M216: Exercise sheet 3

    Warmup questions

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

  • 2. Recall (from Algebra 1A) that if \(f:X\to Y\) is a map of sets and \(A\subset Y\), the inverse image of \(A\) by \(f\) is the subset \(f^{-1}(A)\subset X\) given by

    \begin{equation*} f^{-1}(A):=\set {x\in X\mathrel {|}f(x)\in A}. \end{equation*}

    Now let \(\phi :V\to W\) be a linear map of vector spaces and \(U\leq W\). Prove that \(\ker \phi \leq \phi ^{-1}(U)\leq V\).

  • 3. Let \(U\leq V\) and \(v\in V\). Let \(\lst {v}1k\in v+U\).

    Show that \(\lc {\lambda }v1k\in v+U\) whenever \(\plus {\lambda _1}{\lambda _k}=1\).


  • 4. Let \(\lst {v}1k\in V\). The affine span of \(\lst {v}1k\) is the subset

    \begin{equation*} A(\lst {v}1k):= \set {\lc {\lambda }v1k\mathrel {|}\plus {\lambda _1}{\lambda _k}=1}\subset V. \end{equation*}

    • (a) Show that each \(v_i\in A(\lst {v}1k)\).

    • (b) Show that \(A(\lst {v}1k)\) is a coset of some subspace \(U\leq V\).

      Hint: If this was true, each \(v_i-v_1\in U\): use this to define \(U\).

    Together with question 3, this shows that the affine span is the smallest affine subspace containing \(\lst {v}1k\).

  • 5. 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*}

    Extra questions

  • 6. 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.

  • 7. Let \(U\leq V\) and suppose that \(V/U\) is finite-dimensional. Show that \(U\) has a complement.

Please hand in at 4W level 1 by NOON on Friday 27th October

M216: Exercise sheet 3—Solutions

  • 1.

    • 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 \(v\equiv u\mod U\).

  • 2. First, if \(v\in \ker \phi \) then \(\phi (v)=0\in U\) so \(v\in \phi ^{-1}(U)\). Thus \(\ker \phi \subset \phi ^{-1}(U)\) and, in particular, \(\phi ^{-1}(U)\) is non-empty.

    Next we see that \(\phi ^{-1}(U)\) is closed under addition and scalar multiplication: if \(v,w\in \phi ^{-1}(U)\) and \(\lambda \in \F \), then \(\phi (u),\phi (w)\in U\) so that \(\phi (v+w)=\phi (v)+\phi (w)\in U\) and \(\phi (\lambda v)=\lambda \phi (v)\in U\), since \(U\) is closed under addition and scalar multiplication. Otherwise said, \(v+w,\lambda v\in \phi ^{-1}(U)\) and we are done.

  • 3. Since \(v_i\in v+U\), there is \(u_i\in U\) so that \(v_i=v+u_i\). Now, if \(\plst {\lambda }1k=1\), we have

    \begin{align*} \lc {\lambda }v1k&=\plus {\lambda _1(v+u_1)}{\lambda _k(v+u_k)}\\ &= (\plst {\lambda }1k)v+(\lc {\lambda }u1k)=v+(\lc {\lambda }u1k)\in v+U. \end{align*}

  • 4.

    • (a) Take \(\lambda _i=1\) and \(\lambda _j=0\), for \(i\neq j\), to get \(v_i=\lc {\lambda }v1k\) with \(\plst {\lambda }1k=1\) so that \(v_i\in A(\lst {v}1k)\).

    • (b) If \(A(\lst {v}1k)\) was a coset of a subspace \(U\), we would have to have

      \begin{equation*} A(\lst {v}1k)=v_1+U \end{equation*}

      and then each \(v_i-v_1\), \(2\leq i\leq k\), would lie in \(U\). So set

      \begin{equation*} U:=\Span {v_i-v_1\st 2\leq i\leq k}. \end{equation*}

      Then each \(v_i-v_1\in U\) so that \(v_i\in v_1+U\). Now question 3 tells us that \(A(\lst {v}1k)\subset v+U\).

      Conversely, any element of \(v_1+U\) can be written

      \begin{equation*} v_1+\sum _{i\geq 2}\mu _i(v_i-v_1)=\lc {\lambda }{v}1k \end{equation*}

      with \(\lambda _1=1-\sum _{i\geq 2}\mu _i\) and \(\lambda _j=\mu _j\), for \(j\geq 2\). Thus \(\plst {\lambda }1k=1\) and we conclude that \(v+U\subset A(\lst {v}1k)\).

  • 5.

    • (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.

  • 6. 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.

  • 7. Let \(q:V\to V/U\) be the quotient map and \(q(v_1),\dots ,q(v_k)\) be a basis for \(V/U\), for some \(\lst {v}1k\in V.\) Set \(W=\Span {\lst {v}1k}\leq V\). I claim that \(W\) is a complement to \(U\). So let \(v\in V\). Then there are \(\lst {\lambda }1k\in \F \) so that

    \begin{equation*} q(v)=\plus {\lambda _1q(v_1)}{\lambda _kq(v_k)}=q(\lc \lambda {v}1k). \end{equation*}

    Otherwise said, \(v-(\lc \lambda {v}1k)\in \ker q=U\) so that

    \begin{equation*} v=\lc \lambda {v}1k+u, \end{equation*}

    for some \(u\in U\) and we have \(V=U+W\).

    Now suppose \(v\in U\cap W\). Then we can write \(v=\lc \lambda {v}1k\) since \(v\in W\) but \(v\in \ker q\) so that

    \begin{equation*} 0=q(v)=\plus {\lambda _1q(v_1)}{\lambda _kq(v_k)}. \end{equation*}

    Since the \(q(v_i)\) are linearly independent, we get that each \(\lambda _i=0\) and so \(v=0\).