§4.5 Eigenvalues and eigenvectors · A rough guide to linear algebra

Eigenvectors and eigenvalues are useful tools for analyzing how a linear map acts on a vector space.

Let $V$ be a vector space over $k$ and let $T : V \to V$ be a $k$-linear map. We say that $\lambda \in k$ is an eigenvalue of $T$ if there exists a vector $v \in V$ with $v \neq 0$ such that $Tv = \lambda v$. In this case, we say that $v$ is an eigenvector for $\lambda$.

If $v \in V$ is an eigenvector for $T$ with eigenvalue $\lambda$, then we inductively have

$$ T^n v = \lambda^n v $$

for all $n \ge 0$. Moreover, for any polynomial $p(t) \in k[t]$ with coefficients in $k$, we have

$$ p(T) v = p(\lambda) v. $$

An eigenvector need not exist. For instance, consider the matrix $T : \mathbb{R}^2 \to \mathbb{R}^2$ given by

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

Verify that $T$ does not have an eigenvector.

But if the base field $k$ is algebraically closed, the theorem about Jordan normal form implies that every vector space (of positive finite dimension) has an eigenvector. It moreover tells us exactly what the eigenvectors and eigenvalues are.

Let $V$ be a finite-dimensional vector space over $k$, where $k$ is algebraically closed. Show that the set of eigenvalues of $A : V \to V$ is the same as the set of diagonal entries of any Jordan normal form of $A$.

Before analyzing the situation for algebraically closed field, let us set up a bit of notation. Fix a finite-dimensional vector space $V$ over a field $k$ (not necessarily algebraically closed), and fix a linear map $T : V \to V$.

For each scalar $\lambda \in k$, define the eigenspace as

$$ V_\lambda = \ker (\lambda \cdot \id - T) = \lbrace v \in V : Tv = \lambda v\rbrace . $$

Check that $V_\lambda$ is a subspace of $V$, as a $k$-vector space. Verify also that $\lambda$ is an eigenvalue if and only if $\dim_k V_\lambda > 0$.

The interesting fact is that the spaces $V_\lambda$ are all linearly independent from each other. Let me make this statement precise.

Fix a finite-dimensional $k$-vector space $V$ and a linear map $T : V \to V$. Consider the eigenspaces $V_\lambda \subseteq V$ for each $\lambda \in k$.

(a)
Show that if $\lambda_1, \ldots, \lambda_n \in k$ are different scalars, and $v_i \in V_{\lambda_i}$ for all $1 \le i \le n$, then

$$ v_1 + v_2 + \cdots + v_n = 0 $$

implies $v_1 = v_2 = \cdots = v_n = 0$. (Hint: use Lagrange interpolation)
(b)
The family of inclusion maps $V_\lambda \hookrightarrow V$ induce a $k$-linear map

$$ \bigoplus_{\lambda \in k} V_\lambda \to V. $$

Show that this map is injective.

It will be great if the linear map $\bigoplus_{\lambda \in k} V_\lambda \to V$ is an isomorphism. However, we should not expect this to hold in generality even if $k$ is algebraically closed. For instance,

$$ T : \mathbb{C}^2 \to \mathbb{C}^2, \quad T = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$

has $V_\lambda = 0$ for $\lambda \neq 0$ and $V_0 = \vspan((1, 0))$. So $\dim \bigoplus_{\lambda \in k} V_\lambda = 1$ while $\dim V = 2$. On the other hand, in Chapter 5, we prove a theorem that $\bigoplus_{\lambda \in k} V_\lambda \to V$ is indeed an isomorphism if $T$ is self-adjoint. But without an assumption on the linear map $T$, we do not have such a result.

The solution to this problem is to look at a bigger subspace, called the generalized eigenspace.

Let $V$ be a finite-dimensional vector space and $T : V \to V$ be a linear map. For each $\lambda \in k$, define the generalized eigenspace as

$$ V_{(\lambda)} = \bigcup_{n \ge 1} \ker ((\lambda \cdot \id - T)^n) = \lbrace v \in V : (\lambda - T)^n v = 0 \text{ for some } n \ge 1 \rbrace . $$

Check that $V_{(\lambda)}$ is a linear subspace of $V$. Also verify that $\lambda$ is an eigenvalue if and only if $\dim V_\lambda > 0$ if and only if $\dim V_{(\lambda)} > 0$.

Although we have enlarged each of the spaces $V_\lambda$ to $V_{(\lambda)}$, generalized eigenspaces still satisfy the linear independence property.

Let $V$ be a finite-dimensional space, and $T : V \to V$ be a linear map.

(a)
Show that if $0 \neq v \in V_{(\lambda)}$ and $\lambda \neq \kappa \in k$, then $(\kappa - T) v \in V_{(\lambda)}$ and $(\kappa - T) v \neq 0$.
(b)
Show that the linear map

$$ \bigoplus_{\lambda \in k} V_{(\lambda)} \to V $$

induced by the inclusions $V_{(\lambda)} \hookrightarrow V$ is injective. (Hint: suppose it is not injective, and look at nonzero vectors $v_1, \ldots, v_n$ that add up to $0$ and contained in different generalized eigenspaces, such that $n$ is smallest as possible. Derive a contradiction by applying a suitable $(T - \lambda)^n$ to it.)

Find an example of a vector space $V$ and a linear map $T : V \to V$ such that the linear map $\bigoplus_{\lambda \in k} V_{(\lambda)} \to V$ is not an isomorphism.

Let us now assume that $k$ is an algebraically closed field. In this case, we will show that the map $\bigoplus_{\lambda \in k} V_{(\lambda)} \to V$ is an isomorphism.

Let $V$ be a finite-dimensional vector space over $k$, where $k$ is algebraically closed. Consider a linear map $T : V \to V$ and the generalized eigenspaces $V_{(\lambda)}$ with respect to $T$. Then the linear map

$$ \bigoplus_{\lambda \in k} V_{(\lambda)} \to V $$

induced by the inclusion maps $V_{(\lambda)} \hookrightarrow V$ is an isomorphism.

Proof.

Consider the vector space $V$ with linear map $T$ as a $k[t]$-module, where multiplication by $t$ acts as applying $T$. By the discussion around the Jordan normal form, we see that $V$ has a decomposition

$$ V \cong \bigoplus_{i=1}^{n} k[t] / ((t - \alpha_i)^{d_i}) $$

as $k[t]$-modules, where $\alpha_i \in k$ and $d_i \ge 1$. For each $i$, we may consider $k[t] / ((t - \alpha_i)^{d_i}) \subseteq V$ as a subspace. Then for any $v \in k[t] / ((t - \alpha_i)^{d_i})$, applying $(T - \alpha_i)^{d_i}$ to $v$ gives

$$ (T - \alpha_i)^{d_i} v = (t - \alpha_i)^{d_i} v = 0 \in k[t] / ((t - \alpha_i)^{d_i}). $$

This implies that $k[t] / ((t - \alpha_i)^{d_i}) \subseteq V_{(\alpha_i)}$. Because the subspaces $k[t] / ((t - \alpha_i)^{d_i})$ generate $V$, it follows that the subspaces $V_{(\lambda)}$ also generate $V$. Combining with Exercise 4.5.F, we obtain the desired result.

In the above proof, show that actually we can identify $V_{(\lambda)}$ directly as

$$ V_{(\lambda)} = \bigoplus_{\alpha_i = \lambda} k[t] / ((t - \alpha_i)^{d_i}). $$

Hence it immediately follows that $\bigoplus_{\lambda} V_{(\lambda)} \cong V$.

There is an application of the above discussion, called the Jordan–Chevalley decomposition. The theorem is useful in representation theory, but we use it only as a demonstration of the theory we developed so far. To state the theorem, we make some definitions.

Let $k$ be an algebraically closed field, and let $V$ be a finite-dimensional vector space over $k$. A linear map $T : V \to V$ is said to be diagonalizable or semisimple if the linear map

$$ \bigoplus_{\lambda \in k} V_\lambda \to V $$

induced by the inclusions is an isomorphism of vector spaces. ($V_\lambda$ are the ordinary eigenspaces for $T$.)

We have seen an example of linear map that is not semisimple, namely $T = [\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}]$. On the other hand, most matrices (if the entries are chosen randomly) are semisimple.

Show that $T$ is semisimple if and only if, the matrix in Jordan normal form obtained by conjugating $T$ is a diagonal matrix.

Let $T$ be an $n \times n$ matrix with entries in $k$, where $k$ is algebraically closed. If the polynomial $\det(t I - T)$ in the variable $t$ has $n$ distinct roots in $k$, show that $T$ is semisimple.

Let $k$ be an algebraically closed field, and let $V$ be a finite-dimensional vector space over $k$. A linear map $T : V \to V$ is said to be nilpotent if $T^n = 0$ for some $n \ge 1$.

Show that $T : V \to V$ is nilpotent if and only if $V_{(0)} = V$.

We now state the main theorem.

Let $V$ be a finite-dimensional vector space over an algebraically closed field $k$. Then for every linear map $T : V \to V$ there is a decomposition

$$ T = T_{ss} + T_n $$

into a semisimple map $T_{ss}$ and a nilpotent map $T_n$ such that $T_{ss} T_n = T_n T_{ss}$. Moreover, such a decomposition is unique.

We prove this theorem in a series of exercises.

Consider the decomposition $V \cong \bigoplus V_{(\lambda)}^{(T)}$ with respect to $T$. Define $T_{ss}$ by setting

$$ T_{ss} v = \lambda v $$

if $v \in V_{(\lambda)}^{(T)}$, and then extending linearly.

(a)
Show that the eigenspaces with respect to $T_{ss}$ are $V_\lambda^{(T_{ss})} = V_{(\lambda)}^{(T)}$. Deduce that $T_{ss}$ is indeed semisimple.
(b)
Show that $T_{ss} T = T T_{ss}$.
(c)
Define $T_n = T - T_{ss}$. Show that for every $v \in V_{(\lambda)}^{(T)}$, there exists a $k \ge 1$ such that $T_n^k v = 0$. Conclude that $T_n$ is nilpotent.
(d)
Show that $T_{ss} T_n = T_n T_{ss}$.

In the other direction, suppose that $T = T_{ss} + T_n$, where $T_{ss}$ is semisimple, $T_n$ is nilpotent, and they commute.

(a)
Show that $T$ and $T_{ss}$ commute, i.e., $T T_{ss} = T_{ss} T$.
(b)
Show that if $v \in V_{\lambda}^{(T_{ss})}$, then $v \in V_{(\lambda)}^{(T)}$. That is, $V_{\lambda}^{(T_{ss})} \subseteq V_{(\lambda)}^{(T)}$.
(c)
From the decompositions $V \cong \bigoplus_{\lambda} V_{(\lambda)}^{(T)}$ and $V \cong \bigoplus_{\lambda} V_\lambda^{(T_{ss})}$, show that $V_{\lambda}^{(T_{ss})} = V_{(\lambda)}^{(T)}$.
(d)
Conclude that $T_{ss}$ has to be the linear map constructed in the previous exercise.

There is a similar multiplicative version of the Jordan–Chevalley decomposition.

Let $k$ be an algebraically closed field, and let $V$ be a finite-dimensional vector space over $k$. A linear map $T : V \to V$ is said to be unipotent if $T - \id$ is nilpotent.

Let $V$ be a finite-dimensional vector space over an algebraically closed field $k$. Prove that for every invertible linear map $T : V \to V$ there is a decomposition

$$ T = T_{ss} T_u $$

into a semisimple map $T_{ss}$ and a unipotent map $T_u$ such that $T_{ss} T_u = T_u T_{ss}$. Moreover, show that such a decomposition is unique.