Andrew Walker Problems

Exercise Show that $\{1+i,1-i\}$ form a linearly independent set of vectors in $\mathbb {C}$, viewed as a vector space over $\mathbb {R}$.

Proof Recall that the set of vectors $\{v_{1},\ldots ,v_{n}\}$ in a vector space $V$ (over a field $\mathbb {F}$) are said to be linearly independent if whenever $c_{1},\ldots ,c_{n}$ are scalars in $\mathbb {F}$ such that $c_{1}v_{1}+\cdots +c_{n}v_{n}=0,$ then $c_{1}=\cdots =c_{n}=0$. So for this problem, since we’re considering the complex numbers $\mathbb {C}$ as a vector space over $\mathbb {R}$, we must show that whenever $c_{1},c_{2}\in \mathbb {R}$ and $c_{1}(1+i)+c_{2}(1-i)=0,$ then $c_{1}=c_{2}=0$. Rearranging the above equation, we obtain $(c_{1}+c_{2})+(c_{1}-c_{2})i=0.$ Now, a complex number is equal to $0$ if and only if its real and imaginary parts are both $0$. So in this case, we conclude that $c_{1}+c_{2}=0{\text{ and }}c_{1}-c_{2}=0.$ This implies $c_{1}=c_{2}$, so that $c_{1}+c_{2}=2c_{2}=0$, which yields $c_{1}=c_{2}=0$. Thus we conclude the vectors $1+i,1-i$ are linearly independent in $\mathbb {C}$ (over $\mathbb {R}$).

Exercise Show that $\{1+i,1-i\}$ form a linearly independent set of vectors in $\mathbb {C}$, viewed as a vector space over $\mathbb {R}$.

Proof Recall that a set of vectors $\{v_{1},\ldots ,v_{n}\}$ in a vector space $V$ (over a field $\mathbb {F}$) is said to be ${\textbf {linearlydependent}}$ if they are not linearly independent. More concretely, these vectors are linearly dependent if we can find scalars $c_{1},\ldots ,c_{n}\in \mathbb {F}$ not all equal to zero such that $c_{1}v_{1}+\cdots +c_{n}v_{n}=0.$

So for this problem, to show that $1+i$ and $1-i$ are not linearly dependent over $\mathbb {C}$, all we need to do is exhibit two complex scalars $c_{1}$ and $c_{2}$ that are not both zero such that $c_{1}(1+i)+c_{2}(1-i)=0.$ There are many choices for $c_{1}$ and $c_{2}$, but one such example is $c_{1}=i$ and $c_{2}=1$.

Exercise Let $V$ be a vector space over a field $\mathbb {F}$. If $\{v_{1},v_{2},v_{3},v_{4}\}\subseteq V$ are a linearly independent set of vectors, then show that $\{v_{1}-v_{2},v_{2}-v_{3},v_{3}-v_{4},v_{4}\}$ also form a linearly independent set of vectors in $V$.

Proof Recall that the set of vectors $\{w_{1},\ldots ,w_{n}\}$ in a vector space $V$ (over a field $\mathbb {F}$) are said to be linearly independent if whenever $c_{1},\ldots ,c_{n}$ are scalars in $\mathbb {F}$ such that $c_{1}w_{1}+\cdots +c_{n}w_{n}=0,$ then $c_{1}=\cdots =c_{n}=0$.

Exercise So for this problem, we must show that whenever $c_{1},c_{2},c_{3},c_{4}\in \mathbb {F}$ and $c_{1}(v_{1}-v_{2})+c_{2}(v_{2}-v_{3})+c_{3}(v_{3}-v_{4})+c_{4}v_{4}=0,$ we have that $c_{1}=c_{2}=c_{3}=c_{4}=0.$ After rearranging terms in the above equation, we have that $c_{1}v_{1}+(c_{2}-c_{1})v_{2}+(c_{3}-c_{2})v_{3}+(c_{4}-c_{3})v_{4}=0.$ Now since the vectors $\{v_{1},v_{2},v_{3},v_{4}\}$ are linearly independent in $V$ by assumption, we have that

$c_{1}=0$

$c_{2}-c_{1}=0$

$c_{3}-c_{2}=0$

$c_{4}-c_{3}=0.$

In other words, $c_{1}=c_{2}=c_{3}=c_{4}=0$, so that $\{v_{1}-v_{2},v_{2}-v_{3},v_{3}-v_{4},v_{4}\}$ form a linearly independent set as desired.