M216: Exercise sheet 9

    Warmup questions

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

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

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


  • 3. Prove at least one of the following assertions:

    • (a) Let \(E,F\leq V^{*}\). Then

      \begin{align*} \sol (E+F)&=(\sol E)\cap (\sol F)\\ (\sol E)+(\sol F)&\leq \sol (E\cap F) \end{align*} with equality if \(V\) is finite-dimensional.

    • (b) Let \(U,W\leq V\). Then

      \begin{align*} \ann (U+W)&=(\ann U)\cap (\ann W)\\ (\ann U)+(\ann W)&\leq \ann (U\cap W) \end{align*} with equality if \(V\) is finite-dimensional.

  • 4. 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

  • 5. 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.

    • (a) Show that \(\iota ^T:V^{*}\to U^{*}\) is the restriction map: thus \(\iota ^T(\alpha )=\alpha _{|U}\) with kernel \(\ann U\).

      If \(V\) is finite-dimensional, show that \(\iota ^{T}\) is surjective and deduce that \(V^{*}/\ann U\cong U^{*}\).

    • (b) Show that \(q^{T}:(V/U)^{*}\to V^{*}\) is injective with \(\im q^T\leq \ann U\). If \(V\) is finite-dimensional, show that \(q^T\) is an isomorphism \((V/U)^{*}\to \ann U\).

  • 6. 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.

Please hand in at 4W level 1 by NOON on Friday December 8th

M216: Exercise sheet 9—Solutions

  • 1. 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 \(\ann U\).

  • 2. 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*}

  • 3.

    • (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) of the notes. 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.

  • 4. 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 Question 2.

  • 5.

    • (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 2.

    • (b) All we need to know about \(q\) is that it is a linear surjection with kernel \(U\). Then, by Question 4, \(\ker q^T=\ann (\im q)=\ann V/U=\set 0\) (any \(\alpha \in (V/U)^{*}\) that vanishes on \(V/U\) is zero by definition!) so that \(q^T\) injects. Moreover, Question 4 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.

  • 6. 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.

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.