Math 5718 - Solutions 2
Spring 2002
Book problem 2.2.
Let $w_1=v_1-v_2, w_2=v_2-v_3, \ldots w_n=v_n$. Because $\{v_n\}$ is a linearly independent set, we know that

\begin{displaymath} a_1v_1+ \cdots + a_nv_n = 0 \quad \mbox{only if} \quad a_1=a_2=\cdots = a_n=0. \end{displaymath}

Now let

\begin{displaymath} b_1w_1+\cdots + b_nw_n=0. \end{displaymath}

Replacing $w_i$ by $v_i$ in the previous expression, we have

\begin{displaymath} b_1(v_1-v_2) + b_2(v_2-v_3)+\cdots + b_nv_n= b_1v_1 + (b_2-b_1)v_2 + \cdots + (b_n-b_{n-1})v_n=0. \end{displaymath}

Because $\{v_i\}$ is a linearly independent set, we can conclude that $b_1=b_2-b_1=\cdots= b_n-b_{n-1}=0$. These relations imply that $b_1=b_2=\cdots=b_n=0$. Thus, $\{w_i\}$ is a linearly independent set.
Book problem 2.9.
The given space has dimension four, so we must find four basis vectors. Each degree ($0$ through $3$) must be represented in the basis, but the basis vectors needn't have precisely degrees $0,1,2$, and $3$ (remember that degree means the highest power that appears in the polynomial). For example, the set $\{1,x,x^3+x^2,x^3\}$ is linearly independent. Furthermore, any polynomial $p \in P_4({\bf R})$ can be expressed in terms of this basis:

\begin{displaymath} p(x)=a_0+a_1x+a_2x^2+a_3x^3 = a_0 \cdot 1 + a_1 \cdot x + a_2 (x^3+x^2) + (a_3-a_2)x^3. \end{displaymath}

Thus, the set $\{1,x,x^3+x^2,x^3\}$ is a basis.
Book problem 2.10.
To prove this assertion, you must construct the subspaces $U_1,\ldots, U_n$. Let $\{v_1,\ldots,v_n\}$ be a basis for $V$. Now let $U_j = \mbox{ span }\{v_j\}$ for $j=1,\ldots,n$; that is, $U_j=\{av_j:a \in {\bf F}\}$. You should note or prove that each $U_j$ is a subspace. Then, because $\{v_1,\ldots,v_n\}$ is a basis, every vector $v \in V$ can be written uniquely in the form

\begin{displaymath} v=a_1v_1+ \ldots a_nv_n, \end{displaymath}

where $a_1,\ldots,a_n$ are scalars. It follows from the definition of direct sum that $V=U_1 \oplus \cdots \oplus U_n$. Equivalently, Proposition 2.19 may be used.
Book problem 2.13.
By Theorem 2.13,

\begin{displaymath} 8 = \mbox{ dim } {\bf R}^8 = \mbox{ dim } (U+V) = \mbox{ di... ...V - \mbox{ dim } (U \cap V) = 5+3 - \mbox{ dim } (U \cap V). \end{displaymath}

It follows that $\mbox{ dim } (U \cap V)=0$; therefore, $U \cap V=\{0\}$.
Book problem 2.15.
Perhaps the simplest counterexample can be found in ${\bf R}^2$. Let $U_1=\{(x,0): x \in {\bf R}\}$ (the $x$-axis), $U_2=\{(0,y): y \in {\bf R}\}$ (the $y$-axis), and $U_3=\{(z,z): z \in {\bf R}\}$ (the line $y=x$). Then $\mbox{ dim } (U_1+U_2+U_3)= \mbox{ dim } {\bf R}^2=2$ and $\mbox{ dim } U_i=1$ for $i=1,2,3$. The intersection of all pairs of subspaces is $\{0\}$. Thus, the given relationship is not true in this case.
Book problem 2.16.
Let $\{v_i^j\}$ represent a basis of $U_j$ for $j=1,2,\ldots,m$; these bases may have different lengths. The entire set $\{v_i^j\}$ spans $U_1+\cdots + U_m$, so a basis of $U_1+\cdots U_m$ can have a length of at most $\mbox{ dim } U_1 + \cdots + \mbox{ dim } U_m$. This says that

\begin{displaymath} \mbox{ dim } (U_1+\cdots + U_m) \le \mbox{ dim } U_1 + \cdots + \mbox{ dim } U_m. \end{displaymath}

The result can also be proved by induction using the fact that

\begin{displaymath} \mbox{ dim } (U_1+U_2) = \mbox{ dim } U_1 + \mbox{ dim } U_... ...im } (U_1 \cap U_2) \le \mbox{ dim } U_1 + \mbox{ dim } U_2. \end{displaymath}

3. Practice with vectors.
With apologies for the typos in the statement of the problem ....
(a)
The set is linearly independent because it is not possible to find nonzero scalars $a$ and $b$ such that

\begin{displaymath} a(1,0,1,0)+b(0,1,0,1)=0. \end{displaymath}

(b)
Recall the general facts that if $m>n$, then $m$ vectors in ${\bf R}^n$ must be linearly dependent and fewer than $n$ vectors in ${\bf R}^n$ cannot span ${\bf R}^n$. In this example, two vectors cannot span ${\bf R}^4$. For example, the vector $(1,2,3,4)$ cannot be written as a linear combination of $v_1$ and $v_2$.
(c)
By the observations in part (b), five vectors in ${\bf R}^3$ cannot be linearly independent. It must be possible to express at least two of them as linear combinations of the other vectors (see parts (d) and (f)).
(d,f)
It's probably easier to answer part (f) first. Note that $w_3=-2w_1$, so $w_3$ (or $w_1$) can be removed from the set without changing the span of the set (by the Throw-Away Lemma). Now we need to find another vector that can be expressed as a linear combination of the other three vectors. One possibility is that $w_2 = -\frac 12 w_4 + \frac 12 w_5$, which means that $w_2$ can be removed from the set without changing its span. The remaining vectors $\{w_1,w_4,w_5\}$ are linearly independent and span ${\bf R}^3$, forming a basis for ${\bf R}^3$. It follows that the entire set $\{w_1,w_2,w_3,w_4,w_5\}$ must also span ${\bf R}^3$.
(e)
We expect to add two new vectors to the set $\{v_1,v_2\}$ to form a basis of ${\bf R}^4$. In $v_1$, the first and third components are coupled (dependent), so one of the new vectors must make either the first or third component independent; the vector $(1,0,0,0)$ would work. Similarly, in $v_2$, the second and fourth components are dependent, so one of the new vectors must make either the second or fourth component independent; the vector $(0,1,0,0)$ would work. There are many other possible ways to extend this set to a basis.
4. Practice with bases and dimensions.
(a)
When working in ${\bf R}^n$, the dimension of a subspace can always be determined by counting the number of independent parameters needed to define the subspace. For example, the subspace $U_1$ is defined by two independent parameters, $x_1$ and $x_2$; the subspace $U_2$ is defined by two independent parameters, $x_3$ and $x_5$ (components 3 and 4 are linked); and the subspace $U_3$ is defined by two independent parameters, $x_1$ and $x_4$. Thus, the dimension of all these subspaces is 2. A basis of $U_1$ is $\{(1,0,0,0,0),(0,1,0,0,0\}$; a basis for $U_2$ is $\{(0,0,1,2,0),(0,0,0,0,1\}$; and a basis for $U_3$ is $\{(1,0,0,0,0),(0,0,0,1,-1\}$. Note that in each case, the basis vectors must match the definition of the corresponding subspace!
(b)
You might be able to find the intersection of $U_2$ and $U_3$ by inspection. A more systematic way is to let $u \in U_2 \cap U_3$. Using the bases from part (a), and letting $a,b,c,d$ be scalars, $u$ must be representable as a vector in $U_2$ and a vector in $U_3$:

\begin{displaymath} u=a(0,0,1,2,0)+b(0,0,0,0,1) = c(1,0,0,0,0)+d(0,0,0,1,-1). \end{displaymath}

Matching components on both sides we see, in order, that $c=0$; then $a=0$; then $2a+d=0$, which implies that $d=0$; then $b-d=0$, which implies that $b=0$. Thus $a=b=c=d=0$, which means that $u=0$ and $u \in U_2 \cap U_3 = \{0\}$.
(c)
Recall that unions of subspaces are rarely subspaces. In this case, $(1,1,0,0,0)$ and $(0,0,0,0,1)$ belong to $U_1 \cup U_2$. However, $(1,1,0,0,0)+(0,0,0,0,1) \not\in U_1 \cup U_2$. Thus, $U_1 \cup U_2$ is not closed under addition, and it is not a subspace.
(d)
We have computed all the dimensions we need except $\mbox{ dim } (U_1 \cap U_2)$ and $\mbox{ dim } (U_1 \cap U_3)$. The vectors of $U_1$ are nonzero in the first two components and the vectors of $U_2$ are nonzero in the last three components. Therefore, $U_1 \cap U_2=\{0\}$, and $\mbox{ dim } (U_1 \cap U_2)=0$. The vectors of $U_1$ and $U_3$ both have a nonzero first component. Therefore vectors of the form $a(1,0,0,0,0)$ belong to $U_1 \cap U_3$, and $\mbox{ dim } (U_1 \cap U_3)=1$. By Theorem 2.18,

\begin{displaymath} \mbox{ dim } (U_1+U_2) = \mbox{ dim } U_1 + \mbox{ dim } U_2 - \mbox{ dim } (U_1 \cap U_2)= 2+2-0=4. \end{displaymath}

Similarly,

\begin{displaymath} \mbox{ dim } (U_1+U_3) = \mbox{ dim } U_1 + \mbox{ dim } U_3 - \mbox{ dim } (U_1 \cap U_3)= 2+2-1=3. \end{displaymath}

(e)
It is not true that ${\bf R}^5 = U_1 \oplus U_2$, because ${\bf R}^5 \ne U_1+ U_2$. For example, $(1,2,3,4,5) \ne U_1+U_2$.
(f)
It is not true that ${\bf R}^5 = U_1 \oplus U_3$, because ${\bf R}^5 \ne U_1+ U_3$. For example, $(0,0,1,0,0) \ne U_1+U_3$