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3.13 Invariant

  1. \(f\)-invariant

    Example: \(f;\R^4\to \R^4\) and \([f]_{e}= \begin{pmatrix} -2 & 0 & 0 & 0 \\ 1 & -2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\)

    We know that \(f(e_1)=-2e_1+e_2,f(e_2)=-2e_2,f(e_3)=2e_3,f(e_4)=e_4\)

    Here is the \(f\)-invariant subspace with different dimension

    1. \(\dim=0\), \(\{0\}\) is \(f\)-invariant

    2. \(\dim =1\),

      • \(\text{span} \{e_2\}\), \(\text{span} \{e_3\}\), and \(\text{span} \{e_4\}\) are \(f\)-invariant subspaces of dim 1.
      • \(\text{span} \{e_1\}\) is not \(f\)-invariant. \(f(e_1) = -2e_1 + e_2 \notin \text{span} \{e_1\}\).
        1. \(\dim =2\), \(\text{span} \{e_1, e_2\}\), \(\text{span} \{e_3, e_4\}\), \(\text{span} \{e_2, e_3\}\), \(\text{span} \{e_2, e_4\}\).
        2. \(\dim =3\), \(\text{span} \{e_1, e_2, e_3\}\), \(\text{span}\{e_{1}, e_{2}, e_{4}\}\), \(\text{span} \{e_2, e_3, e_4\}\).
        3. \(\dim = 4\), \(V = \mathbb{R}^4\).

  2. Proposition:

    • \(\ker f\), \(\operatorname{Im}f\) are \(f\)-invariant.
    • \(W\) is a 1-dimensional \(f\)-invariant subspace iff \(W = \operatorname{span} \{w\}\) with \(w\) an eigenvector.
    • \(W, U\) \(f\)-invariant \(\Rightarrow\) \(W + U\), \(W \cap U\) are \(f\)-invariant.

    Ex 2: Find all \(f\)-invariant subspaces in \(\mathbb{R}^3\) with \(f_{A}(v)=Av\) with \(A = \begin{bmatrix} 4 & -2 & 2 \\ 2 & 0 & 2 \\ -1 & 1 & 1 \end{bmatrix}\)
    For \(f_A\), find an infinite family of 2-dimensional \(f\)-invariant subspaces.

    \(\chi_A (X) = \begin{vmatrix} X-4 & 2 & -2 \\ -2 & X & -2 \\ 1 & -1 & X-1 \end{vmatrix}\) \(= X^3 - 5X^2 + 8X - 4\) \(= (X - 1)(X - 2)^2\)

    \(E_{1}=\text{Ker}(A-I)=\text{Ker}\left( \begin{bmatrix} 3 & -2 & 2 \\ 2 & -1 & 2 \\ -1 & 1 & 0 \end{bmatrix}\right)=\text{Ker}\left( \begin{bmatrix} 0 & 1 & 2 \\ 0 & 1 & 2 \\ -1 & 1 & 0 \end{bmatrix}\right)\)

    \(\begin{cases}-x + y = 0 \\y + 2z = 0\end{cases}\Rightarrow x = y = -2z\), \((x,y,z)\in\text{Ker}(A-I)\Leftrightarrow(x,y,z)=z(-2,-2,1)\)

    Thus \(\text{Ker}(A - I) = \text{span} \{ (-2, -2, 1) \}\)

    \(E_{2}=\text{Ker}(A-2I)=\text{Ker} \begin{bmatrix} 2 & -2 & 2 \\ 2 & -2 & 2 \\ -1 & 1 & -1 \end{bmatrix}\), then \((x, y, z) \in \text{Ker}(A - 2I) \Leftrightarrow -x + y - z = 0 \Leftrightarrow x = y - z\)​\(\iff\)​\((x,y,z)=y(1,1,0)+z(-1,0,1)\)

    \(E_{2}=\text{span}\{(1,1,0),(-1,0,1)\}\)

    \(A\text{ is diagonalizable and }A= \begin{bmatrix} -2 & 1 & -1 \\ -2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{pmatrix} 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & 2 & \placeholder{} \\ \placeholder{} & \placeholder{} & 2 \end{pmatrix} \begin{bmatrix} -2 & 1 & -1 \\ -2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}^{-1}\)

    If \(S \leq \mathbb{R}^3\) is \(f\)-invariant.

    i) \(\dim S = 0 \rightarrow S = \{ 0 \}\).

    ii) \(\dim S=1\rightarrow S=\text{span}\{(-2,-2,1)\}\) or \(S = \text{span} \{ (1, 1, 0) \}\) or \(S = \text{span} \{ (-1, 0, 1) \}\).

    iii) \(\dim S = 3 \rightarrow S = \mathbb{R}^3\).

    iv) \(\dim S = 2 \rightarrow S = \text{span}\{\underbrace{ (-2, -2, 1)}_{x'},\underbrace{ \alpha (1, 1, 0) + \beta ( -1, 0, 1)}_{u}\}\).

    \(\mathbb{R}^3 = E_1 \oplus E_2\)

    image​

    image​

  3. \(V\) is an \(n\)-dimensional vector space over \(\mathbb{K}\) (where \(\mathbb{K} = \mathbb{R} \text{ or } \mathbb{C}\)).

    If \(f\) is a nilpotent linear operator with index \(n\).

    Prove that there exists an \(f\)-invariant proper subspace that does not admit an \(f\)-invariant complement.

    Definition: \(f: V \to V\) is a linear operator.

    \(f\) is nilpotent with index \(K\) if \(f^{k} = 0\) and \(f^i \neq 0\) for all \(1 \leq i \leq k - 1\).

    We want a subspace \(S \subsetneq V\) such that:

    i) \(S\) is \(f\)-invariant.

    ii) There does not exist \(T \leq V\) such that \(V = S \oplus T\).

    Remark:
    ​\(\text{Ker}\,f\subset\text{Ker}\,f^{2}\subset\text{Ker}\,f^{3}\subset\ldots\subset \text{Ker}\,f^{n-1}\subset\text{Ker}\,f^{n}=V\)

    Let \(v \not\in \text{Ker} \, f^{n-1}\) and consider \(B = \{ v, f(v), f^2(v), \ldots, f^{n-1}(v) \}\).

    i) \(B\) is a linearly independent set.

    Consider \(x = \alpha_0 v + \alpha_1 f(v) + \alpha_2 f^2(v) + \ldots + \alpha_{n-1} f^{n-1}(v) = 0\).

    \(0 = f^{n-1}(x) = \alpha_0 f^{n-1}(v) + \alpha_1 f^n(v) + \ldots + \alpha_{n-1} f^{2n-2}(v)\)

    Since terms involving \(f^n(v) \ldots f^{2n-2}(v)\) are zero, we have \(= \alpha_0 f^{n-1}(v) = 0, \, \text{but } v \not\in \text{Ker} \, f^{n-1} \Rightarrow \alpha_0 = 0\)

    \(\therefore \, x = \alpha_1 f(v) + \alpha_2 f^2(v) + \ldots + \alpha_{n-1} f^{n-1}(v)\)

    Since \(f^{n-2}(x)=\alpha_{1}f^{n-1}(v)=0\Rightarrow\alpha_{1}=0\).

    \(\ldots\)

    Thus, \(\alpha_0 = \alpha_1 = \ldots = \alpha_{n-1} = 0\).

    \(\therefore B\) is a linearly independent set.

    \(B\) is a basis for \(V\):

    \(\#(B) = n \text{ and } B \text{ is l.i.} \Rightarrow B \text{ is a basis for } V.\)

    \(V \xrightarrow{f} f(v) \xrightarrow{f} f^2(v) \xrightarrow{f} \ldots \xrightarrow{f} f^{n-1}(v) \xrightarrow{f} f^n(v) \xrightarrow{f} 0\).

    Define: \(S = \text{span} \{ f(v), f^2(v), \ldots, f^{n-1}(v) \}\) and \(\dim S = n - 1\)

    \(f(S)=\text{span}\{f^{2}(v),f^{3}(v),\ldots,f^{n-1}(v)\}\subseteq S.\)​

    Hence, \(S\) is \(f\)-invariant!

    ii) Does \(S\) admit an \(f\)-invariant complement?

    Consider \(x \in V \setminus S\) such that \(\{f(v), f^2(v), \ldots, f^{n-1}(v), x\}\) is a basis for \(V\) (\(V = S \oplus T\)).

    Let \(T = \text{span} \{ x \}\).

    If \(T\) is \(f\)-invariant, then \(f(x) = \alpha x\) where \(\alpha \in \mathbb{K}\).

    \(\therefore T = E_{\alpha} \quad (x \text{ is an eigenvector relative to } \alpha)\)

    Since \(f\) is nilpotent, \(0 = f^n(x) = \alpha f^{n-1}(x) = \alpha^2 f^{n-2}(x) = \ldots = \alpha^n x\).

    \(\Rightarrow \alpha = 0 \Rightarrow f(x) = 0 \Rightarrow x \in \text{Ker} \, f\)

    Notice that: \(\dim \text{Ker} \, f = 1\).

    And \(f(f^{n-1}(v)) = 0\).

    \(\therefore \text{span} \{ x \} = \text{Ker} \, f = \text{span} \{ f^{n-1}(v) \}\)

    Thus, \(T\) is not \(f\)-invariant.

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