M216: Exercise sheet 5

    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 \(\im \phi =\oplst {\im \phi }1k\).

    Homework

  • 3. 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 \(n\in \N \).

  • 4. Compute the characteristic and minimum polynomials of

    \begin{equation*} A= \begin{pmatrix*}[r] 1&-5&-7\\1&4&2\\0&1&4 \end {pmatrix*}. \end{equation*}

    Additional questions

  • 5. Let \(\phi \in L(V)\) be a linear operator on a vector space \(V\) and \(v\in V\), \(k\in \N \) such that \(\phi ^{k+1}(v)=0\) but \(\phi ^{k}(v)\neq 0\).

    Show that \(v,\phi (v),\dots ,\phi ^k(v)\) are linearly independent.

    Hint: Induct on \(k\).

  • 6. 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 \(m_{\phi _1},\dots ,m_{\phi _k}\).

Please hand in at 4W level 1 by NOON on Friday 10th November

M216: Exercise sheet 5—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. 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.

  • 4. We compute the characteristic polynomial of \(A\) to be

    \begin{equation*} \Delta _A=-x^3+9x^2-27x+27=-(x-3)^3. \end{equation*}

    We learn from the Cayley–Hamilton theorem that \(m_A=(x-3)^k\), for some \(k\) with \(\bw {k}13\). Clearly \(k=1\) is out, since \(A\) is not diagonal, so we try \(k=2\):

    \begin{equation*} (A-3I)^2= \begin{pmatrix*}[r] -2&-5&-7\\1&1&2\\0&1&1 \end {pmatrix*} \begin{pmatrix*}[r] -2&-5&-7\\1&1&2\\0&1&1 \end {pmatrix*} = \begin{pmatrix*}[r] -1&-2&-3\\ -1&-2&-3\\ 1&2&3 \end {pmatrix*}, \end{equation*}

    which is non-zero. This means we must have \(m_A=(x-3)^3\).

  • 5. We follow the hint and induct on \(k\). For \(k=0\), the assertion is that if \(v\neq 0\) and \(\phi (v)=0\) then the one element list \(v\) is linearly independent which is certainly true (whether or not \(\phi (v)=0\)).

    Suppose now that the result holds for \(\ell <k\) so that \(\phi ^{\ell +1}(w)=0\) and \(\phi ^{\ell }(w)\neq 0\) forces \(w,\dots ,\phi ^{\ell }(w)\) to be linearly independent. Now suppose that \(\phi ^{k+1}(v)=0\) and \(\phi ^k(v)\neq 0\). If there are \(\lambda _i\in \F \) with

    \begin{equation} \label {eq:1} \lambda _1v+\dots +\lambda _k\phi ^k(v)=0, \end{equation}

    then taking \(\phi \) of this gives

    \begin{equation*} \lambda _1\phi (v)+\dots +\lambda _{k-1}\phi ^{k-1}(\phi (v))=0. \end{equation*}

    But with \(w=\phi (v)\), we have \(\phi ^k(w)=0\) and \(\phi ^{k-1}(w)\neq 0\) so the induction hypothesis applies with \(\ell =k-1<k\) to give that \(w,\dots ,\phi ^{k-1}(w)\) are linearly independent. Therefore \(\lst \lambda 1{k-1}\) all vanish. Plugging this back into (1) gives \(\lambda _k\phi ^k(v)=0\) whence \(\lambda _k=0\) also. We conclude that \(v,\dots ,\phi ^k(v)\) are linearly independent and so the result holds for all \(k\in \N \) by induction.

  • 6.

    • (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 \(p\).