📂Linear Algebra

Invariant Subspaces of Vector Spaces

Overview

Let $\beta = v_{1}, \dots, v_{k}$ be the set of eigenvectors of the linear transformation $T : V \to V$. Then, it can be understood that $T$ maps $\span{\beta}$ to $\span{\beta}$. A subspace that maps itself to itself in this manner is defined as an invariant subspace.

Definition¹

Let $V$ be a vector space, and $T : V \to V$ a linear transformation. A subspace $W$ is called an $T$-invariant subspace if it satisfies the following condition:

$$T(W) \subset W$$

In other words,

$$T(v) \in W\quad \forall v \in W$$

$W$ is an $T$-invariant subspace.

Description

For the linear transformation $T : V \to V$, the following are examples of $T$-invariant subspaces:

1. $\left\{ 0 \right\}$
2. $V$
3. Range $R(T)$
4. Null space $N(T)$
5. Eigenspace $E_{\lambda}$

1 and 2 are trivial. For any subset $A \subset V$, since $T(A) \subset R(T)$, $R(T)$ is $T$-invariant. Because $0 \in N(T)$, $T(N(T)) \subset N(T)$. Since $T(\lambda x) = \lambda (\lambda x)$, $T(E_{\lambda}) \subset E_{\lambda}$.

If $W$ is an invariant subspace of $T : V \to V$, a restriction map $T|_{W} : W \to W$ can naturally be defined. In this case, $T|_{W}$ inherits the properties of $T$, and the following theorem shows one relationship between $T$ and $T|_{W}$. Simply put, the characteristic polynomial of $T|_{W}$ is a factor of the characteristic polynomial of $T$. The conclusion itself can also be obtained as a corollary of another theorem.

Theorem

Let $V$ be a dimension vector space of dimension $n$, $T : V \to V$ a linear transformation, and $W$ an $T$-invariant. Then, the characteristic polynomial of $T|_{W}$ divides the characteristic polynomial of $T$.

Proof

Choose an ordered basis $\gamma = \left\{ v_{1} ,\dots, v_{k} \right\}$ of $W$. Then, extend it to an ordered basis $\beta = \left\{ v_{1}, \dots, v_{k}, v_{k+1}, \dots, v_{n} \right\}$ of $V$. Let them be $A = \begin{bmatrix} T \end{bmatrix}_{\beta}$ and $B_{1} = \begin{bmatrix} T|_{W} \end{bmatrix}_{\gamma}$, respectively. Then, the matrix $A$ can be represented as the following block matrix.

$$A = \begin{bmatrix} B_{1} & B_{2} \\ O & B_{3} \end{bmatrix}$$

Let $f(t)$ be the characteristic polynomial of $T$, and $g(t)$ the characteristic polynomial of $T|_{W}$. Then, by the determinant formula for block matrices (where $I$ is an identity matrix of appropriate dimension for matrix calculation), the following is obtained:

$$f(t) = \det(A-tI) = \det \begin{bmatrix} B_{1}-tI & B_{2} \\ O & B_{3}-tI \end{bmatrix} = g(t) \det(B_{3}-tI)$$

Therefore, $g(t)$ divides $f(t)$.

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See Also

• Cyclic subspace