Math 5718 - Solutions 1
Spring 2002
Book problem 1.5.
In each case, let $U$ be the subspace in question.
a.
$U$ is a subspace of ${\bf F}^3$. (i) Clearly, $0 \in U$. (ii) If $u,v \in U$, then $w=u+v$ satisfies $(u_1+v_1) + 2(u_2+v_2) + 3(u_3+v_3)=w_1+2w_2+3w_3=0$, where $u_i,v_i,w_i \in {\bf F}$. This implies that $u+v \in U$. (iii) If $u \in U$ and $a \in {\bf F}$, then $au$ satisfies $au_1+2au_2+3au_3=0$, implying that $au \in U$. (With ${\bf F}= {\bf R}$, $U$ is a plane through the origin, which always forms a subspace.)
b.
$U$ is not a subspace of ${\bf F}^3$ because $0 \not\in U$.
c.
$U$ is not a subspace of ${\bf F}^3$ because it is not closed under addition; for example, $u=(0,1,2) \in U$ and $v= (1,1,0) \in U$, but $u+v \not\in U$.
d.
$U$ is a subspace of ${\bf F}^3$. (i) Clearly, $0 \in U$. (ii) If $u,v \in U$, then $w=u+v=(u_1+v_1,u_2+v_2,u_3+v_3)=(5(u_3+v_3),u_2+v_2,u_3+v_3)=(5w_3,w_2,w_3)$. This implies that $u+v \in U$. (iii) If $u \in U$ and $a \in {\bf F}$, then $au=(au_1,au_2,au_3)=(5au_3,au_2,au_3)$, implying that $au \in U$. (With ${\bf F}= {\bf R}$, $U$ is a plane through the origin, which always forms a subspace.)
Book problem 1.6.
The set $U \subset {\bf R}^2$ given by $U = \{(m,n):m,n \in {\bf Z}\}$ (vectors in the plane with integer coordinates) meets the requirements: it is closed under addition, it contains additive inverses of every vector, but it is not closed under scalar multiplication.
Book problem 1.7.
A set consisting of any two lines in ${\bf R}^2$ passing through the origin works. For example, $U = \{(x,y): x= \pm y\}$ meets the requirements: it is closed under scalar multiplication, but not under addition.
Book problem 1.8.
Let $N=\{1,2,3,\ldots,n\}$ and let $\cap_i U_i$ denote $U_1 \cap \ldots \cap U_n$. (i) Because each $U_i$ is a vector space, $0 \in U_i$ for all $i \in N$. Thus, $0 \in \cap_i U_i$. (ii) Let $u,v \in U_i$ for all $i \in N$. Because each $U_i$ is a vector space, $u+v \in U_i$ for all $i \in N$. Thus, $u+v \in \cap_i U_i$. (iii) Let $a \in {\bf F}$ and $u \in U_i$ for all $i \in N$. Because each $U_i$ is a vector space, $au \in U_i$ for all $i \in N$. Thus, $au \in \cap_i U_i$. We see that $\cap_i U_i$ contains the zero vector and is closed under addition and scalar multiplication; thus, it is a vector space.
Book problem 1.13.
Here is a counterexample in ${\bf R}^3$. Let

\begin{displaymath} U_1 = \{(x,y,0): x,y \in {\bf R}\}\quad U_2 = \{(x,0,0):x \in {\bf R}\} \quad W = \{(0,y,z):y,z \in {\bf R}\}. \end{displaymath}

We see that ${\bf R}^3 = U_1+W$ and ${\bf R}^3 = U_2+W$, but $U_1 \ne U_2$.
Book problem 1.15.
Any three lines in ${\bf R}^2$ through the origin provide a counterexample. For example, taking

\begin{displaymath} U_1 = \{(x,0): x \in {\bf R}\} \quad U_2 = \{(0,y):y \in {\bf R}\} \quad W = \{(x,x):x \in {\bf R}\}, \end{displaymath}

we see that ${\bf R}^3 = U_1 \oplus W$ and ${\bf R}^3 = U_2 \oplus W$, but $U_1 \ne U_2$.
Practice with subspaces.
a.
$U$ is ${\bf R}^2$ viewed as a subset of ${\bf R}^3$. Because ${\bf R}^2$ is itself a vector space, it is a subspace of ${\bf R}^3$.
b.
$U$ is not a subspace of ${\bf R}^3$ because $0 \not\in U$.
c.
$U$ is a not a subspace. It consists of the half-plane above the line $y=-x$ and is not closed under scalar multiplication.
d.
$U$ is a subspace. Note that symmetry of a matrix implies that $m=n$. $U$ contains a zero element, namely the $n \times n$ matrix filled with zeros. Adding two symmetric matrices produces a symmetric matrix. Multiplying a symmetric matrix by a scalar also produces a symmetric matrix. Therefore, the set is closed under addition and scalar multiplication, and is a subspace.
e.
$U$ consists of even polynomials of degree at most four and it is a subspace of ${\cal P}({\bf R})$. First, the zero polynomial $p(x)=0$ belongs to $U$. The sum of two even polynomials of degree at most four is another even polynomial of degree at most four, and a scalar multiple of an even polynomial of degree at most four is another even polynomial of degree at most four.
f.
$U$ is a subspace. (i) Clearly, $0 \in U$ if we take $c_1=c_2=0$. (ii) If $u= c_1(1,1,2) + c_2(0,3,1)$ and $v = d_1(1,1,2) + d_2(0,3,1)$ are two vectors in $U$, then $u+v = (c_1+d_1)(1,1,2) + (c_2+d_2)(0,3,1)$ is also an element of $U$. Similarly, for $u \in U$ and $a \in {\bf F}$, we have $au \in U$.
g.
$U$ is a subspace of ${\bf R}^\infty$. Either note that $U={\bf R}^{22}$ which is a vector space or check the subspace properties: Clearly, $0 \in U$. Adding two such sequences produces another such sequence (we cannot create nonzero elements beyond $x_{23}$ by addition). Multiplying such a sequence by a scalar also produces another such sequence.
h.
$U$ fails to be a subspace because $0 \not\in U$.
i.
$U$ is the space of all convergent sequences and is a subspace of ${\bf R}^\infty$. (i) Certainly the zero sequence $\{0,0,0 \ldots \}$ is convergent. (ii) Adding two convergent sequences results in a convergent sequence, because $\lim_{n \rightarrow \infty}(u_n+v_n) =\lim_{n \rightarrow \infty}u_n + \lim_{n \rightarrow \infty} v_n )$. (iii) Multiplying a convergent sequence by a scalar produces a convergent sequence, because $\lim_{n \rightarrow \infty}(au_n) =a\lim_{n \rightarrow \infty}u_n$, where $a \in {\bf R}$.
j.
I will interpet $U$ to be the set of all arithmetic sequences; it is a subspace of ${\bf R}^\infty$. (i) The zero sequence is an arithmetic sequence. (ii) Suppose we have two sequences that satisfy $x_j-x_{j-1}=c$ and $y_j-y_{j-1}=d$, where $c$ and $d$ are fixed real numbers. Then the sequence $z=x+y$ has elements $\{x_1+y_1, x_2+y_2, \ldots, x_n+y_n, \ldots\}$. The terms of $z$ satisfy $z_j-z_{j-1} = c+d$. Thus, $z$ is also an arithmetic sequence, and $U$ is closed under addition. Similarly, the sequence $z=ax$, where $a \in {\bf F}$ has elements $\{ax_1, ax_2, \ldots ax_n \ldots \}$. We see that the elements of $z$ satisfy $z_j-z_{j-1} = ac$. Thus, $z$ is also an arithmetic sequence, and $U$ is closed under scalar multiplication.
k.
I will interpet $U$ to be the set of all geometric sequences; it is not a subspace of ${\bf R}^\infty$. (i) The zero sequence is a geometric sequence. (ii) Suppose we have two sequences that satisfy $x_j=k_1x_{j-1}$ and $y_j=k_2y_{j-1}$, where $k_1 \ne k_2$ are fixed real numbers. Then the sequence $z=x+y$ has a typical element of the form $z_j = x_j+y_j = k_1x_{j-1}+k_2y_{j-1}$, which cannot be expressed as $k(x_{j-1}+y_{j-1})=kz_{j-1}$, for some fixed real number $k$. The set is not closed under addition and is not a subspace.
Practice with sums and direct sums.
The direct sum problems are easier if you use Proposition 2.19 of the text that gives a condition for direct sums in terms of the dimensions of the subspaces. We didn't have this result at the time these problems were assigned, so I will not use it. Instead, we can use the definition of direct sum or Proposition 1.8 for direct sums of more than two subspaces and we can use Proposition 1.9 only for direct sums of two subspaces.
a.
First observe that ${\bf R}^3 = U+V+Z$. To prove that the sum is a direct sum, it suffices to show that $0=u+v+z$ only when $u=v=z=0$, where $u \in U, v \in V$, and $z \in Z$. Letting $(0,0,0)= (x,0,0)+(0,y,0)+(2z,0,z)$, we see that $y=0$ and $z=0$, which implies that $x=0$.
b.
The statement is not true because $U+V+Y \ne {\bf R}^3$ (for example, (0,0,1) cannot be represented as $u+v+y$).
c.
The statement is true because $W+Z ={\bf R}^3$ and $W \cap Z = \{0\}$. By Proposition 1.9, $W \oplus Z = {\bf R}^3$.
d.
The statement is true because $U+W ={\bf R}^3$ and $U \cap W = \{0\}$. By Proposition 1.9, $U \oplus W ={\bf R}^3$.
e.
The statement is not true because $V+Z \ne {\bf R}^3$; for example, $(1,2,3)$ cannot be represented as $v+z=(2p,q,p)$.
f.
The statement is true. The arbitrary vector $(p,q,r)$ can be expressed as $w+y=(0,q-p,r)+(p,p,0)$, where $w \in W$ and $y \in Y$.
g.
The only direct sum representations I can find, in addition to (a), (c), and (d) above, are $W \oplus Y$, $U \oplus Y \oplus Z$, and $V \oplus Y \oplus Z$.