MA22020: Exercise sheet 2

    Warmup questions

  • 1. Let \(U_1,U_2,U_3\leq \R ^3\) be the \(1\)-dimensional subspaces spanned by \((1,2,0)\), \((1,1,1)\) and \((2,3,1)\) respectively.

    Which of the following sums are direct?

    • (a) \(U_i+U_j\), for \(1\leq i<j\leq 3\).

    • (b) \(U_1+U_2+U_3\).

  • 2. Let \(V_i\leq V\), for \(\bw 1ik\). Prove the converse of Corollary 2.8: if

    \begin{equation*} \dim \plst {V}1k=\plst {\dim V}1k \end{equation*}

    then the sum \(\plst {V}1k\) is direct.

  • 3. Let \(U\leq V\). Show that congruence modulo \(U\) is an equivalence relation.

  • 4. Let \(U=\Span {(1,-1,0),(0,1,-1)}\leq \R ^{3}\). Determine which, if any, of the following cosets are equal:

    \begin{equation*} (1,2,3)+U,\qquad (3,3,0)+U,\qquad (1,1,1)+U. \end{equation*}

  • 5. Let \(U\leq V\) and \(q:V\to V/U\) the quotient map. Let \(W\) be a complement to \(U\).

    Show that \(q_{|W}:W\to V/U\) is an isomorphism.

    Homework

  • 6. Let \(V\) be a vector space. A linear map \(\pi :V\to V\) is called a projection if \(\pi \circ \pi =\pi \).

    In this case, prove that \(\ker \pi \cap \im \pi =\set {0}\) and deduce that \(V=\ker \pi \oplus \im \pi \).

  • 7. Let \(U,W\leq V\). Define a linear map \(\phi :U\to (U+W)/W\) by \(\phi (u)=u+W\).

    • (a) Use the first isomorphism theorem, applied to \(\phi \), to prove the second isomorphism theorem:

      \begin{equation*} U/(U\cap W)\cong (U+W)/W. \end{equation*}

    • (b) Deduce that, when \(V\) is finite-dimensional,

      \begin{equation*} \dim (U+W)=\dim U+\dim W-\dim (U\cap W). \end{equation*}

Please hand in at 4W level 1 by NOON on Thursday 31st October 2024

MA22020: Exercise sheet 2—Solutions

  • 1.

    • (a) All these sums are direct as each \(U_i\cap U_j=\set 0\).

    • (b) Note that \((2,3,1)=(1,2,0)+(1,1,1)\) and so can be written in two different ways as a sum \(u_1+u_2+u_3\), with each \(u_{i}\in U_{i}\):

      \begin{gather*} (1,2,0)+(1,1,1)+(0,0,0)\\ (0,0,0)+(0,0,0)+(2,3,1). \end{gather*} Thus \(U_1+U_2+U_3\) is not a direct sum.

    This shows us that \(U_i\cap U_j=\set 0\), \(i\neq j\), is not enough to force \(U_1+U_2+U_3\) to be direct.

  • 2. Let \(\cB _i\) be a basis for \(V_i\) and let \(\cB =\lst {\cB }1k\) be the concatenation of these. By Corollary 2.7, it suffices to see that \(\cB \) is a basis for \(\plst {V}1k\). However, \(\cB \) clearly spans and, by hypothesis,

    \begin{equation*} \abs {\cB }=\plst {\dim V}1k=\dim \plst {V}1k. \end{equation*}

    We know from last year that, for any vector space \(W\), a list of \(\dim W\) vectors that span is a basis and so we are done.

  • 3.

    • Reflexive \(v-v=0\in U\) so \(v\equiv v\mod U\).

    • Symmetric If \(v \equiv w\mod U\) then \(v-w\in U\) so that \(w-v=-(v-w)\in U\) and \(w\equiv v\mod U\).

    • Transitive If \(v\equiv w\mod U\) and \(w\equiv u\mod U\), then \(v-w,w-u\in U\) whence \(v-u=(v-w)+(w-u)\in U\) and so \(v\equiv u\mod U\).

  • 4. There are two ways to proceed. The first is to work straight from the definitions: we know that \(v_1+U=v_2+U\) if and only if \(v_1-v_2\in U\), that is, \(v_1-v_2\) is a linear combination of \((1,-1,0)\) and \((0,1,-1)\). Now,

    \begin{equation*} (3,3,0)-(1,2,3)=(2,1,-3)=2(1,-1,0)+3(0,1,-1)\in U \end{equation*}

    so that \((3,3,0)+U=(1,2,3)+U\). On the other hand, trying to solve

    \begin{equation*} (3,3,0)-(1,1,1)=(2,2,-1)=\lambda (1,-1,0)+\mu (0,1,-1), \end{equation*}

    for \(\lambda ,\mu \in \R \), leads to inconsistent equations:

    \begin{equation*} \lambda =2;\qquad \mu =-1;\qquad \mu -\lambda =2. \end{equation*}

    Thus \((2,2,-1)\notin U\) and \((3,3,0)+U\neq (1,1,1)+U\).

    A slicker approach is to observe that \(U=\ker \phi \) where \(\phi :\R ^3\to \R \) is the linear map \(\phi (x_1,x_2,x_3)=x_1+x_2+x_3\) (indeed, \(U\leq \ker \phi \) and \(\dim \ker \phi =2\) by rank-nullity). Now \(v_1+U=v_2+U\) if and only if \(\phi (v_1)=\phi (v_2)\) and we simply compute:

    \begin{equation*} \phi (1,2,3)=\phi (3,3,0)=6, \qquad \phi (1,1,1)=3 \end{equation*}

    to learn that

    \begin{equation*} (1,2,3)+U=(3,3,0)+U\neq (1,1,1)+U. \end{equation*}

  • 5. Let \(v\in V\). Since \(V=U+W\), we write \(v=u+w\) with \(u\in U\) and \(w\in W\). Then, since \(\ker q =U\), \(q(v)=q(u+w)=q(w)\) so that \(\im q_{|W}=\im q=V/U\). Thus \(q_{|W}\) surjects.

    Further, \(\ker q_{|W}=\ker q\cap W=U\cap W=\set {0}\) since \(\ker q=U\). Thus \(q_{|W}\) has trivial kernel and so injects.

  • 6. Let \(v\in \ker \pi \cap \im \pi \). Then there is \(w\in V\) such that \(v=\pi (w)\) since \(v\in \im \pi \). But \(v\in \ker \pi \) also so that

    \begin{equation*} 0=\pi (v)=\pi (\pi (w))=\pi (w)=v. \end{equation*}

    Thus \(\ker \pi \cap \im \pi =\set {0}\) so it remains to show that \(V=\ker \pi +\im \pi \). For this, write \(v=(v-\pi (v))+\pi (v)\). The second summand is certainly in \(\im \pi \) while

    \begin{equation*} \pi (v-\pi (v))=\pi (v)-\pi (\pi (v))=\pi (v)-\pi (v)=0 \end{equation*}

    so the first is in \(\ker \pi \) and we are done.

  • 7.

    • (a) Let \(q:U+W\to (U+W)/W\) be the quotient map. Then \(\phi \) is simply the restriction \(q_{|U}\) of \(q\) to \(U\) and so is linear. Moreover, \(\ker \phi =U\cap \ker q=U\cap W\). Finally, if \(q(u+w)\in (U+W)/W\), then, since \(q(w)=0\),

      \begin{equation*} q(u+w)=q(u)+q(w)=q(u)=\phi (u) \end{equation*}

      so that \(\phi \) is onto. The first isomorphism theorem now reads

      \begin{equation*} U/(U\cap W)=U/\ker \phi \cong \im \phi =(U+W)/W. \end{equation*}

    • (b) When \(V\) is finite-dimensional, we have

      \begin{equation*} \dim U-\dim (U\cap W)=\dim U/(U\cap W)=\dim (U+W)/W=\dim (U+W)-\dim W \end{equation*}

      and rearranging this gives the result.