Lecture 22 (online) - Invariant Subspaces

Invariant Subspaces

Recall our usual setup:

The idea of invariant subspaces asks whether the subspace mapping from $T : U \to U$ maps into $U$ only. In the picture above we have an invariant, but if we had:

Then we don't have an invariant.

Invariant

Suppose $T \in L (V)$ , and we have subspace $U$ of $V$ . The subspace $U$ is an invariant under $T$ if, for all $u \in U$ we have $T u \in U$ .

As some examples of invariant subspaces:

$V$ is an invariant subspace of $V$ .
The zero subspace $U = {\vec{0}}$ is an invariant since $T \vec{0} = \vec{0} \in U$ for any $T$ .

Some cooler ones are as follows. Here $null (T)$ is an invariant since all items get mapped to the zero vector which is in $null (T)$ :

Likewise, the $range (T)$ is also an invariant subspace of $T$ :

Why are we interested?

Suppose $V = U_{1} \oplus U_{2} \oplus \dots \oplus U_{m}$ , where $U_{i}$ are all invariant under $T$ . Suppose any $v = u_{1} + u_{2} + \dots + u_{m} \in V$ . Then $T v = T u_{1} + \dots + T u_{m}$ where each $T u_{i}$ are in $U_{i}$ . Hence, we can reduce our space into our corresponding smaller subspaces to talk about the ultimate higher ones.

As an example, consider $T \in L (V)$ , using $R$ as our field. Let $v \in V$ with $v \neq {\vec{0}}_{V}$ , where $T v = 2 v$ . We claim that $span (v)$ is an invariant subspace of $V$ .

Proof
We need to show that for any $u \in span (v)$ and show that $T u \in span (v)$ . Let $u \in span (v)$ be arbitrary. Then:

u = α v

for some $α \in R$ . Then:

T u = 2 u = 2 (α v) = (2 α) v

but notice that $β = 2 α \in R$ :

T u = β u \in span (v)

thus $span (v)$ is an invariant under $T$ .
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Notice that this transformation merely scaled our vector. We're going to hence define the related eigenvector, and the amount it gets scaled by is its eigenvalue.

Eigenvalues and Eigenvectors

Given $T \in L (V)$ :

A number $λ \in F$ is called an eigenvalue of $T$ if there exists $v \in V$ such that $v \neq 0$ and $T v = λ v$ .
A vector $v \in V$ is called an eigenvector of $T$ corresponding to eigenvalue $λ$ if $v \neq {\vec{0}}_{V}$ and $T v = λ v$ .

For instance, let $V$ be the vector space having basis $B = {e^{2 x}, x e^{2 x}}$ and let $D$ be the derivative map. Note that $D \in L (V)$ already, and we can see what it does to our vectors:

D (e^{2 x}) = 2 e^{2 x}

notice here that $λ = 2$ is an eigenvalue of $D$ . Furthermore, $v = e^{2 x}$ is its corresponding eigenvector.

Note that:

D (0 v) = 0 v = λ \cdot 0 v

for all $λ \in F$ . Does this mean that $0 v$ is an eigenvector? Or that any $λ$ is an eigenvalue? We don't really want this for any other transformations or spaces, hence why we remove these cases in the definition.

Note

We say that $\vec{0}$ isn't an eigenvector, but $0$ could be a possible eigenvalue.

Notice the following. If $v$ is a (non-zero) eigenvector with e-val $λ$ then:

\begin{aligned} T (v) & = λ v \\ T (v) - λ v & = \vec{0} \\ T (v) - λ I_{V} v & = \vec{0} \\ (T - λ I_{V}) v & = \vec{0} \end{aligned}

so then $v \in null (T)$ so then all eigenvectors are in the nullspace of $T$ .

This gives rise to the following theorem:

Equivalences of Eigenvalues/Eigenvectors

The following are equivalent:

$λ$ is an eigenvalue of $T$
$T - λ I_{V}$ is not injective.
$T - λ I_{V}$ is not surjective.
$T - λ I_{V}$ is not invertible.
where $I_{V} : V \to V$ where for any $v \in V$ we have $I_{v} (v) = v$ .

Proof
This is a baby proof, which is more outlined in the book. Suppose (1). Then there is some non-zero vector $v$ such that $T v = λ v$ . Subtracting on both sides gives $T v - λ v = 0 v$ , so then $(T - λ I_{V}) v = 0 v$ . We've found something that maps a non-zero vector $v$ to the zero-vector, so then we've gotten (2).

If we have (2) we get (3, 4) as we've shown before. If you work backwards, you can suppose non-injectivity and extract the $λ$ as an eigenvalue thing.
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Here we note that $null (T - λ I_{V})$ is the eigenspace of $T$ on $V$ .

Theorem

Suppose we have distinct eigenvalues $λ_{1}, \dots, λ_{m}$ are distinct eigenvalues of $T \in L (V)$ and $v_{1}, . . ., v_{m}$ are corresponding eigenvectors, where all $T (v_{i}) = λ_{i} v_{i}$ . Then $v_{1}, . . ., v_{m}$ is LI in $V$ .

Proof
We prove this by contradiction. Assume instead that $v_{1}, . . ., v_{m}$ is LD. Note that $v_{1}$ itself is just LI always, and as we add more vectors, we then get LD. Hence, let $k$ be the smallest positive integer such that $v_{1}, . . ., v_{k}$ is LD. Then $v_{k}$ can be written as a linear combination of $v_{1}, . . ., v_{k - 1}$ . Then:

T v_{k} = α_{1} T v_{1} + \dots + α_{k - 1} T v_{k - 1}

where since all $v_{i}$ 's are eigenvectors, then:

λ_{k} v_{k} = α_{1} λ_{1} v_{1} + \dots + α_{k - 1} λ_{k - 1} v_{k - 1}

multiply our initial definition of $v_{k}$ by $λ_{k}$ to get:

λ_{k} v_{k} = α_{1} λ_{k} v_{1} + \dots + α_{k - 1} λ_{k} v_{k - 1}

But then equating coefficients says that $λ_{1} = λ_{k}, \dots, λ_{k - 1} = λ_{k}$ , as subtracting both equations above gives

\vec{0} = α_{1} (λ_{k} - λ_{1}) v_{1} + \dots + α_{k - 1} (λ_{k} - λ_{k - 1}) v_{k - 1}

But since $v_{1}, . . ., v_{k - 1}$ is LI, then all $a_{i} (λ_{k} - λ_{i}) = 0$ . So either $α_{i} = 0$ or $λ_{k} - λ_{i} = 0$ . We cannot have the latter since all the $λ_{i}$ 's are unique, so then $α_{i} = 0$ , so then $v_{k} = \vec{0}$ is a contradiction as $v_{k}$ is non-zero.
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This is handy since we can construct bases from our corresponding eigenvalues and vectors.

Theorem

Suppose $V$ is a finite-dimensional vector space. Then each linear transformation $T$ on $V$ has at most $dim (V)$ distinct eigenvalues and corresponding eigenvectors.

Proof
Assume you had more than $dim (V)$ eigenvalues. Then that would give you more than that $n$ number of eigenvectors, which are all LI. But since the dimension is $n$ , then we have a LI list of longer than $n$ eigenvectors, which is a contradiction.
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Intro to Polynomial Operators

Given some polynomial $p = a_{0} + a_{1} z + \dots + a_{n} z^{n}$ and some operator $T \in L (V)$ , we can construct a new operator from this polynomial, namely $p (T)$ is the linear operator in $L (V)$ such that:

p (T) (\vec{v}) = (a_{0} I + a_{1} T + \dots + a_{n} T^{n}) (\vec{v})

where

T^{k} = \underset{k times}{\underset{⏟}{T \circ T \circ \dots \circ T}}

we'll talk more about this in the next lecture.