MA22020: Exercise sheet 1
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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:
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(i) \(0\in U\);
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(ii) For all \(u_1,u_2\in U\) and \(\lambda \in \F \), \(u_1+\lambda u_2\in U\).
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2. Which of the following subsets of \(\R ^3\) are linear subspaces? In each case, briefly justify your answer.
(a) \(U_1:=\set {(x_1,x_2,x_3)\st x_1^2+x_2^2+x_3^2=1}\) (b) \(U_2:=\set {(x_1,x_2,x_3)\st x_1=x_2}\) (c) \(U_3:=\set {(x_1,x_2,x_3)\st x_1+2x_2+3x_3=0}\)
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3. Which of the following maps \(f:\R ^2\to \R ^2\) 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) \(f(x,y)=(3y^{2},x^3)\).
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4. Let \(U,W\leq V\) be subspaces of a vector space \(V\).
When is \(U\cup W\) also a subspace of \(V\)?
Homework
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5. Which of the following subsets of \(\C ^3\) are linear subspaces over \(\C \)? In each case, briefly justify your answer.
(a) \(U_1:=\set {(z_1,z_2,z_3)\st z_1z_2=1}\) (b) \(U_2:=\set {(z_1,z_2,z_3)\st z_1=\bar {z}_2}\) (c) \(U_3:=\set {(z_1,z_2,z_3)\st z_1+\sqrt {-1}z_2+3z_3=0}\)
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6. 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:
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(a) \(\phi \) injects if and only if \(\lst {w}1n\) is linearly independent.
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(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 \(W\).
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Warmup questions
Please hand in at 4W level 1 by NOON on Thursday 16th October 2025
MA22020: Exercise sheet 1—Solutions
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1. First suppose that \(U\leq V\). 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 \(\lambda \in \F \), 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 \(U\leq V\).
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2.
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(a) \(U_1\) is not a subspace as it does not contain \(0\)!
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(b) \(U_2\) is a subspace: in fact, it is \(\ker \phi _A\) where \(A= \begin {pmatrix} 1&-1&0 \end {pmatrix} \).
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(c) \(U_3\) is a subspace. It is \(\ker \phi _A\) for \(A= \begin {pmatrix} 1&2&3 \end {pmatrix} \).
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3.
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(a) Here \(f\) is linear: it is the map \(\phi _A\) corresponding to the matrix
\(\seteqnumber{0}{}{0}\)\begin{equation*} A= \begin{pmatrix} 5&1\\3&-2 \end {pmatrix}. \end{equation*}
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(b) This is not linear (because of that \(+2\) term). In particular \(f(0,0)=(2,0)\neq 0\)!
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(c) Again \(f(0,0)=(1,0)\neq 0\) so this \(f\) cannot be linear. Of course, we already know this because it is certainly not true that \(\cos (y_1+y_2)=\cos y_1+\cos y_2\).
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(d) Another non-linear map: for example \(f(2x,2y)\neq 2f(x,y)\).
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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.
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5.
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(a) \(0\notin U_1\) so \(U_1\) is not a subspace.
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(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.
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(c) \(U_3=\ker \phi _A\) for \(A= \begin {pmatrix} 1&\sqrt {-1}&3 \end {pmatrix} \) and so is a subspace.
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6.
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(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.
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(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 \(\lambda _i\), \(1\leq i\leq n\).
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