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Solution to Linear Algebra Done Wrong Exercise 1.2.6

Is it possible that vectors ${\bf v}_1$, ${\bf v}_2$, ${\bf v}_3$ are linearly dependent, but the vectors ${\bf w}_1={\bf v}_1+{\bf v}_2$, ${\bf w}_2={\bf v}_2+{\bf v}_3$, ${\bf w}_3={\bf v}_3+{\bf v}_1$ are linearly independent?

Solution: It is impossible. Let us argue it as follows.

Since ${\bf v}_1$, ${\bf v}_2$, ${\bf v}_3$ are linearly dependent, by Proposition 2.6, we can assume without loss of generality that $${\bf v}_3=x{\bf v}_1+y{\bf v}_2$$for some scalars $x$ and $y$. We would like to show that there exist not all zero scalars $a$, $b$, $c$ such that $$0=a{\bf w}_1+b{\bf w}_2+c{\bf w}_3,$$namely $$0= a({\bf v}_1+{\bf v_2})+b({\bf v}_2+{\bf v}_3)+c({\bf v}_3+{\bf v}_1).$$Solving for ${\bf v}_3$ we obtain $${\bf v}_3=-\frac{a+c}{b+c}{\bf v}_1-\frac{a+b}{b+c}{\bf v}_2.$$Therefore, we would like to find scalars $a$, $b$, $c$ such that $$-\frac{a+c}{b+c}=x,\quad -\frac{a+b}{b+c}=y.$$We can set $b+c=1$, then we require that $a+c=-x$ and $a+b=-y$. Solving this system of equations, we have $$a=-\frac{-1-x-y}{2},$$ $$b=-\frac{1+x-y}{2},$$ $$c=\frac{1-x+y}{2}.$$It is clear that scalars $a$, $b$, $c$ are not all zero because $b+c=1$. Hence ${\bf w}_1$, ${\bf w}_2$, ${\bf w}_3$ are linearly dependent as well.

Suppose the system ${\bf v}_1$, ${\bf v}_2$, $\cdots$, ${\bf v}_r$, ${\bf v}_{r+1}$ is linearly dependent, we argue it by contradiction. Then there exist scalars $a_1,\cdots,a_{r+1}$ (not all zero) such that \begin{equation}\label{}a_1{\bf v}_1+\cdots+a_{r+1}{\bf v}_{r+1}={\bf 0}.\qquad (*)\end{equation}If $a_{r+1}=0$, then we have \[a_1{\bf v}_1+\cdots+a_{r}{\bf v}_{r}={\bf 0}.\]Because the system of vectors ${\bf v}_1$, ${\bf v}_2$, $\cdots$, ${\bf v}_r$ is linearly independent, we must have $$a_1=\cdots=a_{r}=0.$$This contradicts that not all of $a_1,\cdots,a_{r+1}$ are zero. Therefore, we have $a_{r+1}\ne 0$.

From $(*)$, we then have\[a_{r+1}{\bf v}_{r+1}=-a_1{\bf v}_1-\cdots-a_{r}{\bf v}_{r}.\]Dividing both sides by $a_{r+1}$ (we proved that $a_{r+1}\ne 0$), we obtain\[{\bf v}_{r+1}=-\frac{a_1}{a_{r+1}}{\bf v}_1-\cdots-\frac{a_r}{a_{r+1}}{\bf v}_{r},\]which implies that ${\bf v}_{r+1}$ is a linear combination $\sum_{k=1}\alpha_k{\bf v}_{k}$ (namely $\alpha_k=-a_k/a_{r+1}$). This contradicts the choice of ${\bf v}_{r+1}$. Hence the system ${\bf v}_1$, ${\bf v}_2$, $\cdots$, ${\bf v}_r$, ${\bf v}_{r+1}$ is linearly independent.

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