Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

7.1: Invariant Subspaces

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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

    Tu \in U \quad \text{for all \(u\in U\).}
    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{bmatrix} 1&2&0\\ 1&1&0\\0&0&2 \end{bmatrix}

    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.