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Solution to Linear Algebra Done Wrong Exercise 1.2.2


True or false:

a) Any set containing a zero vector is linearly dependent;
b) A basis must contain ${\bf 0}$;
c) subsets of linearly dependent sets are linearly dependent;
d) subsets of linearly independent sets are linearly independent;
e) If $\alpha_1{\bf v}_1+\alpha_2{\bf v}_2+\cdots+\alpha_n{\bf v}_n=0$ then all scalars $\alpha_k$ are zero.


Solution:

a) Yes. Consider any set $\{{\bf v}_1,\dots,{\bf v}_n\}$ containing a zero vector. We assume for example ${\bf v}_1={\bf 0}$. Then by Exercise 1.1.7 we have\begin{align*}&\ 1{\bf v}_1+0{\bf v}_2+\cdots+0{\bf v}_n\\ =&\ {\bf 0}+{\bf 0}+\cdots+{\bf 0}={\bf 0}.\end{align*}Hence a nontrivial linear combination of ${\bf v}_1,\dots,{\bf v}_n$ equals ${\bf 0}$, which implies that the set $\{{\bf v}_1,\dots,{\bf v}_n\}$ is linearly dependent.

b) No. By a) any set $\{{\bf v}_1,\dots,{\bf v}_n\}$ containing a zero vector is linearly dependent and hence cannot be a basis.

c) No. Take the set $\{{\bf 0},{\bf v}\}$, where ${\bf v}$ is a nonzero vector. The set is linearly dependent by a). But the subset $\{\bf v\}$ is linearly independent. (Why?)

d) Yes. Note that a linear combination of a subset is also a linear combination of a set (consider the elements not in the subset multiplied by zero), therefore subsets of linearly independent sets are linearly independent.

e) No. The example in a) tells us that it may happen that not all $\alpha_k$ are zero.



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