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Mathematics LibreTexts

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 7.1.1. 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 7.1.2. 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 7.1.3. 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.