# 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 $$\plus {\lambda _1}{\lambda _k}=1$$.

## Homework

• 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 $$v=0$$.