MA22020: Exercise sheet 1

Warmup questions

• 1. Let $U$ be a subset of a vector space $V$. Show that $U$ is a linear subspace of $V$ if and only if $U$ satisfies the following conditions:

  • (i) $0\in U$;

  • (ii) For all $u_1,u_2\in U$ and $\lambda \in \mathbb{F}$, $u_1+\lambda u_2\in U$.

• 2. Which of the following subsets of ${\mathbb{R}}^3$ are linear subspaces? In each case, briefly justify your answer.

  (a) $U_1:=\{(x_1,x_2,x_3)|x_1^2+x_2^2+x_3^2=1\}$ (b) $U_2:=\{(x_1,x_2,x_3)|x_1=x_2\}$ (c) $U_3:=\{(x_1,x_2,x_3)|x_1+2x_2+3x_3=0\}$

• 3. Which of the following maps $f:{\mathbb{R}}^2\to {\mathbb{R}}^2$ are linear? In each case, briefly justify your answer.

  (a) $f(x,y)=(5x+y,3x-2y)$ (b) $f(x,y)=(5x+2,7y)$ (c) $f(x,y)=(\cos y,\sin x)$ (d) $f(x,y)=(3y^2,x^3)$.

• 4. Let $U,W\le V$ be subspaces of a vector space $V$.

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

  Homework

• 5. Which of the following subsets of ${\mathbb{C}}^3$ are linear subspaces over $\mathbb{C}$? In each case, briefly justify your answer.

  (a) $U_1:=\{(z_1,z_2,z_3)|z_1{z}_2=1\}$ (b) $U_2:=\{(z_1,z_2,z_3)|z_1={\overline{z}}_2\}$ (c) $U_3:=\{(z_1,z_2,z_3)|z_1+\sqrt{-1}{z}_2+3z_3=0\}$

• 6. Let $V,W$ be vector spaces, $v_1,\dots ,v_n$ a basis of $V$ and $w_1,\dots ,w_n$ a list of vectors in $W$. Let $\varphi :V\to W$ be the unique linear map with

  $$

  $$\varphi (v_i)=w_i,$$

  for all $1\le i\le n$. Show:

  • (a) $\varphi$ injects if and only if $w_1,\dots ,w_n$ is linearly independent.

  • (b) $\varphi$ surjects if and only if $w_1,\dots ,w_n$ spans $W$.

  Deduce that $\varphi$ is an isomorphism if and only if $w_1,\dots ,w_n$ is a basis for $W$.

Please hand in at 4W level 1 by NOON on Thursday 17th October 2024

MA22020: Exercise sheet 1—Solutions

• 1. First suppose that $U\le V$. The $U$ is non-empty so there is some $u\in U$ and then, since $U$ is closed under addition and scalar multiplication, $0=u+(-1)u\in U$ also and condition (i) is satisfied. Now if $u_1,u_2\in U$ and $\lambda \in \mathbb{F}$, then $\lambda u_2\in U$ ($U$ is closed under scalar multiplication) and so $u_1+\lambda u_2\in U$ ($U$ is closed under addition). Thus condition (ii) holds also.

  For the converse, if conditions (i) and (ii) hold, then, first, $0\in U$ so $U$ is non-empty and, second, $U$ is closed under addition (take $\lambda =1$ in condition (ii)) and under scalar multiplication (take $u_1=0$ in condition (ii)). Thus $U\le V$.

• 2.

  • (a) $U_1$ is not a subspace as it does not contain $0$!

  • (b) $U_2$ is a subspace: in fact, it is $\ker {\varphi }_A$ where $A=\left(\begin{array}{ccc}1& -1& 0\end{array}\right)$.

  • (c) $U_3$ is a subspace. It is $\ker {\varphi }_A$ for $A=\left(\begin{array}{ccc}1& 2& 3\end{array}\right)$.

• 3.

  • (a) Here $f$ is linear: it is the map ${\varphi }_A$ corresponding to the matrix

    $$

    $$A=\left(\begin{array}{cc}5& 1\\ 3& -2\end{array}\right).$$

  • (b) This is not linear (because of that $+2$ term). In particular $f(0,0)=(2,0)\ne 0$!

  • (c) Again $f(0,0)=(1,0)\ne 0$ so this $f$ cannot be linear. Of course, we already know this because it is certainly not true that $\cos (y_1+y_2)=\cos y_1+\cos y_2$.

  • (d) Another non-linear map: for example $f(2x,2y)\ne 2f(x,y)$.

• 4. 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.

• 5.

  • (a) $0\notin U_1$ so $U_1$ is not a subspace.

  • (b) $U_2$ is not a subspace because it is not closed under complex scalar multiplication: $(1,1,0)\in U_2$ but $i(1,1,0)=(i,i,0)$ is not (here $i=\sqrt{-1}$). In general, any time you see complex conjugation in the definition of a subset, it is unlikely to be a complex subspace.

  • (c) $U_3=\ker {\varphi }_A$ for $A=\left(\begin{array}{ccc}1& \sqrt{-1}& 3\end{array}\right)$ and so is a subspace.

• 6.

  • (a) ${\lambda }_1{w}_1+\cdots +{\lambda }_n{w}_n=0$ if and only if ${\lambda }_1{v}_1+\cdots +{\lambda }_n{v}_n\in \ker \varphi$. Thus $w_1,\dots ,w_n$ is linearly independent if and only if $\varphi$ has trivial kernel.

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

    $$

    $$w=\varphi ({\lambda }_1{v}_1+\cdots +{\lambda }_n{v}_n)={\lambda }_1{w}_1+\cdots +{\lambda }_n{w}_n,$$

    for some ${\lambda }_i$, $1\le i\le n$.