$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 7.1: Invariant Subspaces

To begin our study, we will look at subspaces $$U$$ of $$V$$ that have special properties under an operator $$T$$ in $$\mathcal{L}(V,V)$$.

Definition $$\PageIndex{1}$$: invariant subspace

Let $$V$$ be a finite-dimensional vector space over $$\mathbb{F}$$ with $$\dim(V)\ge 1$$, and let $$T\in \mathcal{L}(V,V)$$ be an operator in $$V$$. Then a subspace $$U\subset V$$ is called an invariant subspace under $$T$$ if

\begin{equation*}
Tu \in U \quad \text{for all $$u\in U$$.}
\end{equation*}
That is, $$U$$ is invariant under $$T$$ if the image of every vector in $$U$$ under $$T$$ remains within $$U$$. We denote this as $$TU = \{ Tu \mid u\in U \} \subset U$$.

Example $$\PageIndex{1}$$

The subspaces $$\kernel(T)$$ and $$\range(T)$$ are invariant subspaces under $$T$$. To see this, let $$u\in\kernel(T)$$. This means that $$Tu=0$$. But, since $$0\in\kernel(T)$$, this implies that $$Tu=0\in \kernel(T)$$. Similarly, let $$u\in \range(T)$$. Since $$Tv\in \range(T)$$ for all $$v\in V$$, we certainly also have that $$Tu \in \range(T)$$.

Example $$\PageIndex{2}$$

Take the linear operator $$T:\mathbb{R}^3\to\mathbb{R}^3$$ corresponding to the matrix

\begin{equation*}
\begin{bmatrix} 1&2&0\\ 1&1&0\\0&0&2 \end{bmatrix}
\end{equation*}

with respect to the basis $$(e_1,e_2,e_3)$$. Then $$\Span(e_1,e_2)$$ and $$\Span(e_3)$$ are both invariant subspaces under $$T$$.

An important special case of Definition 7.1.1 involves one-dimensional invariant subspaces under an operator $$T$$ in $$\mathcal{L}(V,V)$$. If $$\dim(U) = 1$$, then there exists a nonzero vector $$u$$ in $$V$$ such that

$U = \{ au \mid a \in \mathbb{F} \}.$

In this case, we must have

$T u = \lambda u \quad ~\text{for some $$\lambda \in \mathbb{F}$$}.$

This motivates the definitions of eigenvectors and eigenvalues of a linear operator, as given in the next section.

### Contributors

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