Chapter 2 Finite-Dimensional Vector Spaces

Exercises 2A

1 Find a list of four distinct vectors in \(\mathbf{F}^3\) whose span equals \(V\):

\[\{(x, y, z) \in \mathbf{F}^3 : x + y + z = 0\}.\]

Answer : \(v_1=(1, -1, 0), v_2=(1, 0, -1), v_3=(2, -1, -1), v_4=(0, 1, -1)\)

Proof :

Apparently, all these four vectors obeys the rule \(x+y+z=0\), thus we have \(span(v_1,v_2,v_3,v_4) \subseteq V\).

Suppose \(v \in V\), we have

\[\begin{aligned} v &= (x, y, z) \\ &= (x, y, -x-y) \\ &= x(1, 0, -1) + y (0, 1, -1) \in span(v_1, v_2, v_3, v_4) \end{aligned}\]

Thus we have \(V \subseteq span(v_1,v_2,v_3,v_4)\).

Take them together, we have \(span(v_1,v_2,v_3,v_4) = V\).

2 Prove or give a counterexample: If \(v_1,v_2,v_3,v_4\) spans \(V\) ,then the list

\[v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4\]

also spans \(V\).

Answer :

Since \(v_1,v_2,v_3,v_4\) spans \(V\), thus the vectors \(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4\) are included in \(V\).

For any \(v \in V\), there exists \(a_1, a_2, a_3, a_4 \in \mathbf{F}\) such that \(v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4\)

Take some operations, we have:

\[\begin{aligned} v &= a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 \\ &= a_1v_1 - a_1v_2 + a_1v_2+ a_2v_2 + a_3v_3 + a_4v_4 \\ &= a_1(v_1 - v_2) + (a_1 + a_2)v_2 + a_3v_3 + a_4v_4 \\ &= a_1(v_1 - v_2) + (a_1 + a_2)v_2 -(a_1 + a_2)v_3 +(a_1 + a_2)v_3 + a_3v_3 + a_4v_4 \\ &= a_1(v_1 - v_2) + (a_1 + a_2)(v_2 -v_3) +(a_1 + a_2 + a_3)v_3 + a_4v_4 \\ &= a_1(v_1-v_2) + (a_1+a_2)(v_2-v_3) +(a_1+a_2+a_3)v_3 - (a_1+a_2+a_3)v_4 + (a_1+a_2+a_3)v_4 + a_4v_4 \\ &= a_1(v_1-v_2) + (a_1+a_2)(v_2-v_3) +(a_1+a_2+a_3)(v_3-v_4) + (a_1+a_2+a_3+a_4)v_4 \end{aligned}\]

Thus \(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4\) spans \(V\).

3 Suppose \(v_1, \ldots, v_m\) is a list of vectors in \(V\). For \(k \in \{1, \ldots, m\}\), let

\[w_k = v_1 + \cdots + v_k\]

show that \(span(v_1, \ldots, v_m)=span(w_1, \ldots, w_m)\) .

Proof :

Since \(w_k = v_1 + \cdots + v_k\) , we get \(v_k = w_k - w_{k-1}\).

Any \(v \in span(v_1, \ldots, v_m)\) is a linear combination of \(v_1, \ldots, v_m\) . Replace each \(v_k\) by \(w's\), we have that \(v\) is a linear combination of \(w_1, \ldots, w_m\) . In other words, \(span(v_1, \ldots, v_m) \subseteq span(w_1, \ldots, w_m)\) .

Interchange the role of \(v's\) and \(w's\), we get \(span(v_1, \ldots, v_m) \supseteq span(w_1, \ldots, w_m)\) .

Thus we have \(span(v_1, \ldots, v_m)=span(w_1, \ldots, w_m)\) .

4

• (a) Show that a list of length one in a vector space is linearly independent if and only if the vector in the list is not 0.
• (b) Show that a list of length two in a vector space is linearly independent if and only if neither of the two vectors in the list is a scalar multiple of the other.