Vector Space 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 linearly dependent 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 $c_{1}=\cdots =c_{n}=0$. 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.

Exercise Prove that a vector space $V$ over a field $\mathbb {F}$ is infinite-dimensional if and only if there is a sequence $v_{1},v_{2},\ldots$ in $V$ such that $v_{1},\ldots ,v_{m}$ is linearly independent for every $m\in \mathbb {N}$.

Proof:

Recall that a vector space $V$ is said to be finite dimensional if it is spanned by a finite list of vectors $w_{1},\ldots ,w_{m}\in V.$ In other words, $V$ has finite dimension if every vector in $V$ may be written as a linear combination of some list of vectors $w_{1},\ldots ,w_{m}\in V$. On the other hand, a vector space $V$ is infinite dimensional if it is not finite dimensional, i.e., $V$ cannot be spanned by a finite list of vectors. Now before we proceed in the proof, we will need the following fact: Lemma Suppose $V$ is a vector space over a field $\mathbb {F}$, and $v_{1},\ldots ,v_{n}$ are vectors that span $V$. If $w_{1},\ldots ,w_{m}$ in $V$ are linearly independent, then $m\leq n$. We are ready now to proceed with the proof: ’$\Rightarrow$’: Suppose that $V$ is an infinite dimensional vector space. Then, in particular, $V\neq 0$, so that there is some $v_{1}\neq 0$ in $V$. Then $v_{1}$ is a linearly independent vector in $V$. By way of induction now, suppose that for some $k\geq 1$, we have produced vectors $v_{1},\ldots ,v_{k}\in V$ such that $v_{1},\ldots ,v_{k}$ are linearly independent. Since $V$ is infinite-dimensional, it cannot be spanned by the (finite!) list of vectors $v_{1},\ldots ,v_{k}$. Thus we have that there is some $v_{k+1}\in V$ such that $v_{k+1}\neq c_{1}v_{1}+\cdots +c_{k}v_{k},{\text{ for any }}c_{1},\ldots ,c_{k}\in \mathbb {F} .$ We claim that now that $v_{1},\ldots ,v_{k},v_{k+1}$ form a linearly independent set in $V$. To see this, suppose that $a_{1}v_{1}+\cdots +a_{k}v_{k}+a_{k+1}v_{k+1}=0{\text{ for some }}a_{1},\ldots ,a_{k},a_{k+1}\in \mathbb {F} .$ Now if $a_{k+1}\neq 0$, then we may re-write the above equation as $v_{k+1}={\Big (}{\frac {-a_{1}}{a_{k+1}}}{\Big )}v_{1}+\cdots +{\Big (}{\frac {-a_{k}}{a_{k+1}}}{\Big )}v_{k},$ contradicting the fact that $v_{k+1}$ is not in the span of $v_{1},\ldots ,v_{k}$. So we conclude $a_{k+1}=0$, and thus we have that $a_{1}v_{1}+\cdots +a_{k}v_{k}+a_{k+1}v_{k+1}=a_{1}v_{1}+\cdots +a_{k}v_{k}+(0)v_{k+1}$ $=a_{1}v_{1}+\cdots +a_{k}v_{k}=0.$ Now by induction hypothesis, since $v_{1},\ldots ,v_{k}$ are linearly independent, we must have $a_{1},\ldots ,a_{k}$ are all zero. Thus we’ve shown that $v_{1},\ldots ,v_{k},v_{k+1}$ also form a linearly independent set, completing the induction. Thus we have constructed a sequence of vectors $\{v_{k}\}_{k=1}^{\infty }$ in $V$ so that $v_{1},\ldots ,v_{m}$ is linearly independent for each $m\in \mathbb {N}$. ’$\Leftarrow$’: On the other hand, suppose that $V$ contains a sequence of vectors $\{v_{k}\}_{k=1}^{\infty }$ so that $v_{1},\ldots ,v_{m}$ is linearly independent for each $m\in \mathbb {N}$. By way of contradiction, let’s suppose $V$ is not infinite dimensional, i.e. is finite dimensional. Then $V$ can be spanned by a finite list of vectors $w_{1},\ldots ,w_{n}\in V$. Now, since $V$ contains a linearly independent set of $V$.

Exercise Suppose that $U$ and $W$ are subspaces of a vector space $V$. Prove that $U\cup W$ is a subspace of $V$ if and only if $U\subseteq W$ or $W\subseteq U$.

Proof:

Recall that a subset $A$ of a vector space $V$ is a subspace of $V$ if $A$ itself is a vector space with the same addition and scalar multiplication operations as $V$. ’$\Rightarrow$’: Instead of proving that $U\cup W$ is a subspace of $V$ implies $U\subseteq W$ or $W\subseteq U$, we’ll show the contrapositive of this statement. That is, if $U\not \subseteq W$ and $W\not \subseteq U$, then $U\cup W$ is not a subspace of $V$. So suppose there is some $x\in U$ that is not in $W$, and likewise that there is some $y\in W$ that is not in $U$. We claim that $x+y\notin U\cup W$. For if it were, then $x+y$ would lie in either $U$ or $W$. If $x+y\in U$, then since $U$ is a subspace, this would imply $y=(x+y)-x\in U,$ contradicting our choice of $y$. Likewise, if $x+y\in W$, this would yield $x\in W$, which is again a contradiction. So we conclude that $x+y\notin U\cup W$, and thus $U\cup W$ fails to be closed under addition, so cannot be a subspace of $V$. ’$\Leftarrow$’: Suppose now that $U\subseteq W$ or $W\subseteq U$. Then $U\cup W$ is equal to either $W$ or $U$ respectively, which, by assumption are subspaces of $V$.

Before we begin the next exercise, we will need the following notation: for an arbitrary non-empty set $X$, let $\mathbb {R} ^{X}$ denote the set of all functions $f\colon X\to \mathbb {R}$. Then $\mathbb {R} ^{X}$ is always a vector space, with addition and scalar multiplication defined pointwise.

Exercise Let $b\in \mathbb {R}$ and consider the set $W={\Big \{}f\in \mathbb {R} ^{[0,1]}\colon f{\text{ is continuous and}}\int _{0}^{1}f(x)dx=b{\Big \}}.$ Show that $W$ is a subspace of $\mathbb {R} ^{[0,1]}$ if and only if $b=0$.

Proof:

Recall that a subset $A$ of a vector space $V$ is a subspace of $V$ if $A$ itself is a vector space with the same addition and scalar multiplication operations as $V$. There is a very convenient test that determines if $A$ is a subspace of $V$, sometimes called the subspace test. It says the following: (Subspace Test) Suppose that $A\subseteq V$, where $V$ is a vector space over a field $\mathbb {F}$. Then $A$ is a subspace of $V$ if and only if the following conditions are met: $0_{V}\in A$, $c\in \mathbb {F} ,v\in A\Rightarrow cv\in A$ $v,w\in A\Rightarrow v+w\in A$. We are now ready to proceed with the proof: ’$\Rightarrow$’: Suppose $W$ is a subspace of $\mathbb {R} ^{[0,1]}$. Then by condition $(1)$ of the subspace test, $W$ contains the zero vector of $\mathbb {R} ^{[0,1]}$, which is just the function that maps $x$ to $0$ for all $x\in [0,1]$. We will write this zero vector as ${\textbf {0}}$. Now since ${\textbf {0}}\in W$, by definition of being in $W$, we must have that $\int _{0}^{1}{\textbf {0}}(x)dx=b.$ On the other hand, when we actually integrate ${\textbf {0}}$, we find the integral must be zero. Thus $b=0$ as desired. ’$\Leftarrow$’: Say $b=0$. We will show that $W$ is a subspace of $V$ by showing that it passes all three conditions of the subspace test above. For condition (1), just note that by our previous remark, $\int _{0}^{1}{\textbf {0}}(x)dx=0=b$, and since ${\textbf {0}}$ is continuous, we have that ${\textbf {0}}\in W$. For condition (2), suppose that $c\in \mathbb {R}$ and $f\in W$. We must show that $cf\in W$. Since a continuous function multiplied by a constant is still continuous, $cf$ is still a continuous function. Now, $\int _{0}^{1}(cf)(x)dx=\int _{0}^{1}c[f(x)]dx=c\int _{0}^{1}f(x)dx=c(0)=0,$ so that we conclude $cf\in W$. Lastly, for condition (3), we must show that if $f,g\in W$, then $f+g\in W$. The addition of two continuous functions is always continuous, so that $f+g$ is continuous. Now since $f,g\in W$, we have that $\int _{0}^{1}(f+g)(x)dx=\int _{0}^{1}[f(x)+g(x)]dx=\int _{0}^{1}f(x)dx+\int _{0}^{1}g(x)dx=0+0=0,$ so that $f+g\in W$, and thus $W$ satisfies all three conditions of the subspace test.

Exercise Prove or give a counterexample to the following statement: If $U_{1},U_{2},W$ are subspaces of a vector space $V$ with $V=U_{1}\oplus W{\text{ and }}V=U_{2}\oplus W,$ then $U_{1}=U_{2}$.

Proof:

Let’s think about what it means for two subspaces $A,B$ of a vector space $C$ to satisfy $C=A\oplus B$. This means that $A\cap B=\{0_{C}\}$ and that $C=A+B$. In other words, for any $c\in C$, we may write $c$ uniquely in the form $c=a+b$, where $a\in A$ and $b\in B$. It turns out the statement of the problem is false, so that we must provide a counterexample to this statement: Let $V=\mathbb {R} ^{2}$ and consider its subspaces (one should check that they actually form subspaces first): $W=\{(x,0)\in \mathbb {R} ^{2}\colon x\in \mathbb {R} \}$ $U_{1}=\{(0,y)\in \mathbb {R} ^{2}\colon y\in \mathbb {R} \}$ $U_{2}=\{(z,z)\in \mathbb {R} ^{2}\colon z\in \mathbb {R} \}$ Then $U_{1}$ and $U_{2}$ are not the same subspaces of $V$, so that all we need to check is $U_{1}\oplus W=V=U_{2}\oplus W.$ Suppose that $(x,y)\in V$. Then $(x,y)=(x,0)+(0,y)$, where $(x,0)\in W$ and $(0,y)\in U_{1}$, so that $V=W+U_{1}$. Now if $(a,b)\in W\cap U_{1}$, then $(a,b)=(x,0)\in W$ for some $x\in \mathbb {R}$, and thus $b=0$. Likewise, $(a,b)=(0,y)\in W$ for some $y\in \mathbb {R}$, so that $a=0$. This shows $W\cap U_{1}=\{(0,0)\}$, and hence $V=U_{1}\oplus W$. Now again say $(x,y)\in V$. Then $(x,y)=(x-y,0)+(y,y)$, where $(x-y,0)\in W$ and $(y,y)\in U_{2}$ so that $V=W+U_{2}$. Now suppose $(a,b)\in W\cap U_{2}$. Then $(a,b)=(x,0)\in W$ for some $x\in \mathbb {R}$, so that $b=0$. Likewise, $(a,b)=(a,0)=(z,z)\in U_{2}$ for some $z\in \mathbb {R}$, thus $z=0=a$, so that we conclude $W\cap U_{2}=\{(0,0)\}$, and thus $V=U_{2}\oplus W$.