Exercises 2A — Axler, Linear Algebra Done Right, 4th Edition
Exercise 1. Find a list of four distinct vectors in \(\mathbf{F}^3\) whose span equals \[ \{(x,y,z)\in\mathbf{F}^3 : x+y+z=0\}. \]
Solution. Let \(a\in\mathbf{F}\) with \(a\neq 0\). The following list of four distinct vectors works:
\[ (0,0,0),\quad (a,-a,0),\quad (a,0,-a),\quad (0,a,-a). \qquad\square \]
Exercise 2. Prove or give a counterexample: if \(v_1,v_2,v_3,v_4\) spans \(V\), then the list \[ v_1-v_2,\quad v_2-v_3,\quad v_3-v_4,\quad v_4 \] also spans \(V\).
Proof. The statement is true. Let \(v\in V\). Since \(v_1,v_2,v_3,v_4\) spans \(V\), there exist \(a_1,a_2,a_3,a_4\in\mathbf{F}\) with \[ v=a_1v_1+a_2v_2+a_3v_3+a_4v_4. \]
Define \(b_1=a_1\), \(b_2=a_1+a_2\), \(b_3=a_1+a_2+a_3\), \(b_4=a_1+a_2+a_3+a_4\), so that \(a_1=b_1\), \(a_2=b_2-b_1\), \(a_3=b_3-b_2\), \(a_4=b_4-b_3\). Then
\[ \begin{aligned} v &= a_1v_1+a_2v_2+a_3v_3+a_4v_4 && \text{[}v_1,\dots,v_4 \text{ spans } V\text{]} \\ &= b_1v_1+(b_2-b_1)v_2+(b_3-b_2)v_3+(b_4-b_3)v_4 && \text{[substitution]} \\ &= b_1(v_1-v_2)+b_2(v_2-v_3)+b_3(v_3-v_4)+b_4 v_4 && \text{[regroup]} \\ \Rightarrow\ v &\in \operatorname{span}(v_1-v_2,\,v_2-v_3,\,v_3-v_4,\,v_4). \end{aligned} \]
Since \(v\in V\) was arbitrary, \[ \therefore\ v_1-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4 \text{ spans } V. \qquad\square \]
Exercise 3. Suppose \(v_1,\dots,v_m\) is a list of vectors in \(V\). For \(k\in\{1,\dots,m\}\) let \[ w_k=\sum_{i=1}^{k} v_i. \] Show that \(\operatorname{span}(v_1,\dots,v_m)=\operatorname{span}(w_1,\dots,w_m)\).
Proof. We show each span is contained in the other.
\[ \begin{aligned} w_k=\sum_{i=1}^{k} v_i &\in \operatorname{span}(v_1,\dots,v_m) && \text{[def. of } w_k,\ \forall k\in\{1,\dots,m\}] \\ \Rightarrow\ \operatorname{span}(w_1,\dots,w_m) &\subseteq \operatorname{span}(v_1,\dots,v_m). && \text{[span of a sublist]} \end{aligned} \]
For the reverse inclusion, note \(w_1=v_1\), and for \(k\in\{2,\dots,m\}\),
\[ \begin{aligned} v_k &= \sum_{i=1}^{k} v_i - \sum_{i=1}^{k-1} v_i && \text{[telescoping]} \\ &= w_k - w_{k-1} && \text{[def. of } w_k,\ w_{k-1}] \\ \Rightarrow\ v_k &\in \operatorname{span}(w_{k-1},w_k)\subseteq\operatorname{span}(w_1,\dots,w_m). \end{aligned} \]
Since \(v_1=w_1\) and each \(v_k\in\operatorname{span}(w_1,\dots,w_m)\), \[ \operatorname{span}(v_1,\dots,v_m)\subseteq\operatorname{span}(w_1,\dots,w_m). \]
\[ \therefore\ \operatorname{span}(v_1,\dots,v_m)=\operatorname{span}(w_1,\dots,w_m). \qquad\square \]
Exercise 5. Find a number \(t\) such that \((3,1,4),\ (2,-3,5),\ (5,9,t)\) is not linearly independent in \(\mathbf{R}^3\).
Solution. The list fails to be linearly independent exactly when \((5,9,t)\) is a linear combination of the other two vectors, i.e. when there exist \(x,y\in\mathbf{R}\) with \(x(3,1,4)+y(2,-3,5)=(5,9,t)\). The first two coordinates give
\[ \begin{aligned} 3x+2y &= 5 \\ x-3y &= 9. \end{aligned} \]
Row-reducing the augmented matrix of these two equations:
\[ \left[\begin{array}{cc|c} 3 & 2 & 5 \\ 1 & -3 & 9 \end{array}\right] \longrightarrow \left[\begin{array}{cc|c} 1 & -3 & 9 \\ 0 & 11 & -22 \end{array}\right] \longrightarrow \left[\begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & -2 \end{array}\right] \ \Rightarrow\ x=3,\ y=-2. \]
The third coordinate then forces
\[ t = 4x+5y = 4(3)+5(-2) = 12-10 = 2. \]
\[ \therefore\ t=2 \text{ makes the list linearly dependent.} \qquad\square \]
Exercise 7.
(a) If \(\mathbb{C}\) is regarded as a vector space over \(\mathbb{R}\), show that \(1+i,\ 1-i\) is linearly independent.
(b) If \(\mathbb{C}\) is regarded as a vector space over \(\mathbb{C}\), show that \(1+i,\ 1-i\) is linearly dependent.
Proof of (a). Let \(a,b\in\mathbb{R}\) and suppose \(a(1+i)+b(1-i)=0\). Then
\[ \begin{aligned} 0 = a(1+i)+b(1-i) &= (a+b) + (a-b)i && \text{[collect real/imaginary parts]} \\ \Rightarrow\ a+b=0 \ \text{ and }\ a-b=0 && \text{[a complex number is } 0 \text{ iff both parts are } 0] \\ \Rightarrow\ a=0 \ \text{ and }\ b=0. && \text{[solve]} \end{aligned} \]
\[ \therefore\ 1+i,\ 1-i \text{ is linearly independent over } \mathbb{R}. \qquad\square \]
Proof of (b). Work over \(\mathbb{C}\) and let \(a=a_R+a_I i\) with \(a_R,a_I\in\mathbb{R}\). We solve \(a(1+i)=1-i\):
\[ \begin{aligned} a(1+i) &= (a_R+a_I i)(1+i) && \text{[substitution]} \\ &= (a_R-a_I) + (a_R+a_I)i && \text{[expand, } i^2=-1] \\ \overset{!}{=}\ 1-i \ \Rightarrow\ a_R-a_I=1 \ \text{ and }\ a_R+a_I=-1 && \text{[match parts]} \\ \Rightarrow\ a_R=0,\ a_I=-1 \ \Rightarrow\ a=-i. && \text{[solve]} \end{aligned} \]
Indeed \(-i(1+i) = -i - i^2 = 1-i\), so \(1-i = -i\,(1+i)\) is a nontrivial linear combination.
\[ \therefore\ 1+i,\ 1-i \text{ is linearly dependent over } \mathbb{C}. \qquad\square \]
Exercise 10. Prove or give a counterexample: if \(v_1,\dots,v_m\) is a linearly independent list in \(V\) and \(\lambda\in\mathbf{F}\) with \(\lambda\neq 0\), then \(\lambda v_1,\lambda v_2,\dots,\lambda v_m\) is linearly independent.
Proof. The statement is true. Suppose, for contradiction, that \(\lambda v_1,\lambda v_2,\dots,\lambda v_m\) is linearly dependent. Then there exist \(a_1,\dots,a_{m-1}\in\mathbf{F}\) such that
\[ \begin{aligned} \lambda v_m &= a_1\lambda v_1+a_2\lambda v_2+\dots+a_{m-1}\lambda v_{m-1} && \text{[linear dependence]} \\ v_m &= a_1 v_1+a_2 v_2+\dots+a_{m-1}v_{m-1} && \text{[divide both sides by } \lambda\neq 0\text{]} \\ \Rightarrow\ v_m &\in \operatorname{span}(v_1,v_2,\dots,v_{m-1}). \end{aligned} \]
But this contradicts the assumption that \(v_1,v_2,\dots,v_m\) is linearly independent.
\[ \therefore\ \lambda v_1,\lambda v_2,\dots,\lambda v_m \text{ is linearly independent.} \qquad\square \]
Exercise 11. Prove or give a counterexample: if \(v_1,\dots,v_m\) is linearly independent in \(V\) and \(w_1,\dots,w_m\) is a list of vectors in \(V\), then \(v_1+w_1,\ v_2+w_2,\dots,\ v_m+w_m\) is linearly independent.
Proof. The statement is false. For a counterexample, let \(w_i := \lambda v_i\) where \(\lambda = -1\). By Exercise 10, since \(v_1,\dots,v_m\) is linearly independent and \(\lambda\neq 0\), the list \(w_1,\dots,w_m\) is linearly independent (so this is a legitimate choice of list). But note
\[ \begin{aligned} v_i + w_i &= v_i + (-1)v_i = 0 && \text{[}\forall i\in\{1,2,\dots,m\}] \\ \Rightarrow\ v_1+w_1,\,v_2+w_2,\dots,v_m+w_m &= 0,\,0,\dots,0. \end{aligned} \]
\[ \therefore\ v_1+w_1,\,v_2+w_2,\dots,v_m+w_m \text{ is linearly dependent.} \qquad\square \]
Exercise 18. Prove that \(\mathbf{F}^\infty\) is infinite-dimensional.
Proof. Suppose, for contradiction, that \(\mathbf{F}^\infty\) is finite-dimensional. Then some finite list of vectors spans \(\mathbf{F}^\infty\); let \(m\) be the length of such a spanning list. In a vector space, the length of every linearly independent list is at most the length of every spanning list, so
\[ \text{every linearly independent list in } \mathbf{F}^\infty \text{ has length } \le m. \]
For \(i\in\{1,\dots,m+1\}\) let \(e_i\in\mathbf{F}^\infty\) be the sequence with \(1\) in coordinate \(i\) and \(0\) in every other coordinate. Suppose \(a_1,\dots,a_{m+1}\in\mathbf{F}\) satisfy
\[ \begin{aligned} a_1 e_1+a_2 e_2+\dots+a_{m+1}e_{m+1} &= 0 \\ \Rightarrow\ (a_1,a_2,\dots,a_{m+1},0,0,\dots) &= (0,0,0,\dots) && \text{[def. of the } e_i] \\ \Rightarrow\ a_i &= 0 \quad \forall i\in\{1,\dots,m+1\}. \end{aligned} \]
Hence \(e_1,\dots,e_{m+1}\) is a linearly independent list of length \(m+1>m\), contradicting the bound above.
\[ \therefore\ \mathbf{F}^\infty \text{ is infinite-dimensional.} \qquad\square \]
Exercise 20. Suppose \(p_0,p_1,\dots,p_m\) are polynomials in \(\mathcal{P}_m(\mathbf{F})\) such that \(p_k(2)=0\) for each \(k\in\{0,\dots,m\}\). Prove that \(p_0,p_1,\dots,p_m\) is not linearly independent in \(\mathcal{P}_m(\mathbf{F})\).
Proof. Since \(p_k(2)=0\), the number \(2\) is a root of \(p_k\), so \((z-2)\) divides \(p_k\): for each \(k\in\{0,\dots,m\}\) there is a polynomial \(q_k\) with
\[ p_k(z) = (z-2)\,q_k(z),\qquad q_k\in\mathcal{P}_{m-1}(\mathbf{F}). \]
(The factor \((z-2)\) has degree \(1\), so each \(q_k\) has degree at most \(m-1\).) Now \(q_0,q_1,\dots,q_m\) is a list of \(m+1\) vectors in \(\mathcal{P}_{m-1}(\mathbf{F})\), which is spanned by the \(m\) monomials \(1,z,\dots,z^{m-1}\). Since the length of a linearly independent list cannot exceed the length of a spanning list,
\[ m+1 > m \ \Rightarrow\ q_0,q_1,\dots,q_m \text{ is linearly dependent.} \]
Hence there exist \(b_0,b_1,\dots,b_m\in\mathbf{F}\), not all zero, with
\[ \begin{aligned} b_0 q_0+b_1 q_1+\dots+b_m q_m &= 0 && \text{[linear dependence of the } q_k] \\ (z-2)\big(b_0 q_0+\dots+b_m q_m\big) &= 0 && \text{[multiply by } (z-2)] \\ b_0 p_0+b_1 p_1+\dots+b_m p_m &= 0 && \text{[}p_k=(z-2)q_k]. \end{aligned} \]
This is a nontrivial linear combination of \(p_0,\dots,p_m\) equal to \(0\).
\[ \therefore\ p_0,p_1,\dots,p_m \text{ is not linearly independent in } \mathcal{P}_m(\mathbf{F}). \qquad\square \]