Vector Space Problems

Exercise Show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{1+i,1-i \}} form a linearly independent set of vectors in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} , viewed as a vector space over Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} .

Proof:

Recall that the set of vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{v_{1},\ldots, v_{n} \}} in a vector space Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} (over a field Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{F}} ) are said to be linearly independent if whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1},\ldots,c_{n}} are scalars in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{F}} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1}v_{1} + \cdots + c_{n}v_{n} = 0,} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} = \cdots = c_{n} = 0} . So for this problem, since we’re considering the complex numbers Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} as a vector space over Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} , we must show that whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1},c_{2} \in \mathbb{R}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1}(1+i) + c_{2}(1-i) = 0,} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} = c_{2} = 0} . Rearranging the above equation, we obtain Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (c_{1} + c_{2}) + (c_{1} - c_{2}) i = 0.} Now, a complex number is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} if and only if its real and imaginary parts are both Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} . So in this case, we conclude that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} + c_{2} = 0 \text{ and } c_{1} - c_{2} = 0.} This implies Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} = c_{2}} , so that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} + c_{2} = 2c_{2} = 0} , which yields Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} = c_{2} = 0} . Thus we conclude the vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1+i,1-i} are linearly independent in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} (over Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} ).

Exercise Show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{1+i,1-i \}} form a linearly independent set of vectors in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} , viewed as a vector space over Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} .

Proof:

Recall that a set of vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{v_{1},\ldots,v_{n}\}} in a vector space Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} (over a field Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1},\ldots, c_{n} \in \mathbb{F}} not all equal to zero such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1}v_{1} + \cdots + c_{n}v_{n} = 0.} So for this problem, to show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1+i} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1-i} are not linearly dependent over Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} , all we need to do is exhibit two complex scalars Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{2}} that are not both zero such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1}(1+i) + c_{2}(1-i) = 0.} There are many choices for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{2}} , but one such example is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{1} = i} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_{2} = 1} .

Exercise Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} be a vector space over a field Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{F}} . If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{v_{1},v_{2},v_{3},v_{4}\} \subseteq V} are a linearly independent set of vectors, then show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{v_{1} - v_{2}, v_{2} - v_{3}, v_{3} - v_{4},v_{4}\}} also form a linearly independent set of vectors in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} .

Proof:

Recall that the set of vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{w_{1},\ldots, w_{n} \}} in a vector space Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 $V=U_{2}\oplus W$.