M216: Exercise sheet 2

Warmup questions

• 1. Let \(U,W\leq V\) be subspaces of a vector space \(V\).

  When is \(U\cup W\) also a subspace of \(V\)?

• 2. Let \(V,W\) be vector spaces, \(\lst {v}1n\) a basis of \(V\) and \(\lst {w}1n\) a list of vectors in \(W\). Let \(\phi :V\to W\) be the unique linear map with

  \(\seteqnumber{0}{}{0}\)

  \begin{equation*} \phi (v_i)=w_i, \end{equation*}

  for all \(1\leq i\leq n\). Show:

  • (a) \(\phi \) injects if and only if \(\lst {w}1n\) is linearly independent.

  • (b) \(\phi \) surjects if and only if \(\lst {w}1n\) spans \(W\).

  Deduce that \(\phi \) is an isomorphism if and only if \(\lst {w}1n\) is a basis for \(W\).

  Homework

• 3. Let \(V\) be a vector space. A linear map \(\pi :V\to V\) is called a projection if \(\pi \circ \pi =\pi \).

  In this case, prove that \(\ker \pi \cap \im \pi =\set {0}\) and deduce that \(V=\ker \pi \oplus \im \pi \).

• 4. Let \(U_1,U_2,U_3\leq \R ^3\) be the \(1\)-dimensional subspaces spanned by \((1,2,0)\), \((1,1,1)\) and \((2,3,1)\) respectively.

  Which of the following sums are direct?

  • (a) \(U_i+U_j\), for \(1\leq i<j\leq 3\).

  • (b) \(U_1+U_2+U_3\).

  Additional questions

• 5. Let \(V_1,V_2,V_3\leq V\). Which of the following statements are true? (In each case, give a proof or a counter-example.)

  • (a) \(V_1+(V_2\cap V_3)=(V_1+V_2)\cap (V_1+V_3)\).

  • (b) \(V_1\cap (V_2+V_3)=(V_1\cap V_2)+(V_1\cap V_3)\).

  • (c) \((V_1\cap V_2)+(V_1\cap V_3)\subseteq V_1\cap (V_2+V_3)\).

• 6. Let \(V_1,V_1',V_2\leq V\) and suppose that \(V=V_1\oplus V_2\) and \(V=V_1'\oplus V_2\).

  • (a) Must \(V_1=V_1'\)?

  • (b) Are \(V_1\) and \(V_1'\) isomorphic?

Please hand in at 4W level 1 by NOON on Friday 20th October

M216: Exercise sheet 2—Solutions

• 1. If \(U\subseteq W\) then \(U\cup W=W\) is a subspace and similarly if \(W\subseteq U\). In any other case, \(U\cup W\) is not a subspace: we can find \(u\in U\setminus W\) and \(w\in W\setminus U\) and then \(u+w\notin U\) (else \(w=(u+w)-u\in U\)) and similarly \(u+w\notin W\). Thus \(U\cup W\) is not closed under addition.

• 2.

  • (a) \(\lc {\lambda }w1n=0\) if and only if \(\lc {\lambda }v1n\in \ker \phi \). Thus \(\lst {w}1n\) is linearly independent if and only if \(\phi \) has trivial kernel.

  • (b) \(\phi \) surjects if and only if any \(w\in W\) can be written \(w=\phi (v)\), or equivalently,

    \(\seteqnumber{0}{}{0}\)

    \begin{equation*} w=\phi (\lc {\lambda }v1n)=\lc {\lambda }{w}1n, \end{equation*}

    for some \(\lambda _i\), \(1\leq i\leq n\).

• 3. Let \(v\in \ker \pi \cap \im \pi \). Then there is \(w\in V\) such that \(v=\pi (w)\) since \(v\in \im \pi \). But \(v\in \ker \pi \) also so that

  \(\seteqnumber{0}{}{0}\)

  \begin{equation*} 0=\pi (v)=\pi (\pi (w))=\pi (w)=v. \end{equation*}

  Thus \(\ker \pi \cap \im \pi =\set {0}\) so it remains to show that \(V=\ker \pi +\im \pi \). For this, write \(v=(v-\pi (v))+\pi (v)\). The second summand is certainly in \(\im \pi \) while

  \(\seteqnumber{0}{}{0}\)

  \begin{equation*} \pi (v-\pi (v))=\pi (v)-\pi (\pi (v))=\pi (v)-\pi (v)=0 \end{equation*}

  so the first is in \(\ker \pi \) and we are done.

• 4.

  • (a) All these sums are direct as each \(U_i\cap U_j=\set 0\).

  • (b) Note that \((2,3,1)=(1,2,0)+(1,1,1)\) and so can be written in two different ways as a sum \(u_1+u_2+u_3\), with each \(u_{i}\in U_{i}\):

    \(\seteqnumber{0}{}{0}\)

    \begin{gather*} (1,2,0)+(1,1,1)+(0,0,0)\\ (0,0,0)+(0,0,0)+(2,3,1). \end{gather*} Thus \(U_1+U_2+U_3\) is not a direct sum.

  This shows us that \(U_i\cap U_j=\set 0\), \(i\neq j\), is not enough to force \(U_1+U_2+U_3\) to be direct.

• 5.

  • (a) This is false: take \(V_1,V_2,V_3\leq \R ^2\) to be the subspaces spanned at \((1,0)\), \((0,1)\) and \((1,1)\) respectively. Then any \(V_i+V_j=\R ^2\) and \(V_i\cap V_j=\set {0}\), for \(i\neq j\). Now the left side is \(V_1+\set {0}=V_1\) while the right is \(\R ^2\cap \R ^2=\R ^2\).

  • (b) This is also false. With the same \(V_i\) as in part (a), the left side is \(V_1\cap \R ^2=V_1\) while the right is \(\set {0}+\set {0}=\set {0}\).

  • (c) This is true: \(V_2,V_3\leq V_2+V_3\) so that \(V_1\cap V_2, V_1\cap V_3\leq V_1\cap (V_2+V_3)\). It now follows from Proposition 2.1 that \((V_1\cap V_2)+(V_1\cap V_3)\subseteq V_1\cap (V_2+V_3)\).

• 6.

  • (a) No: a given \(V_2\) has many complements. For example, take \(V=\R ^2\), \(V_2\) to be spanned by \((1,0)\) and then \(V_1,V_1'\) to be spanned by \((0,1)\) and \((1,1)\) respectively.

  • (b) This is true. For example, consider the projection \(\pi _1\) with image \(V_1\) and kernel \(V_2\) and restrict this to \(V_1'\) to get a linear map \(V_1'\to V_1\). Then \(\ker (\pi _1{}_{|V_1'})=\ker \pi _1\cap V_1'=V_2\cap V_1'=\set {0}\) so that \(\pi _1{}_{|V_1'}\) injects. Moreover, for \(v_1\in V_1\), write \(v_1=v_1'+v_2\) with \(v_1'\in V_1'\) and \(v_2\in V_2\). Then \(v_1=\pi (v_1)=\pi _1(v_1'+v_2)=\pi _1(v_1')\) so that \(\pi _1{}_{|V_1'}:V_1'\to V_1\) surjects also and so is an isomorphism.