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M216: Exercise sheet 9

    Warmup questions

  1. Let \(\lst \alpha 1k\) span \(E\leq V^{*}\). Show that

    \begin{equation*} \sol E=\bigcap _{i=1}^k\ker \alpha _i. \end{equation*}

  2. Let \(U\leq V\). Show that \(\ann U\leq V^{*}\).

  3. Let \(V\) be finite-dimensional and \(U\leq V\). Show that

    \begin{equation*} \dim \ann U+\dim U=\dim V. \end{equation*}

    Homework

  4. Prove at least one of the following assertions:

  5. Let \(\phi \in L(V,W)\) be a linear map of vector spaces. Show that

    \begin{align*} \ker \phi ^T&=\ann (\im \phi )\\ \im \phi ^T&\leq \ann (\ker \phi ) \end{align*} with equality if \(V,W\) are finite-dimensional.

    Extra questions

  6. Let \(U\leq V\) and let \(\iota :U\to V\) be the inclusion map (so that \(\iota (u)=u\), for all \(u\in U\)) and \(q:V\to V/U\) the quotient map.

  7. Recall the linear injection \(\ev :V\to V^{**}\). For \(U\leq V\), show that \(\ev (U)\leq \ann (\ann U)\) with equality if \(V\) is finite-dimensional.


M216: Exercise sheet 9—Solutions

  1. Let \(v\in \sol E\) so that \(\alpha (v)=0\), for all \(\alpha \in E\). Then, in particular, each \(\alpha _i(v)=0\) so that \(v\in \ker \alpha _i\), for \(\bw 1ik\). That is, \(v\in \bigcap _{i=1}^k\ker \alpha _i\) and \(\sol E\leq \bigcap _{i=1}^k\ker \alpha _i\).

    Conversely, let \(v\in \bigcap _{i=1}^k\ker \alpha _i\) so that \(\alpha _i(v)=0\), for \(\bw 1ik\). Let \(\alpha \in E\). Then \(\alpha =\sum _{i=1}^k\lambda _i\alpha _i\), for some \(\lst \lambda 1k\in \F \), since the \(\alpha _i\) span \(E\), and

    \begin{equation*} \alpha (v)=\sum _{i=1}^k\lambda _i\alpha _i(v)=0 \end{equation*}

    so that \(v\in \sol E\). Thus \(\bigcap _{i=1}^k\ker \alpha _i\leq \sol E\) and we are done.

  2. Firstly, \(0\in \ann U\) so \(\ann U\neq \emptyset \). So we just check that \(\ann U\) is closed under addition and scalar multiplication. Let \(\alpha _1,\alpha _2\in \ann U\) and \(u\in U\). Then, \(\alpha _1(u)=\alpha _2(u)=0\) so that \((\alpha _1+\alpha _2)(u)=0+0=0\) whence \(\alpha _1+\alpha _2\in \ann U\) also. Similarly, for \(\alpha \in \ann U\) and \(\lambda \in \F \), \((\lambda \alpha )(u)=\lambda \alpha (u)=\lambda 0=0\) so that \(\lambda \alpha \in \ann U\).

    Alternatively, note that restriction to \(U\), \(\alpha \mapsto \alpha _{|U}\) is a linear map \(V^{*}\to U^{*}\) with kernel \(\ann U\).

  3. Let \(\lst {v}1k\) be a basis of \(U\) and extend to a basis \(\lst {v}1n\) of \(V\). Let \(\dlst {v}1n\) be the dual basis. Now observe that \(\alpha \in V^{*}\) is in \(\ann U\) if and only if \(\alpha (v_j)=0\), for \(\bw 1jk\). Thus, writing \(\alpha =\sum _{i=1}^n\alpha (v_i)v^{*}_i\), we see that \(\alpha \in \ann U\) if and only if \(\alpha \in \Span {v^{*}_i\st \bw {k+1}in}\). Thus \(\ann U=\Span {v^{*}_i\st \bw {k+1}in}\) so that

    \begin{equation*} \dim \ann U=n-k=\dim V-\dim U. \end{equation*}

  4. (a) \(E,F\leq E+F\) so \(\sol (E+F)\leq \sol E,\sol F\) whence \(\sol (E+F)\leq (\sol E)\cap (\sol F)\). Conversely, if \(v\in (\sol E)\cap (\sol F)\) then \(\alpha (v)=\beta (v)=0\), for all \(\alpha \in E\) and \(\beta \in F\). Thus, for \(\alpha +\beta \in E+F\), \((\alpha +\beta )(v)=0+0=0\) so that \(v\in \sol (E+F)\). We conclude that \((\sol E)\cap (\sol F)\leq \sol (E+F)\) and so \((\sol E)\cap (\sol F)=\sol (E+F)\).

    For the second statement, \(E\cap F\leq E,F\) so that \(\sol E,\sol F\leq \sol (E\cap F)\) whence \((\sol E) + (\sol F)\leq \sol (E\cap F)\) by Proposition 2.1(2). For equality when \(V\) is finite-dimensional, we show that both subspaces have the same dimension using the first part, the formula for \(\sol E\) and the dimension formula1. The dimension formula gives

    \begin{align*} \dim ((\sol E)+(\sol F))&=\dim \sol E+\dim \sol F-\dim ((\sol E)\cap (\sol F))\\ &=\dim \sol E+\dim \sol F-\dim \sol (E+F), \intertext {using the first part,} &=\dim V-\dim E+\dim V-\dim F-(\dim V-\dim (E+F))\\ &=\dim V-\dim (E\cap F), \intertext {by the dimension formula again,} &=\dim \sol (E\cap F). \end{align*}

    (b) First we note that if \(X\leq Y\leq V\) then \(\ann Y\leq \ann X\): if \(\alpha \in \ann Y\), then \(\alpha _{|Y}=0\) and so, in particular, \(\alpha _{|X}=0\), that is \(\alpha \in \ann X\).

    We now put this to work: \(U,W\leq U+W\) so \(\ann (U+W)\leq \ann U,\ann W\) whence \(\ann (U+W)\leq (\ann U)\cap (\ann W)\). For the converse, if \(\alpha \in (\ann U)\cap (\ann W)\) we have \(\alpha _{|U}=0\) and \(\alpha _{|W}=0\). So if \(v=u+w\in U+W\) then \(\alpha (v)=\alpha (u)+\alpha (w)=0+0=0\) so that \(v\in \ann (U+W)\). Thus \(\ann (U+W)=(\ann U)\cap (\ann W)\).

    For the second statement, \(U\cap W\leq U,W\) so that \(\ann U,\ann W\leq \ann (U\cap W)\) and then \((\ann U)+(\ann W)\leq \ann (U\cap W)\) by Proposition 2.1(2). For equality when \(V\) is finite-dimensional, we argue as in part (a). The dimension formula says

    \begin{align*} \dim ((\ann U)+(\ann W))&=\dim \ann U+\dim \ann W-\dim ((\ann U)\cap (\ann W))\\ &=\dim \ann U+\dim \ann W-\dim \ann (U+W), \intertext {using the first part,} &=\dim V-\dim U+\dim V-\dim W-(\dim V-\dim (U+W))\\ &=\dim V-\dim (U\cap W), \intertext {by the dimension formula again,} &=\dim \ann (U\cap W). \end{align*} Notice that the arguments for part (b) are essentially identical to those for part (a): the key points are that \(\ann \) and \(\sol \) reverse inclusions and take subspaces to ones of complementary dimension.

  5. Let \(\alpha \in W^{*}\). Then \(\alpha \in \ker \phi ^T\) if and only if \(\alpha \circ \phi =0\) if and only if \(\alpha (\im \phi )=\set 0\), that is \(\alpha \in \ann (\im \phi )\). Thus \(\ker \phi ^T=\ann (\im \phi )\).

    For the second statement, suppose that \(\beta \in \im \phi ^T\) so that \(\beta =\phi ^T(\alpha )=\alpha \circ \phi \), for some \(\alpha \in W^{*}\). Then if \(v\in \ker \phi \), \(\beta (v)=\alpha (\phi (v))=0\) so that \(\beta \in \ann (\ker \phi )\). Thus \(\im \phi ^T\leq \ann (\ker \phi )\).

    For equality when \(V\) is finite-dimensional, recall that we already know from lectures that \(\rank \phi =\rank \phi ^T\) from which we see from rank-nullity that

    \begin{equation*} \dim \im \phi ^T=\rank \phi =\dim V-\dim \ker \phi =\dim \ann (\ker \phi ), \end{equation*}

    where the last equality comes from 3.

  6. (a) For \(\alpha \in V^{*}\) and \(u\in U\), \(\iota ^T(\alpha )(u)=\alpha (\iota (u))=\alpha (u)=\alpha _{|U}(u)\). Thus \(\iota ^T(\alpha )=\alpha _{|U}\) and \(\iota ^T\) is the restriction map. Now \(\ker \iota ^T=\set {\alpha \in V^{*}\st \alpha _{|U}=0}=\ann U\).

    Proposition 2.11 tells us2 that any \(\beta \in U^{*}\) is the restriction of some \(\alpha \in V^{*}\) so that \(\iota ^T\) surjects: \(\im \iota ^{T}=U^{*}\). Thus, the First Isomorphism Theorem, applied to \(\iota ^{T}\), tells us that

    \begin{equation*} V^{*}/\ann U=V^{*}/\ker \iota ^{T}\cong \im \iota ^T=U^{*}. \end{equation*}

    This gives us another approach to Question 3.

    (b) All we need to know about \(q\) is that it is a linear surjection with kernel \(U\). Then, by Question 5, \(\ker q^T=\ann (\im q)=\ann V/U=\set 0\) (any \(\alpha \in (V/U)^{*}\) that vanishes on \(V/U\) is zero by definition!) so that \(q^T\) injects. Moreover, Question 5 tells us that \(\im q^T\leq \ann (\ker q)=\ann U\) with equality when \(V\) is finite-dimensional. Thus, in that case, \(q^T\) is a linear bijection \((V/U)^{*}\to \ann U\) and so an isomorphism.

  7. This is just a matter of not panicking! Let \(f\in \ev (U)\) so that \(f=\ev (u)\), for some \(u\in U\). Let \(\alpha \in \ann U\). We need \(f(\alpha )=0\). But

    \begin{equation*} f(\alpha )=\ev (u)(\alpha )=\alpha (u)=0, \end{equation*}

    since \(\alpha \in \ann U\).

    When \(V\) is finite-dimensional, we know that \(\ev \) is an isomorphism so that \(\dim \ev (U)=\dim U\). Meanwhile

    \begin{equation*} \dim (\ann (\ann U))=\dim V^{*}-\dim \ann U=\dim V-(\dim V-\dim U)=\dim U \end{equation*}

    so that \(\ev (U)\) and \(\ann (\ann U)\) have the same dimension and so coincide.

  8. 1 If \(X,Y\leq W\) then \(\dim (X+Y)+\dim (X\cap Y)=\dim X+\dim Y\).

    2 This is where we use that \(V\) is finite-dimensional.