# 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*}

## Homework

• 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.