

(Solution:) First, we determine the eigenvalues via the characteristic polynomial.

	$\displaystyle c_{A}(x)=\det\begin{bmatrix}3-x&2&1\\ 0&3-x&1\\ -1&-4&-1-x\end{bmatrix}$	$\displaystyle=(3-x)\left((3-x)(-1-x)+4\right)-(2-(3-x))$
		$\displaystyle=-x^{3}+5x^{2}-8x+4=(2-x)^{2}(1-x).$

So the eigenvalues are $\lambda=2$ and $\lambda=1$ .

$\lambda=1$ : We compute the dimensions of the generalized eigenspaces as follows:

(A-I_{3})=\begin{bmatrix}2&2&1\\ 0&2&1\\ -1&-4&-2\end{bmatrix}.

Since $\lambda=1$ is an eigenvalue, we have $\det(A-I_{3})=0$ , and so $\operatorname{rank}(A-I_{3})\neq 3$ . But two of its rows are clearly linearly independent, and so $\operatorname{rank}(A-I_{3})\geq 2$ by Theorem 4.25. This proves that $\operatorname{rank}(A-I_{3})=2$ . By the dimension theorem, $\dim\ker(A-I_{3})=1$ , in other words $\dim V_{1}^{(1)}=1$ .

(A-I_{3})^{2}=\begin{bmatrix}2&2&1\\ 0&2&1\\ -1&-4&-2\end{bmatrix}\begin{bmatrix}2&2&1\\ 0&2&1\\ -1&-4&-2\end{bmatrix}=\begin{bmatrix}3&4&2\\ -1&0&0\\ 0&-2&-1\end{bmatrix}.

This shows two rows of $(A-I_{3})^{2}$ are linearly independent, and we can use that $\det BC=\det B\det C$ to see that $\det[(A-I_{3})^{2}]=0$ . So $\operatorname{rank}(A-I_{3})^{2}=2$ . By the dimension theorem again, $V_{1}^{(2)}=\dim\ker(A-I)^{2}=1.$ Now by Theorem 6.21 all of the generalized eigenspaces for $\lambda=1$ are equal to each other:

V_{1}^{(i)}=\operatorname{span}_{{\mathbb{C}}}\{(0,1,-2)\},

for all $i\geq 1$ . So, in this case, a basis for each generalized eigenspace is the vector $(0,1,-2)$ .

$\lambda=2$ : The computation is similar to the previous case:

(A-2I_{3})=\begin{bmatrix}1&2&1\\ 0&1&1\\ -1&-4&-3\end{bmatrix}.

This matrix is rank 2, and so $\dim V_{2}^{(1)}=\dim\ker(A-2I_{3})=1$ , by the dimension theorem. Moreover, the eigenspace is

V_{2}^{(1)}=\operatorname{span}_{{\mathbb{C}}}\{(1,-1,1)\}.

To determine the next generalized eigenspace, we compute:

(A-2I_{3})^{2}=\begin{bmatrix}1&2&1\\ 0&1&1\\ -1&-4&-3\end{bmatrix}\begin{bmatrix}1&2&1\\ 0&1&1\\ -1&-4&-3\end{bmatrix}=\begin{bmatrix}0&0&0\\ -1&-3&-2\\ 2&6&4\end{bmatrix}.

Observe that the rows are scalar multiples of each other, and therefore the row space is 1-dimensional; in other words $\operatorname{rank}(A-2I_{3})^{2}=1$ . So by the dimension theorem, $\dim V_{2}^{(2)}=\dim\ker(A-2I_{3})^{2}=2$ . We can express the generalized eigenspace as the span of 2 vectors as follows:

V_{2}^{(2)}=\ker\begin{bmatrix}0&0&0\\ -1&-3&-2\\ 2&6&4\end{bmatrix}=\operatorname{span}_{{\mathbb{C}}}\{(3,-1,0),(2,0,-1)\}.

The next generalized eigenspace is the kernel of the following matrix:

(A-2I_{3})^{3}=\begin{bmatrix}1&2&1\\ 0&1&1\\ -1&-4&-3\end{bmatrix}\begin{bmatrix}0&0&0\\ -1&-3&-2\\ 2&6&4\end{bmatrix}=\begin{bmatrix}0&0&0\\ 1&3&2\\ -2&-6&-4\end{bmatrix}.

So $\ker(A-2I_{3})^{r}=\ker(A-2I_{3})^{2}$ for every $r\geq 3$ . In particular,

V_{2}^{(r)}=\operatorname{span}_{{\mathbb{C}}}\{(3,-1,0),(2,0,-1)\},

for all $r\geq 3$ .

Since the sequence $(3,-1,0),(2,0,-1)$ is linearly independent (it consists of two vectors which are not multiples of each other) and spans this generalized eigenspace, it forms a basis.

To summarize: the generalized eigenspaces for the eigenvalue $\lambda=1$ are of dimension $1,1,1,1,\cdots$ , and they each have a basis $(0,1,-2)$ . The generalized eigenspaces for the eigenvalue $\lambda=2$ are of dimension $1,2,2,2,\cdots$ , and the first one has a basis $(1,-1,1)$ , while each of the others has a basis $(3,-1,0),(2,0,-1)$ .

Exercise 6.24:

For each of the matrices in Exercise 6.15, in $\operatorname{M}_{{n}}({{\mathbb{C}}})$ , and for each eigenvalue $\lambda$ , compute the dimensions and find a basis of each generalized eigenspace $V_{\lambda}^{(i)}$ for $i\geq 1$ .

[End of Exercise]

In all of the above exercises and examples, the observant reader may have noticed that generalized eigenvectors for different eigenvalues are always linearly independent. This is always true (as stated in the following theorem), and the proof uses an induction argument which we omit.

Theorem 6.25.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , and assume $\vec{v_{1}},\cdots,\vec{v_{r}}$ are generalized eigenvectors for different eigenvalues. Then $\vec{v_{1}},\cdots,\vec{v_{r}}$ are linearly independent.

As we will see, the dimensions of the generalized eigenspaces will be used to deduce the Jordan normal form; no other information is needed. But if you want to find a specific change of basis matrix $P$ such that $P^{-1}AP$ is in Jordan normal form, this is equivalent to finding a Jordan basis, which is the purpose of the next section.