Chapter 5 Eigenvalues and Eigenvectors

Section 5A Invariant Subspaces

5.1 definition: operator

A linear map from a vector space to itself is called an operator.

The reason we want to study invariant subspaces is because we want to express the original space $V$ by the direct sum of some smaller subspace. But we also want to each smaller subspace $V_k$ maps to it self under $T$.

5.2 definition: invariant subspace

Suppose $𝑇 ∈ ℒ(𝑉$). A subspace $𝑈$ of $𝑉$ is called invariant under $𝑇$ if $𝑇𝑢 ∈ 𝑈$ for every $𝑢 ∈ 𝑈$.

5.4 Examples

$\{0\}, V, \text{range } T, \text{null } T$ are all invariant under $T$.

We investigate the simplest possible nontrivial invariant subspaces—invariant subspaces of dimension one.

5.5 definition: eigenvalue

Suppose $𝑇 ∈ ℒ(𝑉)$. A number $𝜆 ∈ 𝐅$ is called an eigenvalue of $𝑇$ if there exists $𝑣 ∈ 𝑉$ such that $𝑣 ≠ 0$ and $𝑇𝑣 = 𝜆𝑣$.

5.7 equivalent conditions to be an eigenvalue

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝜆 ∈ 𝐅$. Then the following are equivalent.

(a) $𝜆$ is an eigenvalue of $𝑇$.

(b) $𝑇 − 𝜆𝐼$ is not injective.

(c) $𝑇 − 𝜆𝐼$ is not surjective.

(d) $𝑇 − 𝜆𝐼$ is not invertible.

Proof:

If $\lambda$ is an eigenvalue of $𝑇$, then we can find $u \neq 0$ such that $T u = \lambda u$, then $(T - \lambda I) (u) = 0$ so $\text{null } (𝑇 − \lambda 𝐼)$ is not $\{0\}$, so $𝑇 − \lambda 𝐼$ is not injective.

Conditions (b), (c), and (d) are equivalent by 3.65.

$\square$

5.8 definition: eigenvector

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝜆 ∈ 𝐅$ is an eigenvalue of $𝑇$. A vector $𝑣 ∈ 𝑉$ is called an eigenvector of $𝑇$ corresponding to $𝜆$ if $𝑣 ≠ 0$ and $𝑇𝑣 = 𝜆𝑣$.

a vector $𝑣 ∈ 𝑉$ with $𝑣 ≠ 0$ is an eigenvector of $𝑇$ corresponding to $𝜆$ if and only if $𝑣 ∈ \text{null } (𝑇 − \lambda 𝐼)$.

5.11 linearly independent eigenvectors

Suppose $𝑇 ∈ ℒ(𝑉)$. Then every list of eigenvectors of $𝑇$ corresponding to distinct eigenvalues of $𝑇$ is linearly independent.

Summary of Proof

Assume they are linearly dependent. Then find the minimal number of $v_i$ such that

$$a_1 v_1 + \cdots + a_m v_m = 0$$

And $a_i \neq 0$. Then apply $T- \lambda_m I$ to it we can reach a contradition.

$\square$

5.12 operator cannot have more eigenvalues than dimension of vector space

Suppose $𝑉$ is finite-dimensional. Then each operator on $𝑉$ has at most $\dim 𝑉$ distinct eigenvalues.

5.17 multiplicative properties

$$(𝑝𝑞)(𝑇) = 𝑝(𝑇)𝑞(𝑇); \\ 𝑝(𝑇)𝑞(𝑇) = 𝑞(𝑇)𝑝(𝑇). \\ $$

5.18 null space and range of $𝑝(𝑇)$ are invariant under $𝑇$

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑝 ∈ 𝒫(𝐅)$. Then $\text{null } 𝑝(𝑇)$ and $\text{range } 𝑝(𝑇)$ are invariant under $𝑇$.

Proof:

If $u \in \text{range } p(T)$, then we can find $v$ such that $p(T)(v) = u$, then $T(u) = T(p(T)(v)) = P(T)(T(v)) \in \text{range } P(T)$.

If $u \in \text{null } P(T)$, then $P(T)(T(u)) = T(P(T)(u))= T(0)$. So $T(u) \in \text{null } P(T)$.

$\square$

Section 5B The Minimal Polynomial

5.19 existence of eigenvalues

Every operator on a finite-dimensional nonzero complex vector space has an eigenvalue.

Proof:

Let $n = \dim V$.

Consider $v \neq 0$ and $v, Tv, \cdots, T^nv$ is a group of $n+1$ vectors. Since $\dim V = n$, then $v, Tv, \cdots, T^nv$ are linear dependent.

Then we can find a $k \leq n$, such that $v, Tv, \cdots, T^{k-1}v$ are linear independent, but $v, Tv, \cdots, T^kv$ is linear dependent.

So we can find

$$a_0v + a_1 T^1v + \cdots + a_k T^kv = 0$$

In particular, $a_k \neq 0$.

Then consider $p(z) = a_0 + a_1 z^1 + \cdots + a_k z^k$. From the fundamental theorem of algebra, it has a zero $\lambda$.

Then $p(z) = (z-\lambda ) q(z)$ where $q(z) = b_0 + b_1 z^1 + \cdots + b_{k-1} z^{k-1}$. In particular, $b_{k-1} = a_k \neq 0$.

Then

$$\begin{align*} q(T)(v) &= (b_0 + b_1 T^1 + \cdots + b_{k-1} T^{k-1})(v) \\ &= b_0v + b_1 T^1v + \cdots + b_{k-1} T^{k-1}v \\ &\neq 0 & \\ (&\because v, Tv, \cdots, T^{k-1}v \text{ are linear independent and } b_{k-1} \neq 0) \end{align*}$$

Thus $0 = p(T)(v) = (T-\lambda I)(q(T)(v))$, so $T$ has an eigenvalue $\lambda$ and $q(T)(v)$ is an eigenvector.

$\square$

5.20 example

an operator on a complex vector space with no eigenvalues

$$T(p(z)) = zp(z)$$

5.21 definition: monic polynomial

A monic polynomial is a polynomial whose highest-degree coefficient equals 1.

5.22 existence, uniqueness, and degree of minimal polynomial

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Then there is a unique monic polynomial $𝑝 ∈ 𝒫(𝐅)$ of smallest degree such that $𝑝(𝑇) = 0$. Furthermore, $\deg 𝑝 ≤ \dim 𝑉$.

Proof:

We use induction on the $\dim V$. When $\dim V = 0$, then $p(z) = 1$, so $p(T) = I$ which is a zero operator on $V$.

Assume when $\dim V \leq n-1$, it is true. Now assume $\dim V = n$.

Given $v \neq 0$, then $v, Tv, \cdots, T^nv$ are linear dependent. Then we can find a $1 \leq k \leq n$, such that $v, Tv, \cdots, T^{k-1}v$ are linear independent, but $v, Tv, \cdots, T^kv$ is linear dependent.

So we can find $a_0, \cdots, a_k$ which are not all zero such that

$$a_0v + a_1 T^1v + \cdots + a_k T^kv = 0$$

In particular, $a_k \neq 0$, otherwise $a_0v + a_1 T^1v + \cdots + a_{k-1} T^{k-1}v = 0$ then $a_0 = \cdots = a_k = 0$.

multiply by $a^{-1}_k$, we get

$$b_0v + b_1 T^1v + \cdots + T^kv = 0$$

Let $p(z) = z^k + \cdots + b_1z + b_0$.

Consider the $\text{null } p(T)$, note

$$p(T)(T^i(v)) = T^i(p(T)(v)) = T^i(0) = 0$$

So $v, Tv, \cdots, T^{k-1}v$ are a group of linear independent vectors in $\text{null } p(T)$. Thus $\text{dim null } P(T) \geq k$.

From the fundamental theorem of linear mapping

$$\text{dim range } p(T) = \dim V - \text{dim null } p(T) \leq n-k$$

From 5.18 null space and range of $𝑝(𝑇)$ are invariant under $𝑇$, we know $\text{range } p(T)$ is invariant under $T$. Then we limit $T$ on $\text{range } p(T)$ and use induction to find a monic polynomial $q(z)$, such that $q(T|_{\text{range } p(T)}) = 0$.

Then consider the monic polynomial $q(z)p(z)$. Given any $v$

$$\begin{align*} q(T) p(T) (v) &= q(T) (p(T)(v)) \\ &= q(T|_{\text{range } p(T)}) (p(T)(v)) \\ &\because p(T)(v) \in \text{range } p(T) \\ &= 0 \end{align*}$$

Furthermore $\deg q(z) \leq n-k, \deg p(z) = k$, so $\deg q(z)p(z) \leq n$.

So we finished the induction.

$\square$

5.24 definition: minimal polynomial

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Then the minimal polynomial of $𝑇$ is the unique monic polynomial $𝑝 ∈ 𝒫(𝐅)$ of smallest degree such that $𝑝(𝑇) = 0$.

5.27 eigenvalues are the zeros of the minimal polynomial

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$.

(a) The zeros of the minimal polynomial of $𝑇$ are the eigenvalues of $𝑇$.

Proof:

Assume $\lambda$ is a zero of $p(z)$ which is the minimal polynomial of $𝑇$. Then from the fundamental theorem of algebra, we have $p(z) = (z-\lambda)q(z)$, since $\deg q < \deg p$ then $q(z)$ is not the minimal polynomial of $𝑇$, then we can find $v$ such that $q(T)(v) \neq 0$. Then

$$\begin{align*} 0 = p(T)(v) = (T-\lambda I)(q(T)(v)) \end{align*}$$

So $\lambda$ is an eigenvalue and $q(T)(v)$ is an eigenvector.

On the other hand, if $\lambda$ is an eigenvalue and $v$ is an eigenvector, then

$$\begin{align*} 0 = p(T)(v) &=(a_0 + a_1 T + \cdots + a_{n} T^{n})(v) \\ &= a_0v + a_1 Tv + \cdots + a_{n} T^{n}v \\ &= a_0v + a_1 \lambda v + \cdots + a_{n} \lambda ^{n}v \\ &= (a_0 + a_1 \lambda + \cdots + a_{n} \lambda ^{n})v \\ \end{align*}$$

Since $v \neq 0$, then $a_0 + a_1 \lambda + \cdots + a_{n} \lambda ^{n} = 0$. So $p(\lambda) = 0$, then $\lambda$ is a zero of the minimal polynomial $T$.

$\square$

(b) If $𝑉$ is a complex vector space, then the minimal polynomial of $𝑇$ has the form

$$(𝑧 − \lambda_1)⋯(𝑧 − \lambda_𝑚),$$

where $\lambda_1, …, \lambda_𝑚$ is a list of all eigenvalues of $𝑇$, possibly with repetitions.

Proof:

Using the fundamental theorem of algebra, we can represent the minimal polynomial of $T$ as

$$p(z) = (𝑧 − \lambda_1)⋯(𝑧 − \lambda_𝑚).$$

Then use (a), we know $\lambda_1, \cdots, \lambda_m$ are eigenvalues.

$\square$

5.29 $𝑞(𝑇) = 0 ⟺ 𝑞$ is a polynomial multiple of the minimal polynomial

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑞 ∈ 𝒫(𝐅)$. Then $𝑞(𝑇) = 0$ if and only if $𝑞$ is a polynomial multiple of the minimal polynomial of $𝑇$.

Proof:

$\Rightarrow$

Assume $p(z)$ is the minimal polynomial of $T$, then $\deg p \leq \deg q$. Then we can use the division algorithm of polynomial to get $q(z) = s(z)p(z) + r(z)$ where $\deg r < \deg p$.

Then

$$\begin{align*} r(T)(v) &= (q(T) - s(T)p(T))(v) \\ &= q(T)(v) - s(T)p(T)(v) \\ &= 0 - 0 \\ &= 0 \\ \end{align*}$$

So $r(T) = 0$, then $𝑞$ is a polynomial multiple of $p$.

$\square$

$\Leftarrow$

Assume $q(z) = s(z)p(z)$, then

$$\begin{align*} q(T)(v) &= (s(T)p(T))(v) \\ &= s(T)(p(T)(v)) \\ &= s(T)(0) \\ &= 0 \\ \end{align*}$$

$\square$

5.31 minimal polynomial of a restriction operator

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑈$ is a subspace of $𝑉$ that is invariant under $𝑇$. Then the minimal polynomial of $𝑇$ is a polynomial multiple of the minimal polynomial of $𝑇|𝑈$.

Proof:

Assume $p(z)$ is the minimal polynomial of $𝑇$. Then given $v \in V$, we have $p(T)(v) = 0$.

In particular, for any $u \in U$, $p(T)(u) = 0$.

Since for any $u \in U$, $T|_U(u) = T(u)$. $U$ is invariant under $T$, so we have $p(T|_U)(u) = p(T)(u) = 0$.

Then apply 5.29, we have $p(z)$ is a polynomial multiple of the minimal polynomial of $𝑇|𝑈$.

$\square$

5.32

$𝑇$ not invertible $⟺$ constant term of minimal polynomial of $𝑇$ is $0$.

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Then $𝑇$ is not invertible if and only if the constant term of the minimal polynomial of $𝑇$ is $0$.

Proof:

$$\begin{align*} T &\text{ not invertible } \\ & ⟺ \\ \text{null } T &\neq \{0\} \\ & ⟺ \\ 0 &\text{ is an eigenvalue of } T \\ & ⟺ \\ 0 &\text{ is a zero of the minimal polynomial of } T \\ & ⟺ \\ &\text{the constant term of the minimal polynomial of } 𝑇 \text{ is } 0 \end{align*}$$

$\square$

5.33 even-dimensional null space

Suppose $𝐅 = 𝐑$ and $𝑉$ is finite-dimensional. Suppose also that $𝑇 ∈ ℒ(𝑉)$ and $𝑏, 𝑐 ∈ 𝐑$ with $𝑏^2 < 4𝑐$. Then $\text{dim null } 𝑇^2 + 𝑏𝑇 + 𝑐𝐼$ is an even number.

Proof:

From "5.18 null space and range of $𝑝(𝑇)$ are invariant under $𝑇$", we know $\text{null } 𝑇^2 + 𝑏𝑇 + 𝑐𝐼$ is invariant under $T$.

So we can restrict $T$ to $\text{null } 𝑇^2 + 𝑏𝑇 + 𝑐𝐼$.

We reuse the symbol $V$ and let it denote $\text{null } 𝑇^2 + 𝑏𝑇 + 𝑐𝐼$.

We just need to prove $\dim V$ is an even number.

First, we will show $T$ does not have an eigenvalue. We prove by contradition. Assume $k$ is an eigenvalue and $v$ is an eigenvector. Then

$$\begin{align*} 0 &= (𝑇^2 + 𝑏𝑇 + 𝑐𝐼)(v) \\ &= 𝑇^2v + 𝑏𝑇v + 𝑐𝐼v \\ &= (k^2 + bk + c)v \\ \end{align*}$$

Since $𝑏^2 < 4𝑐$ and $k \in \mathbb{R}$, so $k^2 + bk + c \neq 0$, so $v = 0$, then $T$ does not have an eigenvalue.

Now let $U$ be the subspace that is invariant under $T$ with the largest even degree.

If $U \neq V$, we can find $w \not\in U$.

Consider $W = \text{span}(w, Tw)$, $W$ is invariant under $T$. This is because $T^2w = -bTw - cw \in W$.

Now

$$\dim (U+W) = \dim U + \dim W - \dim (U \cap W)$$

Since both $U, W$ are invariant under $T$, so is $U \cap W$.

If $\dim (U \cap W) = 1$, then $U \cap W$ is an one dimensional invariant subspace, then we can find an eigenvalue which is contradictory to the face that $T$ does not have an eigenvalue.

So $\dim (U \cap W) = 0$, then

$$\dim (U+W) = \dim U + 2$$

So we produce an subspace that is invariant under $T$ with bigger even degree, and we reach a contradition.

So $U = V$ and we finished the proof.

$\square$

Section 5C Upper-Triangular Matrices

5.39 conditions for upper-triangular matrix

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑣_1, …, 𝑣_𝑛$ is a basis of $𝑉$. Then the following are equivalent.

(a) The matrix of $𝑇$ with respect to $𝑣_1, …, 𝑣_𝑛$ is upper triangular.

(b) $\text{span}(𝑣_1, \cdots, 𝑣_k)$ is invariant under $𝑇$ for each $𝑘 = 1, …, 𝑛$.

(c) $Tv_k \in \text{span}(𝑣_1, \cdots, 𝑣_k)$ for each $𝑘 = 1, …, 𝑛$.

5.40 equation satisfied by operator with upper-triangular matrix

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑉$ has a basis with respect to which $𝑇$ has an upper-triangular matrix with diagonal entries $𝜆_1, …, 𝜆_𝑛$. Then

$$(𝑇 − 𝜆_1𝐼)⋯(𝑇 − 𝜆_𝑛𝐼) = 0$$

Proof:

Assume the matrix of $𝑇$ with respect to $𝑣_1, …, 𝑣_𝑛$ is upper triangular.

Then $(T - 𝜆_1𝐼) v_1 = Tv_1 - 𝜆_1 v_1 = 𝜆_1 v_1 - 𝜆_1 v_1 = 0$.

So using 5.17(b) the commutativity of multiplication we have

$(𝑇 − 𝜆_1𝐼)⋯(𝑇 − 𝜆_l𝐼)v_1 = 0$ for $l = 1, 2, \cdots, n$.

Now since $T v_2 = a_{12} v_1 + 𝜆_2 v_2$, then $(T-𝜆_2 I) v_2 = a_{12} v_1 \in \text{span}(𝑣_1)$, so $(T-𝜆_1 I)(T-𝜆_2 I) v_2 = 0$.

Again, using 5.17(b) the commutativity of multiplication we have

$(𝑇 − 𝜆_1𝐼)⋯(𝑇 − 𝜆_l𝐼)v_2 = 0$ for $l = 2, \cdots, n$.

We can continue this process, $T v_k = a_{1k} v_1 + \cdots + a_{(k-1)k} v_{k-1} + 𝜆_k v_k$, so $(T - 𝜆_kI) v_k \in \text{span}(𝑣_1, \cdots, 𝑣_{k-1})$

So $(T-𝜆_1 I)\cdots(T-𝜆_k I) v_k = 0$. Then

$(𝑇 − 𝜆_1𝐼)⋯(𝑇 − 𝜆_l𝐼)v_k = 0$ for $l = k, \cdots, n$.

So for any $v_i, i = 1, \cdots , n$, $(𝑇 − 𝜆_1𝐼)⋯(𝑇 − 𝜆_𝑛𝐼)v_i = 0$.

So $(𝑇 − 𝜆_1𝐼)⋯(𝑇 − 𝜆_𝑛𝐼) = 0$.

$\square$

5.41 determination of eigenvalues from upper-triangular matrix

Suppose $𝑇 ∈ ℒ(𝑉)$ has an upper-triangular matrix with respect to some basis of $𝑉$. Then the eigenvalues of $𝑇$ are precisely the entries on the diagonal of that upper-triangular matrix.

Proof:

Assume under basis $v_1, \cdots, v_n$, the matrix of $T$

$$\mathcal{M}(T) = \begin{pmatrix} \lambda_1 & & * \\ & \ddots & \\ 0 & & \lambda_n \\ \end{pmatrix}$$

Since $T v_1 = \lambda_1 v_1$, then $\lambda_1$ is an eigenvalue.

For $k > 1$, let $V_k$ denote $\text{span}(𝑣_1, \cdots, 𝑣_k)$.

We know $(T-\lambda_k I) v_k \in V_{k-1}$. Furthermore, for since $V_{k-1}$ is invariant under $T$, then $(T-\lambda_k I) v_i \in V_{k-1}$ for $i = 1, \cdots, k-1$.

If we restrict $(T-\lambda_k I)$ on $V_k$, then $\text{range } T-\lambda_k I \subseteq V_{k-1}$. Since $\dim V_k = k, \dim V_{k-1} = k-1$, from the fundamental theorem of linear mapping, we have $\text{null } T-\lambda_k I \neq \{0\}$. So we can find $u_k \neq 0$, such that $(T-\lambda_k I) u_k = 0$, so $\lambda_k$ is an eigenvalue.

On the other hand, if $\lambda$ is an eigenvalue of $T$, and $p(z)$ is the minimal polynomial of $T$, then $p(\lambda) = 0$.

Since from 5.40, $(𝑇 − 𝜆_1𝐼)⋯(𝑇 − 𝜆_𝑛𝐼) = 0$, then $q(z) = (z − 𝜆_1)⋯(z − 𝜆_𝑛)$ is a multiple of $p$ from 5.29.

Then $q(\lambda) = 0$, so $\lambda \in \{\lambda_1, \cdots, \lambda_n\}$.

$\square$

5.44 necessary and sufficient condition to have an upper-triangular matrix

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Then $𝑇$ has an upper-triangular matrix with respect to some basis of $𝑉$ if and only if the minimal polynomial of $𝑇$ equals $(𝑧 − 𝜆_1)⋯(𝑧 − 𝜆_𝑚)$ for some $𝜆_1, …, 𝜆_𝑚 ∈ 𝐅$.

Proof:

$\Rightarrow$

Assume $T$ has has an upper-triangular matrix with respect to some basis $v_1, \cdots, v_n$.

And the elements on the diagonal are $\alpha_1, \cdots, \alpha_n$.

Also assume $p(z)$ is the minimal polynomial of $T$.

From 5.40, $(T − \alpha_1 I)⋯(T − \alpha_n I) = 0$, then $q(z) = (z − \alpha_1 )⋯(z − \alpha_n )$ is a multiple of $p(z)$.

So $p(z)$ has to take the form of $(𝑧 − 𝜆_1)⋯(𝑧 − 𝜆_𝑚)$, where $\lambda_1, \cdots, \lambda_m \in \{\alpha_1, \cdots, \alpha_n\}$.

$\Leftarrow$

We use induction for $m$.

If $m = 1$, then $p(z) = z - \lambda_1$ is the minimal polynomial.

Then every $v_k$ is a eigenvector, so $\text{span}(𝑣_1, \cdots, 𝑣_k)$ is invariant for $k = 1, \cdots, n$.

From "5.39 conditions for upper-triangular matrix", given any basis $v_1, \cdots, v_n$, the matrix of $T$ is upper-triangular.

Now consider when $m = 1, \cdots, l-1$, it all holds, then let $m = l$.

Consider $U = \text{range } T-\lambda_l I$. Then from 5.18 $U$ is invariant under $T$ so we can consider $T|_U$.

Since $(T|_U − 𝜆_1 I)⋯(T|_U − 𝜆_{l-1} I) = 0$, So $(z − 𝜆_1)⋯(z − 𝜆_{l-1})$ is a multiple of the minimal polynomial of $T|_U$. Then with our induction, we can find a basis $u_1, \cdots, u_M$ under which the matrix of $T|_U$ is an upper-triangular matrix.

Then from "5.39 conditions for upper-triangular matrix" again, we have $T u_k = T|_U (u_k) \in \text{span}(u_1, \cdots, u_k)$.

Extend $u_1, \cdots, u_M$ to a basis of $V$ by adding $v_1, \cdots, v_N$.

Note $T v_k = (T - \lambda_kI) v_k + \lambda_k v_k \in \text{span}(u_1, \cdots, u_M, 𝑣_1, \cdots, 𝑣_k)$ because $(T - \lambda_kI) v_k \in U = \text{span}(u_1, \cdots, u_M)$.

Then again use 5.39, we have 𝑇 has an upper-triangular matrix with respect to the basis $u_1, \cdots, u_M, v_1, \cdots, v_N$

$\square$

Section 5D

5.48 definition: diagonal matrix

A diagonal matrix is a square matrix that is 0 everywhere except possibly on the diagonal.

5.50 definition: diagonalizable

An operator on $𝑉$ is called diagonalizable if the operator has a diagonal matrix with respect to some basis of $𝑉$.

5.52 definition: eigenspace, $𝐸(𝜆, 𝑇)$

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝜆 ∈ 𝐅$. The eigenspace of $𝑇$ corresponding to $𝜆$ is the subspace $𝐸(𝜆, 𝑇)$ of $𝑉$ defined by

$$𝐸(𝜆, 𝑇) = \text{null } (𝑇 − 𝜆𝐼) = \{𝑣 ∈ 𝑉 ∶ 𝑇𝑣 = 𝜆𝑣\}.$$

Hence $𝐸(𝜆, 𝑇)$ is the set of all eigenvectors of $𝑇$ corresponding to $𝜆$, along with the $0$ vector.

For $𝑇 ∈ ℒ(𝑉)$ and $𝜆 ∈ 𝐅$, the set $𝐸(𝜆, 𝑇)$ is a subspace of $𝑉$ because the null space of each linear map on $𝑉$ is a subspace of $𝑉$. The definitions imply that $𝜆$ is an eigenvalue of $𝑇$ if and only if $𝐸(𝜆, 𝑇) ≠ \{0\}$.

5.54 sum of eigenspaces is a direct sum

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝜆_1, …, 𝜆_𝑚$ are distinct eigenvalues of $𝑇$. Then

$$𝐸(𝜆_1, 𝑇) + ⋯ + 𝐸(𝜆_𝑚, 𝑇)$$

is a direct sum. Furthermore, if $𝑉$ is finite-dimensional, then

$$\dim 𝐸(𝜆_1, 𝑇) +⋯ + \dim 𝐸(𝜆_𝑚, 𝑇) ≤ \dim 𝑉.$$

Proof:

Assume

$$v_1 + \cdots + v_m = 0$$

Since eigenvectors corresponding to distinct eigenvalues are linearly independent (by 5.11), this implies that each $𝑣_𝑘$ equals $0$. Thus $𝐸(𝜆_1, 𝑇) +⋯ + 𝐸(𝜆_𝑚, 𝑇)$ is a direct sum (by 1.45), as desired.

$$\begin{align*} \dim 𝐸(𝜆_1, 𝑇) +⋯ + \dim 𝐸(𝜆_𝑚, 𝑇) &= \dim(𝐸(𝜆_1, 𝑇) ⊕ ⋯ ⊕ 𝐸(𝜆_𝑚, 𝑇)) \\ &≤ \dim 𝑉 \end{align*}$$

where the first line follows from 3.94 and the second line follows from 2.37.

$\square$