MA22020: Exercise sheet 4

    Warmup questions

  • 1. Write down matrices \(A\in M_n(\R )\) of the following forms:

    • (a) \(A_1\oplus A_2\oplus A_3\) with each \(A_i\in M_2(\R )\).

    • (b) \(\oplst {A}15\) with each \(A_i\in M_1(\R )\).

    • (c) \(A\in M_3(\R )\) such that \(A\) is not of the form \(\oplst {A}1k\) with \(k>1\).

  • 2. Let \(\lst {V}1k\leq V\) and \(\phi _i\in L(V_i)\), \(\bw 1ik\). Suppose that \(V=\oplst {V}1k\) and set \(\phi =\oplst \phi 1k\).

    • (a) If \(U_i\leq V_i\), \(\bw 1ik\), show that the sum \(\plst {U}1k\) is direct.

    • (b) Prove that \(\im \phi =\oplst {\im \phi }1k\).

  • 3. In the situation of Question 2, prove:

    • (a) \(m_{\phi _i}\) divides \(m_{\phi }\), for each \(\bw 1ik\).

    • (b) If each \(m_{\phi _i}\) divides \(p\in \F [x]\), then \(p(\phi )=0\).

    Thus \(m_{\phi }\) is the monic polynomial of smallest degree divided by each \(m_{\phi _i}\). Otherwise said, \(m_{\phi }\) is the least common multiple of \(m_{\phi _1},\dots ,m_{\phi _k}\).

  • 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 \(\phi \).

    Homework questions

  • 5. Let \(\phi \in L(V)\) be a linear operator on a vector space \(V\).

    Prove that \(\im \phi ^{k}\geq \im \phi ^{k+1}\), for all \(k\in \N \). Moreover, if \(\im \phi ^{k}=\im \phi ^{k+1}\) then \(\im \phi ^k=\im \phi ^{k+n}\), for all \(n\in \N \).

  • 6. 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 \(\phi \).

Please hand in at 4W level 1 by NOON on Thursday 28th November

MA22020: Exercise sheet 4—Solutions

  • 1. There are a gazillion possibilities.

    • (a)

      \begin{equation*} \begin{pmatrix} 1&2&0&0&0&0\\ 2&4&0&0&0&0\\ 0&0&5&6&0&0\\ 0&0&7&8&0&0\\ 0&0&0&0&9&0\\ 0&0&0&0&1&2 \end {pmatrix}= \begin{pmatrix} 1&2\\3&4 \end {pmatrix}\oplus \begin{pmatrix} 5&6\\7&8 \end {pmatrix}\oplus \begin{pmatrix} 9&0\\1&2 \end {pmatrix}. \end{equation*}

    • (b) Any \(5\times 5\) diagonal matrix will do:

      \begin{equation*} \begin{pmatrix} \lambda _1&0&0&0&0\\ 0&\lambda _2&0&0&0\\ 0&0&\lambda _3&0&0\\ 0&0&0&\lambda _4&0\\ 0&0&0&0&\lambda _5 \end {pmatrix}= (\lambda _1)\oplus \dots \oplus (\lambda _5). \end{equation*}

    • (c) Any block matrix with more than one block will have zeros so

      \begin{equation*} A= \begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1 \end {pmatrix} \end{equation*}

      cannot be written \(\oplst {A}1k\) with \(k>1\).

  • 2.

    • (a) Let \(u\in \plst {U}1k\) so we can write \(u=\plst {u}1k\), with \(u_i\in U_i\leq V_i\). However, since the sum of the \(V_i\) is direct, there is only one way to write \(u\) as a sum of elements of the \(V_i\) and so, in particular, as a sum of elements of the \(U_i\). Thus the sum of the \(U_i\) is direct.

    • (b) Let \(v\in \im \phi \) so that \(v=\phi (w)\), for some \(w\in V\). Then, writing \(w=\plst {w}1k\) with each \(w_i\in V_i\), we have

      \begin{equation*} v=\phi (w)=\phi _1(w_1)+\dots +\phi _k(w_k)\in \oplst {\im \phi }1k. \end{equation*}

      Thus \(\im \phi \leq \oplst {\im \phi }1k\).

      For the converse, let \(v\in \oplst {\im \phi }1k\) so that \(v=\phi _1(w_1)+\dots +\phi _k(w_k)\) with \(w_i\in V_i\), \(\bw 1ik\). Since each \(\phi _i=\phi \restr {V_i}\), this reads

      \begin{equation*} v=\phi (w_1)+\dots +\phi (w_k)=\phi (\plst {w}1k)\in \im \phi . \end{equation*}

      Thus \(\oplst {\im \phi }1k\leq \im \phi \) and we are done.

  • 3.

    • (a) We have that \(m_{\phi }(\phi )=0\) so that \(0=m_{\phi }(\phi )\restr {V_i}=m_{\phi }(\phi _i)\). It follows that \(m_{\phi _i}\) divides \(m_{\phi }\).

    • (b) Since \(m_{\phi _i}\) divides \(p\), \(p(\phi _i)=0\) for each \(i\). But then

      \begin{equation*} p(\phi )=p(\phi _1)\oplus \dots \oplus p(\phi _k)=0 \end{equation*}

      so that \(m_{\phi }\) divides \(p\).

  • 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 question 3 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. Let \(v\in \im \phi ^{k+1}\) so that \(v=\phi ^{k+1}(w)\), for some \(w\in V\). Then \(v=\phi ^k(\phi (w))\in \im \phi ^k\). Thus \(\im \phi ^{k}\geq \im \phi ^{k+1}\).

    Suppose now that \(\im \phi ^k=\im \phi ^{k+1}\). We prove that \(\im \phi ^k=\im \phi ^{k+n}\) by induction on \(n\). We are given that this holds for \(n=1\) so we suppose this holds for some \(n\) (\(\im \phi ^k=\im \phi ^{k+n}\)) and prove it then holds for \(n+1\). Thus, let \(v\in \im \phi ^{k}=\im \phi ^{k+1}\) so that \(v=\phi (\phi ^k(w))\), for some \(w\in V\). Then \(\phi ^k(w)\in \im \phi ^k=\im \phi ^{k+n}\), by the induction hypothesis, so that \(\phi ^k(w)=\phi ^{k+n}(u)\), some \(u\in V\), whence \(v=\phi (\phi ^{k+n}(u))=\phi ^{k+n+1}(u)\in \im \phi ^{k+n+1}\). We conclude that \(\im \phi ^k\leq \im \phi ^{k+n+1}\). The converse inclusion always holds so we have equality. Induction now bakes the cake.

  • 6.

    • (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 \((1,0,2)\).

      Finally,

      \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*}