M216: Exercise sheet 6

    Warmup questions

  • 1. Let \(f:X\to X\) be a map of sets. Show that if \(f\) is injective then \(f^k\) is injective for each \(k\in \N \).

  • 2. Let \(U_1\leq U_2\leq \dots \leq V\) be an increasing sequence of subspaces of \(V\), so that \(U_m\leq U_n\) whenever \(m\leq n\).

    Show that \(\bigcup _{n\in \N }U_n\leq V\).


  • 3. Let \(\phi =\phi _A\in L(\C ^3)\) where \(A\) is given by

    \begin{equation*} \begin{pmatrix*}[r] 0&1&-1\\-10&-2&5\\-6&2&1 \end {pmatrix*}. \end{equation*}

    • (a) Compute the characteristic and minimum polynomials of \(\phi \).

    • (b) Find bases for the eigenspaces and generalised eigenspaces of \(\phi \).

  • 4. Let \(\phi =\phi _A\in L(\C ^3)\) where \(A\) is given by

    \begin{equation*} \begin{pmatrix*} 0&0&0\\4&0&0\\0&0&5 \end {pmatrix*}. \end{equation*}

    • (a) Compute the characteristic and minimum polynomials of \(\phi \).

    • (b) Find bases for the eigenspaces and generalised eigenspaces of \(\phi \).

    Extra questions

  • 5. Let \(\phi \in L(V)\) be an invertible linear operator on a finite-dimensional vector space and \(\lambda \) an eigenvalue of \(\phi \). Show that \(G_{\phi }(\lambda )=G_{\phi ^{-1}}(\lambda ^{-1})\).

  • 6. Let \(\lambda \in \F \) and define \(J(\lambda ,n)\in M_n(\F )\) by

    \begin{equation*} J(\lambda ,n)= \begin{pmatrix} \lambda &1&0&\dots &0\\ &\ddots &\ddots &\ddots &\vdots \\ &&\ddots &\ddots &0\\ &&&\ddots &1\\ 0&&&&\lambda \end {pmatrix}. \end{equation*}

    Set \(J_n:=J(0,n)\).


    • (a) \(\ker J_n^k=\Span {\lst {e}1k}\).

    • (b) \(\im J_n^k=\Span {\lst {e}1{n-k}}\).

    • (c) \(m_{J(\lambda ,n)}=\pm \Delta _{J(\lambda ,n)}=(x-\lambda )^n\).

    • (d) \(\lambda \) is the only eigenvalue of \(J(\lambda ,n)\) and \(E_{J(\lambda ,n)}(\lambda )=\Span {e_1}\), \(G_{J(\lambda ,n)}(\lambda )=\F ^{n}\).

Please hand in at 4W level 1 by NOON on Friday 17th November

M216: Exercise sheet 6—Solutions

  • 1. Let \(x,y\in X\) be such that \(f^k(x)=f^k(y)\). Thus \(f(f^{k-1}(x))=f(f^{k-1}(y))\) whence, since \(f\) is injective, \(f^{k-1}(x)=f^{k-1}(y)\). Repeat the argument to eventually conclude that \(x=y\) so that \(f^k\) is injective. For a more formal argument, induct on \(k\).

  • 2. Let \(U=\bigcup _{n\in \N }U_n\) and let \(v,w\in U\). Then there are \(n,m\in \N \) with \(v\in U_n\) and \(w\in U_m\). Without loss of generality, assume that \(m\leq n\) so that \(U_m\leq U_n\) whence \(v,w\in U_n\). Since \(U_n\) is a subspace, \(v+\lambda w\in U_n\subseteq U\), for any \(\lambda \in \F \), so that \(U\) is indeed a subspace.

  • 3.

    • (a) We compute the characteristic polynomial: \(\Delta _{\phi }=\Delta _A=-x^{3}-x^{2}+8x+12=(3-x)(x+2)^{2}\). Consequently, \(m_{\phi }\) is either \((x-3)(x+2)^{2}\) or \((x-3)(x+2)\). We try the latter:

      \begin{align*} A-3I_3&= \begin{pmatrix*}[r] -3&1&-1\\-10&-5&5\\-6&2&-2 \end {pmatrix*}& A+2I_{3}&= \begin{pmatrix*}[r] 2&1&-1\\-10&0&5\\-6&2&3 \end {pmatrix*} \end{align*} so that

      \begin{equation*} (A-3I_3)(A+2I_3)= \begin{pmatrix*}[r] -10&-5&5\\0&0&0\\-20&-10&10 \end {pmatrix*}\neq 0. \end{equation*}

      Thus \(m_{\phi }=m_A=(x-3)(x+2)^2\).

    • (b) We deduce that \(G_{\phi }(3)=E_{\phi }(3)=\ker (A-3I_3)\) while \(E_{\phi }(-2)=\ker (A+2I_3)\) and \(G_{\phi }(-2)=\ker (A+2I_3)^2\). We compute these: an eigenvector \(x\) with eigenvalue 3 solves

      \begin{align*} -3x_1+x_3-x_3&=0\\ -2x_1-x_2+x_3&=0 \end{align*} which rapidly yields \(x_1=0\) and \(x_2=x_3\). Thus the \(3\)-eigenspace is spanned by \((0,1,1)\).

      An eigenvector \(x\) with eigenvalue \(2\) solves

      \begin{align*} 2x_1+x_2-x_3&=0\\ -2x_1+0x_2+x_3&=0 \end{align*} giving \(x_2=0\) and \(2x_1=x_3\) so the eigenspace is spanned by \((1,0,2)\).


      \begin{equation*} (A+2I_3)^{2}= \begin{pmatrix*}[r] 0&0&0\\-50&0&25\\-50&0&25 \end {pmatrix*} \end{equation*}

      with kernel spanned by \((1,0,2)\) and \((0,1,0)\).

      To summarise:

      \begin{align*} E_{\phi }(3)=G_{\phi }(3)&=\Span {(0,1,1)}\\ E_{\phi }(-2)&=\Span {(1,0,2)}\\ G_{\phi }(-2)&=\Span {(1,0,2),(0,1,0)}. \end{align*}

  • 4.

    • (a) Since \(A\) is lower triangular, we immediately see that \(\Delta _{\phi }=\Delta _A=x^{2}(x-5)\). So the only possibilities for \(m_{\phi }=x(x-5)\) and \(x^2(x-5)\). However

      \begin{equation*} A-5I_3= \begin{pmatrix*}[r] -5&0&0\\4&-5&0\\0&0&0 \end {pmatrix*} \end{equation*}

      so that

      \begin{equation*} A(A-5I_3)= \begin{pmatrix*}[r] 0&0&0\\-20&0&0\\0&0&0 \end {pmatrix*}\neq 0. \end{equation*}

      We conclude that \(m_{\phi }=x^2(x-5)\).

      Alternatively, \(A\) is block diagonal:

      \begin{equation*} A= \begin{pmatrix} 0&0\\4&0 \end {pmatrix}\oplus \begin{pmatrix} 5 \end {pmatrix} \end{equation*}

      and the summands clearly have minimum polynomials \(x^2\) and \(x-5\) respectively. It follows from a previous sheet that \(m_{\phi }=x^2(x-5)\).

    • (b) We have \(E_{\phi }(5)=G_{\phi }(5)=\Span {(0,0,1)}\), \(E_{\phi }(0)=\ker A=\Span {(0,1,0)}\) and finally \(G_{\phi }(0)=\ker A^2=\Span {(1,0,0),(0,1,0)}\) since

      \begin{equation*} A^{2}= \begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&0&25 \end {pmatrix}. \end{equation*}

  • 5. Note that \((\phi -\lambda \id _V)^{n}(v)=0\) if and only if \(\lambda ^{-n}\phi ^{-n}(\phi -\lambda ^n\id _V)^n(v)=0\), that is \((\lambda ^{-1}\id _V-\phi )^n(v)=0\). Thus \(G_{\phi }(\lambda )=G_{\phi ^{-1}}(\lambda ^{-1})\).

    Here, of course, we need \(\lambda \neq 0\) but, since \(\phi \) is invertible, zero is not an eigenvalue.

  • 6. Note that \(\phi _{J_n}(x)=(x_2,\dots ,x_n,0)\) so that \(\phi _{J_n}^k(x)=(x_{k+1},\dots ,x_{n},0,\dots ,0)\), \(k<n\) and \(\phi _{J_n}^{n}=0\).

    • (a) It is clear from the above that \(\ker J_n^k=\set {x\in \F ^{n}\st x_{k+1}=\dots =x_n=0}=\Span {\lst {e}1k}\).

    • (b) Similarly, \(\im J_n^k=\set {y\in \F ^n\st y_{n-k+1}=\dots =y_n=0}=\Span {\lst {e}1{n-k}}\).

    • (c) \(J(\lambda ,n)\) is upper triangular so that \(\Delta _{J(\lambda ,n)}=(\lambda -x)^n\). Therefore \(m_{J(\lambda ,n)}=(x-\lambda )^{s}\), for some \(s\leq n\). However \((J(\lambda ,n)-\lambda I_n)^{k}=J_n^k\neq 0\), for \(k<n\), so that \(m_{J(\lambda ,n)}=(x-\lambda )^{n}\).

    • (d) Finally, it is clear that \(\lambda \) is the only eigenvalue and the eigenspace is \(\ker (J(\lambda ,n)-\lambda I_{n})=\ker J_n=\Span {e_1}\) by part (a). Similarly, \(G_{J(\lambda ,n)}(\lambda )=\ker J_n^n=\F ^n\).