# Vector Space Problems

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

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

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

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

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

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

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

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

${\displaystyle c_{1}=0}$

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

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

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

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

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

Proof:
Recall that a vector space ${\displaystyle V}$ is said to be finite dimensional if it is spanned by a finite list of vectors ${\displaystyle w_{1},\ldots ,w_{m}\in V.}$ In other words, ${\displaystyle V}$ has finite dimension if every vector in ${\displaystyle V}$ may be written as a linear combination of some list of vectors ${\displaystyle w_{1},\ldots ,w_{m}\in V}$. On the other hand, a vector space ${\displaystyle V}$ is infinite dimensional if it is not finite dimensional, i.e., ${\displaystyle V}$ cannot be spanned by a finite list of vectors. Now before we proceed in the proof, we will need the following fact:

Lemma Suppose ${\displaystyle V}$ is a vector space over a field ${\displaystyle \mathbb {F} }$, and ${\displaystyle v_{1},\ldots ,v_{n}}$ are vectors that span ${\displaystyle V}$. If ${\displaystyle w_{1},\ldots ,w_{m}}$ in ${\displaystyle V}$ are linearly independent, then ${\displaystyle m\leq n}$.

We are ready now to proceed with the proof:
${\displaystyle \Rightarrow }$’: Suppose that ${\displaystyle V}$ is an infinite dimensional vector space. Then, in particular, ${\displaystyle V\neq 0}$, so that there is some ${\displaystyle v_{1}\neq 0}$ in ${\displaystyle V}$. Then ${\displaystyle v_{1}}$ is a linearly independent vector in ${\displaystyle V}$. By way of induction now, suppose that for some ${\displaystyle k\geq 1}$, we have produced vectors ${\displaystyle v_{1},\ldots ,v_{k}\in V}$ such that ${\displaystyle v_{1},\ldots ,v_{k}}$ are linearly independent. Since ${\displaystyle V}$ is infinite-dimensional, it cannot be spanned by the (finite!) list of vectors ${\displaystyle v_{1},\ldots ,v_{k}}$. Thus we have that there is some ${\displaystyle v_{k+1}\in V}$ such that ${\displaystyle v_{k+1}\neq c_{1}v_{1}+\cdots +c_{k}v_{k},{\text{ for any }}c_{1},\ldots ,c_{k}\in \mathbb {F} .}$ We claim that now that ${\displaystyle v_{1},\ldots ,v_{k},v_{k+1}}$ form a linearly independent set in ${\displaystyle V}$. To see this, suppose that ${\displaystyle a_{1}v_{1}+\cdots +a_{k}v_{k}+a_{k+1}v_{k+1}=0{\text{ for some }}a_{1},\ldots ,a_{k},a_{k+1}\in \mathbb {F} .}$ Now if ${\displaystyle a_{k+1}\neq 0}$, then we may re-write the above equation as ${\displaystyle v_{k+1}={\Big (}{\frac {-a_{1}}{a_{k+1}}}{\Big )}v_{1}+\cdots +{\Big (}{\frac {-a_{k}}{a_{k+1}}}{\Big )}v_{k},}$ contradicting the fact that ${\displaystyle v_{k+1}}$ is not in the span of ${\displaystyle v_{1},\ldots ,v_{k}}$. So we conclude ${\displaystyle a_{k+1}=0}$, and thus we have that ${\displaystyle a_{1}v_{1}+\cdots +a_{k}v_{k}+a_{k+1}v_{k+1}=a_{1}v_{1}+\cdots +a_{k}v_{k}+(0)v_{k+1}}$ ${\displaystyle =a_{1}v_{1}+\cdots +a_{k}v_{k}=0.}$ Now by induction hypothesis, since ${\displaystyle v_{1},\ldots ,v_{k}}$ are linearly independent, we must have ${\displaystyle a_{1},\ldots ,a_{k}}$ are all zero. Thus we’ve shown that ${\displaystyle v_{1},\ldots ,v_{k},v_{k+1}}$ also form a linearly independent set, completing the induction. Thus we have constructed a sequence of vectors ${\displaystyle \{v_{k}\}_{k=1}^{\infty }}$ in ${\displaystyle V}$ so that ${\displaystyle v_{1},\ldots ,v_{m}}$ is linearly independent for each ${\displaystyle m\in \mathbb {N} }$.

${\displaystyle \Leftarrow }$’: On the other hand, suppose that ${\displaystyle V}$ contains a sequence of vectors ${\displaystyle \{v_{k}\}_{k=1}^{\infty }}$ so that ${\displaystyle v_{1},\ldots ,v_{m}}$ is linearly independent for each ${\displaystyle m\in \mathbb {N} }$. By way of contradiction, let’s suppose ${\displaystyle V}$ is not infinite dimensional, i.e. is finite dimensional. Then ${\displaystyle V}$ can be spanned by a finite list of vectors ${\displaystyle w_{1},\ldots ,w_{n}\in V}$.
Now, since ${\displaystyle V}$ contains a linearly independent set of ${\displaystyle V}$.

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

Proof:
Recall that a subset ${\displaystyle A}$ of a vector space ${\displaystyle V}$ is a subspace of ${\displaystyle V}$ if ${\displaystyle A}$ itself is a vector space with the same addition and scalar multiplication operations as ${\displaystyle V}$.

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

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

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

Proof:
Recall that a subset ${\displaystyle A}$ of a vector space ${\displaystyle V}$ is a subspace of ${\displaystyle V}$ if ${\displaystyle A}$ itself is a vector space with the same addition and scalar multiplication operations as ${\displaystyle V}$. There is a very convenient test that determines if ${\displaystyle A}$ is a subspace of ${\displaystyle V}$, sometimes called the subspace test. It says the following:

(Subspace Test) Suppose that ${\displaystyle A\subseteq V}$, where ${\displaystyle V}$ is a vector space over a field ${\displaystyle \mathbb {F} }$. Then ${\displaystyle A}$ is a subspace of ${\displaystyle V}$ if and only if the following conditions are met:

1. ${\displaystyle 0_{V}\in A}$,
2. ${\displaystyle c\in \mathbb {F} ,v\in A\Rightarrow cv\in A}$
3. ${\displaystyle v,w\in A\Rightarrow v+w\in A}$.

We are now ready to proceed with the proof:

${\displaystyle \Rightarrow }$’: Suppose ${\displaystyle W}$ is a subspace of ${\displaystyle \mathbb {R} ^{[0,1]}}$. Then by condition ${\displaystyle (1)}$ of the subspace test, ${\displaystyle W}$ contains the zero vector of ${\displaystyle \mathbb {R} ^{[0,1]}}$, which is just the function that maps ${\displaystyle x}$ to ${\displaystyle 0}$ for all ${\displaystyle x\in [0,1]}$. We will write this zero vector as ${\displaystyle {\textbf {0}}}$. Now since ${\displaystyle {\textbf {0}}\in W}$, by definition of being in ${\displaystyle W}$, we must have that ${\displaystyle \int _{0}^{1}{\textbf {0}}(x)dx=b.}$ On the other hand, when we actually integrate ${\displaystyle {\textbf {0}}}$, we find the integral must be zero. Thus ${\displaystyle b=0}$ as desired.
${\displaystyle \Leftarrow }$’: Say ${\displaystyle b=0}$. We will show that ${\displaystyle W}$ is a subspace of ${\displaystyle V}$ by showing that it passes all three conditions of the subspace test above. For condition (1), just note that by our previous remark, ${\displaystyle \int _{0}^{1}{\textbf {0}}(x)dx=0=b}$, and since ${\displaystyle {\textbf {0}}}$ is continuous, we have that ${\displaystyle {\textbf {0}}\in W}$. For condition (2), suppose that ${\displaystyle c\in \mathbb {R} }$ and ${\displaystyle f\in W}$. We must show that ${\displaystyle cf\in W}$. Since a continuous function multiplied by a constant is still continuous, ${\displaystyle cf}$ is still a continuous function. Now, ${\displaystyle \int _{0}^{1}(cf)(x)dx=\int _{0}^{1}c[f(x)]dx=c\int _{0}^{1}f(x)dx=c(0)=0,}$ so that we conclude ${\displaystyle cf\in W}$. Lastly, for condition (3), we must show that if ${\displaystyle f,g\in W}$, then ${\displaystyle f+g\in W}$. The addition of two continuous functions is always continuous, so that ${\displaystyle f+g}$ is continuous. Now since ${\displaystyle f,g\in W}$, we have that ${\displaystyle \int _{0}^{1}(f+g)(x)dx=\int _{0}^{1}[f(x)+g(x)]dx=\int _{0}^{1}f(x)dx+\int _{0}^{1}g(x)dx=0+0=0,}$ so that ${\displaystyle f+g\in W}$, and thus ${\displaystyle W}$ satisfies all three conditions of the subspace test.

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

Proof:
Let’s think about what it means for two subspaces ${\displaystyle A,B}$ of a vector space ${\displaystyle C}$ to satisfy ${\displaystyle C=A\oplus B}$. This means that ${\displaystyle A\cap B=\{0_{C}\}}$ and that ${\displaystyle C=A+B}$. In other words, for any ${\displaystyle c\in C}$, we may write ${\displaystyle c}$ uniquely in the form ${\displaystyle c=a+b}$, where ${\displaystyle a\in A}$ and ${\displaystyle b\in B}$.

It turns out the statement of the problem is false, so that we must provide a counterexample to this statement: Let ${\displaystyle V=\mathbb {R} ^{2}}$ and consider its subspaces (one should check that they actually form subspaces first): ${\displaystyle W=\{(x,0)\in \mathbb {R} ^{2}\colon x\in \mathbb {R} \}}$

${\displaystyle U_{1}=\{(0,y)\in \mathbb {R} ^{2}\colon y\in \mathbb {R} \}}$

${\displaystyle U_{2}=\{(z,z)\in \mathbb {R} ^{2}\colon z\in \mathbb {R} \}}$

Then ${\displaystyle U_{1}}$ and ${\displaystyle U_{2}}$ are not the same subspaces of ${\displaystyle V}$, so that all we need to check is ${\displaystyle U_{1}\oplus W=V=U_{2}\oplus W.}$ Suppose that ${\displaystyle (x,y)\in V}$. Then ${\displaystyle (x,y)=(x,0)+(0,y)}$, where ${\displaystyle (x,0)\in W}$ and ${\displaystyle (0,y)\in U_{1}}$, so that ${\displaystyle V=W+U_{1}}$. Now if ${\displaystyle (a,b)\in W\cap U_{1}}$, then ${\displaystyle (a,b)=(x,0)\in W}$ for some ${\displaystyle x\in \mathbb {R} }$, and thus ${\displaystyle b=0}$. Likewise, ${\displaystyle (a,b)=(0,y)\in W}$ for some ${\displaystyle y\in \mathbb {R} }$, so that ${\displaystyle a=0}$. This shows ${\displaystyle W\cap U_{1}=\{(0,0)\}}$, and hence ${\displaystyle V=U_{1}\oplus W}$.
Now again say ${\displaystyle (x,y)\in V}$. Then ${\displaystyle (x,y)=(x-y,0)+(y,y)}$, where ${\displaystyle (x-y,0)\in W}$ and ${\displaystyle (y,y)\in U_{2}}$ so that ${\displaystyle V=W+U_{2}}$. Now suppose ${\displaystyle (a,b)\in W\cap U_{2}}$. Then ${\displaystyle (a,b)=(x,0)\in W}$ for some ${\displaystyle x\in \mathbb {R} }$, so that ${\displaystyle b=0}$. Likewise, ${\displaystyle (a,b)=(a,0)=(z,z)\in U_{2}}$ for some ${\displaystyle z\in \mathbb {R} }$, thus ${\displaystyle z=0=a}$, so that we conclude ${\displaystyle W\cap U_{2}=\{(0,0)\}}$, and thus ${\displaystyle V=U_{2}\oplus W}$.