Linear Algebra Done Right: Proofs

Proof. Zero vector.

$0+2\ast 0+3\ast 0\ne 4$, thus $0$ is not in the given set, hence it is not a subspace of ${𝐅}^3$.

Proof. Additive identity.

$0=(0,0,0)$. Substituting the corresponding coordinates $0\ast 0\ast 0=0$; thus the given set includes the additive identity.

Closure under addition.

Let’s take $𝑢=(0,0,1)$ and $𝑣=(1,1,0)$ so that $𝑢+𝑣=(0+1,0+1,0+1)=(1,1,1)$. Since $𝑢,𝑣\in 𝑉$ but $𝑢+𝑣\notin 𝑉$, it does not satisfy the closure under addition. Thus it is not a subspace of ${𝐅}^3$.

Proof. Let $𝐷=\{𝑓:(-4,4)\to 𝐑|𝑓\text{is differentiable},𝑓\prime (-1)=3𝑓(2)\}$. $𝐷\subseteq {𝐑}^{(-4,4)}$.

Additive identity: let $0$ denote the zero function; it is differentiable, since $0\prime =0$. It is also in the $𝐷$, since $0\prime (-1)=3\cdot 0(2)=0$.

Closed under addition. $𝐷\ne \varnothing$, for at least the zero function is in it. Let $𝑓,𝑔\in 𝐷$. Let $𝑠(𝑥)=𝑓(𝑥)+𝑔(𝑥)$. $𝑠\prime (-1)=𝑓\prime (-1)+𝑔\prime (-1)$, since the derivative of the sums of two differentiable functions equal to the sum of their derivatives. $𝑓\prime (-1)+𝑔\prime (-1)=3𝑓(2)+3𝑔(2)=3(𝑓(2)+𝑔(2))=3𝑠(2)$; thus $𝑠\in 𝐷$.

Closed under scalar multiplication. Let $𝜆\in 𝐑$ and $𝑓\in 𝐷$. $(𝜆𝑓)(𝑥)=𝜆𝑓(𝑥)$. Let $𝑚(𝑥)=𝜆𝑓(𝑥)$. Then $𝑚\prime (𝑥)=𝜆𝑓\prime (𝑥)$. $𝑚\prime (-1)=𝜆3𝑓(2)=3𝜆𝑓(2)=3𝑚(2)$; thus $𝑚\in 𝐷$.

Hence $𝐷$ is a subspace of ${𝐑}^{(-4,4)}$.

Proof. $𝑉=span({𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑣}_4)=\{{𝑎}_1{𝑣}_1+\cdots +{𝑎}_4{𝑣}_4:{𝑎}_1,\dots ,{𝑎}_4\in 𝐅\}$.

Let $𝑣\in 𝑉$. Then there exist $𝑣={𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3+{𝑎}_4{𝑣}_4$, where ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑎}_4\in 𝐅$. Define ${𝑏}_1={𝑎}_1$, ${𝑏}_2={𝑎}_1+{𝑎}_2$, ${𝑏}_3={𝑎}_1+{𝑎}_2+{𝑎}_3$, and ${𝑏}_4={𝑎}_1+{𝑎}_2+{𝑎}_3+{𝑎}_4$. Then

so $𝑣\in 𝑆$. Hence $𝑉\subseteq 𝑆$. Conversely, $𝑆\subseteq 𝑉$ because ${𝑏}_1,{𝑏}_2-{𝑏}_1,{𝑏}_3-{𝑏}_2,{𝑏}_4-{𝑏}_3\in 𝐅$ and $𝑉$ is closed under scalar multiplication and vector addition. Thus $𝑆=𝑉$.

Proof. Suppose $(𝑣)$ is linearly independent, and suppose $𝑣=0$. Then $1\cdot 𝑣=1\cdot 0=0$. Taking $𝑎=1$ gives a nonzero scalar with $𝑎𝑣=0$, contradicting the linear independence of $(𝑣)$. Therefore $𝑣\ne 0$.

Conversely, suppose $𝑣\ne 0$, and suppose $𝑎\ne 0$. Then there exists ${𝑎}^{-1}$, such that ${𝑎}^{-1}𝑎=1$. Multiplying both sides of $𝑎𝑣=0$ by ${𝑎}^{-1}$, $1𝑣=𝑣=0$, hence contradicting $𝑣\ne 0$. Therefore $𝑎=0$ if $𝑎𝑣=0$ and $𝑣\ne 0$.

Thus $(𝑣)$ is linearly independent if and only if the $𝑣\ne 0$.

Proof. Let $(𝑢,𝑤)$.

Suppose there exists $𝜆\in 𝐅$, such that $𝑢=𝜆𝑤$. Take $𝑎=1$ and $𝑏=-𝜆$. Then $𝑎𝑢+𝑏𝑤=1𝑢+(-𝜆)𝑤=𝜆𝑤+(-𝜆)𝑤=0𝑤=0$. Since $𝑎=1\ne 0$, $(𝑢,𝑤)$ is dependent. Symmetrically, the same is for $𝑤=𝜆𝑢$. Thus if either of the two vectors $𝑢$, $𝑤$ is a scalar multiple of the other, then $(𝑢,𝑤)$ is dependent.

Suppose $(𝑢,𝑤)$ is dependent. Then there exist $𝑎,𝑏\in 𝐹$, not both $0$, such that $𝑎𝑢+𝑏𝑤=0$. Suppose $𝑎\ne 0$. Then there exist ${𝑎}^{-1}$. Multiplying both sides of $𝑎𝑢+𝑏𝑤=0$ by ${𝑎}^{-1}$ gives us ${𝑎}^{-1}(𝑎𝑢)+{𝑎}^{-1}(𝑏𝑤)=({𝑎}^{-1}𝑎)𝑢+({𝑎}^{-1}𝑏)𝑤=1𝑢+({𝑎}^{-1}𝑏)𝑤=𝑢+({𝑎}^{-1}𝑏)𝑤=0$; hence $𝑢=-(({𝑎}^{-1}𝑏)𝑤)=(-{𝑎}^{-1}𝑏)𝑤$. Symmetrically, for $𝑏\ne 0$, $𝑤=(-{𝑏}^{-1}𝑎)𝑢$.

Proof. Let ${𝑣}_1=(2,3,1)$, ${𝑣}_2=(1,-1,2)$, and ${𝑣}_3=(7,3,𝑐)$.

Suppose the list is linearly dependent. Then there exist ${𝑎}_1,{𝑎}_2,{𝑎}_3\in 𝐅$, not all $0$, such that ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3=0$. Suppose ${𝑎}_3=0$. Then ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2=(2{𝑎}_1+{𝑎}_2,3{𝑎}_1-{𝑎}_2,{𝑎}_1-2{𝑎}_2)=(0,0,0)$. Solving the linear system of equations: ${𝑎}_1={𝑎}_2=0$. ${𝑎}_1={𝑎}_2={𝑎}_3=0$, hence contradicting the linear dependence. Hence ${𝑎}_3\ne 0$. Solving the linear system of the first two equations in terms of ${𝑎}_3$ for ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3=(2{𝑎}_1+{𝑎}_2+7{𝑎}_3,3{𝑎}_1-{𝑎}_2+3{𝑎}_3,{𝑎}_1+2{𝑎}_2+𝑐{𝑎}_3)=(0,0,0)$, we have ${𝑎}_1=-2{𝑎}_3$ and ${𝑎}_2=-3{𝑎}_3$. Substituting both in the third equation, $-2{𝑎}_3-6{𝑎}_3+𝑐{𝑎}_3=-8{𝑎}_3+𝑐{𝑎}_3=(-8+𝑐){𝑎}_3=0$. ${𝑎}_3\ne 0$, so there exists ${𝑎}_3^{-1}$. Multiplying both sides of the equation: $-8+𝑐=0$. Hence $𝑐=8$.

Suppose $𝑐=8$. Take ${𝑎}_1=2$, ${𝑎}_2=3$, and ${𝑎}_3=-1$. Then ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3=(2\ast 2+3-7,2\ast 3-3-3,2+6-8)=(0,0,0)$. Hence there exist ${𝑎}_1,{𝑎}_2,{𝑎}_3$, not all $0$, such that ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3=0$. Therefore the list is linearly dependent.

Thus the list is linearly dependent if and only if $𝑐=8$.

Proof. Suppose ${𝑣}_1,\dots ,{𝑣}_{𝑚},𝑤$ is linearly independent, and suppose for contradiction that $𝑤\in span({𝑣}_1,\dots ,{𝑣}_{𝑚})$. Then $𝑤={𝑐}_1{𝑣}_1+\cdots +{𝑐}_{𝑚}{𝑣}_{𝑚}$ for some ${𝑐}_1,\dots ,{𝑐}_{𝑚}\in 𝐅$. Then ${𝑐}_1{𝑣}_1+\cdots +{𝑐}_{𝑚}{𝑣}_{𝑚}-𝑤=0$, contradicting the linear independence of ${𝑣}_1,\dots ,{𝑣}_{𝑚},𝑤$.

Suppose $𝑤\notin span({𝑣}_1,\dots ,{𝑣}_{𝑚})$, and suppose for contradiction that ${𝑣}_1,\dots ,{𝑣}_{𝑚},𝑤$ is linearly dependent. Then there exist ${𝑎}_1,\dots ,{𝑎}_{𝑚},{𝑎}_{𝑚+1}\in 𝐅$, not all $0$, such that ${𝑎}_1{𝑣}_1+\cdots +{𝑎}_{𝑚}{𝑣}_{𝑚}+{𝑎}_{𝑚+1}𝑤=0$. Suppose ${𝑎}_{𝑚+1}\ne 0$. Multiplying both sides by ${𝑎}_{𝑚+1}^{-1}$ gives $\frac{{𝑎}_1}{{𝑎}_{𝑚+1}}{𝑣}_1+\cdots +\frac{{𝑎}_{𝑚}}{{𝑎}_{𝑚+1}}{𝑣}_{𝑚}+𝑤=0$. Then $𝑤=-\frac{{𝑎}_1}{{𝑎}_{𝑚+1}}{𝑣}_1-\cdots -\frac{{𝑎}_{𝑚}}{{𝑎}_{𝑚+1}}{𝑣}_{𝑚}$, contradicting $𝑤\notin span({𝑣}_1,\dots ,{𝑣}_{𝑚})$. Suppose ${𝑎}_{𝑚+1}=0$. Then there exist ${𝑎}_1,\dots ,{𝑎}_{𝑚}$, not all $0$, such that ${𝑎}_1{𝑣}_1+\cdots +{𝑎}_{𝑚}{𝑣}_{𝑚}=0$, thus contradicting the linear independence of ${𝑣}_1,\dots ,{𝑣}_{𝑚}$.

Therefore ${𝑣}_1,\dots ,{𝑣}_{𝑚},𝑤$ is linearly independent $⟺$ $𝑤\notin span({𝑣}_1,\dots ,{𝑣}_{𝑚})$.

Proof. $𝑈\subseteq {𝐑}^5$, so ${𝑣}_1,{𝑣}_2,{𝑣}_3$ is independent in ${𝐑}^5$. By 2.32 we can extend ${𝑣}_1,{𝑣}_2,{𝑣}_3$ to a basis of ${𝐑}^5$. Take ${𝑤}_1=(1,0,0,0,0)$ and ${𝑤}_2=(0,0,1,0,0)$ where ${𝑤}_1,{𝑤}_2\in {𝐑}^5$. Suppose there exist ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑎}_4,{𝑎}_5\in 𝐑$ such that ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3+{𝑎}_4{𝑤}_1+{𝑎}_5{𝑤}_2=(3{𝑎}_1+{𝑎}_4,{𝑎}_1,7{𝑎}_2+{𝑎}_5,{𝑎}_2,{𝑎}_3)=0$. The equation forces ${𝑎}_1={𝑎}_2={𝑎}_3={𝑎}_4={𝑎}_5=0$. Therefore ${𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2$ is independent. Let $𝑢=(𝑎,𝑏,𝑐,𝑑,𝑒)\in {𝐑}^5$. Then

Therefore ${𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2$ spans ${𝐑}^5$. Thus ${𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2$ is a basis of ${𝑅}^5$.

Proof. Let $𝑊=span({𝑤}_1,{𝑤}_2)=\{({𝑥}_1,{𝑥}_2,{𝑥}_3,{𝑥}_4,{𝑥}_5)\in {𝐑}^5:{𝑥}_2={𝑥}_4={𝑥}_5=0\}$. Suppose $𝑢\in {𝐑}^5$. The list ${𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2$ spans ${𝐑}^5$, so there exist ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑎}_4,{𝑎}_5\in 𝐑$ such that $𝑢={𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3+{𝑎}_4{𝑤}_1+{𝑎}_5{𝑤}_2$. We have $𝑢=𝑣+𝑤$, where $𝑣\in 𝑈$ and $𝑤\in 𝑊$. Thus $𝑢\in 𝑈+𝑊$, hence ${𝐑}^5=𝑈+𝑊$.

Suppose $𝑢\in 𝑈\cap 𝑊$. Then there exists $𝑢={𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3={𝑎}_4{𝑤}_1+{𝑎}_5{𝑤}_5$. Thus ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3-{𝑎}_4{𝑤}_1-{𝑎}_5{𝑤}_5=0$. Because ${𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2$ is linearly independent, ${𝑎}_1={𝑎}_2={𝑎}_3={𝑎}_4={𝑎}_5=0$. Thus $𝑢=0$, hence $𝑈\cap 𝑊=\{0\}$.

Therefore ${𝐑}^5=𝑈\oplus 𝑊$.

Proof. Let $𝑢=({𝑧}_1,{𝑧}_2,{𝑧}_3,{𝑧}_4,{𝑧}_5)\in 𝑈$. Then $6{𝑧}_1={𝑧}_2$ and ${𝑧}_3+2{𝑧}_4+3{𝑧}_5=0$, so $𝑢=({𝑧}_1,6{𝑧}_1,-2{𝑧}_4-3{𝑧}_5,{𝑧}_4,{𝑧}_5)$. Splitting that vector into a sum, we have $𝑢={𝑧}_1(1,6,0,0,0)+{𝑧}_4(0,0,-2,1,0)+{𝑧}_5(0,0,-3,0,1)$.

Let ${𝑣}_1=(1,6,0,0,0)$, ${𝑣}_2=(0,0,-2,1,0)$, and ${𝑣}_3=(0,0,-3,0,1)$.

${𝑣}_1,{𝑣}_2,{𝑣}_3\in 𝑈$, because $6\cdot 1=6$ and $0+2\cdot 0+3\cdot 0=0$; $6\cdot 0=0$ and $-2+2\cdot 1+3\cdot 0=0$; $6\cdot 0=0$ and $-3+2\cdot 0+3\cdot 1=0$. Hence $span({𝑣}_1,{𝑣}_2,{𝑣}_3)\subseteq 𝑈$. $𝑢$ is a linear combination of ${𝑣}_1,{𝑣}_2,{𝑣}_3$, hence $𝑈\subseteq span({𝑣}_1,{𝑣}_2,{𝑣}_3)$. Thus $span({𝑣}_1,{𝑣}_2,{𝑣}_3)=𝑈$.

Suppose there exist ${𝑎}_1,{𝑎}_2,{𝑎}_3\in 𝐂$ such that ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3=0$. Then ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3=({𝑎}_1,6{𝑎}_1,-2{𝑎}_2-3{𝑎}_3,{𝑎}_2,{𝑎}_3)=0$. The first, fourth, and fifth coordinates give ${𝑎}_1=0$, ${𝑎}_2=0$, and ${𝑎}_3=0$. Hence ${𝑎}_1={𝑎}_2={𝑎}_3=0$, therefore ${𝑣}_1,{𝑣}_2,{𝑣}_3$ is linearly independent.

Proof. Take ${𝑤}_1=(0,1,0,0,0)$ and ${𝑤}_2=(0,0,1,0,0)$ where ${𝑤}_1,{𝑤}_2\in {𝐂}^5$.

Suppose there exist ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑎}_4,{𝑎}_5\in 𝐂$ such that ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3+{𝑎}_4{𝑤}_1+{𝑎}_5{𝑤}_2=({𝑎}_1,6{𝑎}_1+{𝑎}_4,-2{𝑎}_2-3{𝑎}_3+{𝑎}_5,{𝑎}_2,{𝑎}_3)=0$. The first, fourth, and fifth coordinates force ${𝑎}_1=0$, ${𝑎}_2=0$, and ${𝑎}_3=0$. Substituting corresponding values to other coordinates, $6\cdot 0+{𝑎}_4={𝑎}_4=0$, $-2\cdot 0-3\cdot 0+{𝑎}_5={𝑎}_5=0$; hence ${𝑎}_4={𝑎}_5=0$. Therefore ${𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2$ is linearly independent.

${𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2\in {𝐂}^5$, so $span({𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2)\subseteq {𝐂}^5$, because span is a subspace, so it is closed under linear combination. Suppose $𝑢=({𝑢}_1,{𝑢}_2,{𝑢}_3,{𝑢}_4,{𝑢}_5)\in {𝐂}^5$. Take ${𝑎}_1={𝑢}_1$, ${𝑎}_2={𝑢}_4$, ${𝑎}_3={𝑢}_5$, ${𝑎}_4={𝑢}_2-6{𝑢}_1$, and ${𝑎}_5={𝑢}_3+2{𝑢}_4+3{𝑢}_5$. Then

So ${𝐂}^5\subseteq span({𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2)$. Hence $span({𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2)={𝐂}^5$.

Proof. Let $𝑊=span({𝑤}_1,{𝑤}_2)$ and $𝑣\in {𝐂}^5$. The list $({𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2)$ spans ${𝐂}^5$, so there exist ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑎}_4,{𝑎}_5\in 𝐂$ such that ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3+{𝑎}_4{𝑤}_1+{𝑎}_5{𝑤}_2=𝑣$ where ${𝑣}_1,{𝑣}_2,{𝑣}_3\in 𝑈$ and ${𝑤}_1,{𝑤}_2\in 𝑊$. $𝑣\in 𝑈+𝑊$, hence ${𝐂}^5\subseteq 𝑈+𝑊$. Conversely, $𝑈$ and $𝑊$ are subspaces of ${𝐂}^5$, so $𝑈+𝑊\subseteq {𝐂}^5$.

Suppose $𝑣\in 𝑈\cap 𝑊$. Then there exist ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑎}_4,{𝑎}_5\in 𝐂$ such that $𝑣={𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3={𝑎}_4{𝑤}_1+{𝑎}_5{𝑤}_2$. Hence ${𝑎}_1{𝑣}_1+{𝑎}_2{𝑣}_2+{𝑎}_3{𝑣}_3-{𝑎}_4{𝑤}_1-{𝑎}_5{𝑤}_2=0$. Since ${𝑣}_1,{𝑣}_2,{𝑣}_3,{𝑤}_1,{𝑤}_2$ is linearly independent, so ${𝑎}_1={𝑎}_2={𝑎}_3={𝑎}_4={𝑎}_5=0$. Hence $𝑣=0$, thus $𝑈\cap 𝑊=\{0\}$.

Thus $𝑈\oplus 𝑊={𝐂}^5$.

Proof. $\{0\}$ is a subspace of ${𝐑}^2$: $0\in \{0\}$; $0+0=0\in \{0\}$; $𝜆0=0\in \{0\}$ for every $𝜆\in 𝐑$. Let $𝐿$ be a line in ${𝐑}^2$ containing the origin. Then $𝐿=\{𝑎𝑣:𝑎\in 𝐑\}$ for some nonzero $𝑣\in {𝐑}^2$. Then $𝐿$ is a subspace of ${𝐑}^2$: take $𝑎=0$, so $0𝑣=0\in 𝐿$; let $𝑎𝑣,𝑏𝑣\in 𝐿$, so $𝑎𝑣+𝑏𝑣=(𝑎+𝑏)𝑣\in 𝐿$ since $𝑎+𝑏\in 𝐑$; let $𝜆\in 𝐑$, so $𝜆(𝑎𝑣)=(𝜆𝑎)𝑣$, where $𝜆𝑎\in 𝐑$, by the compatibility of scalar multiplication with field multiplication, hence closed under scalar multiplication. ${𝐑}^2$ is a subspace of itself, as expected.

Let $𝑈$ be a subspace of ${𝐑}^2$.

Case 1. If $𝑈=\{0\}$, done.

Case 2. Suppose $𝑈\ne \{0\}$. $𝑈$ contains some nonzero $𝑣$, and so it contains the line $\{𝑎𝑣:𝑎\in 𝐑\}$. Case 2.1. $𝑈$ is exactly that line, done. Case 2.2. Otherwise, $𝑈$ contains some nonzero $𝑤$ that is not a multiple of $𝑣$, i.e., $𝑤\ne 𝜆𝑣$ for all $𝜆\in 𝐑$ (and vice versa).

Since $𝑈$ is a subspace of ${𝐑}^2$: for any $𝑎,𝑏\in 𝐑$, $𝑎𝑣,𝑏𝑤\in 𝑈$ by closure under scalar multiplication; $𝑎𝑣+𝑏𝑤\in 𝑈$ by closure under vector addition. Let’s define $𝑊=\{𝑎𝑣+𝑏𝑤:𝑎,𝑏\in 𝐑\}$, so $𝑊\subseteq 𝑈$.

Let’s define $𝐷={𝑣}_1{𝑤}_2-{𝑣}_2{𝑤}_1$. Suppose $𝐷=0$. Then ${𝑣}_1{𝑤}_2={𝑣}_2{𝑤}_1$. Suppose ${𝑣}_1\ne 0$, so ${𝑤}_2=\frac{{𝑤}_1}{{𝑣}_1}{𝑣}_2$. Take $𝜆=\frac{{𝑤}_1}{{𝑣}_1}$. Then both $𝜆{𝑣}_1={𝑤}_1$ and $𝜆{𝑣}_2={𝑤}_2$ hold. Hence $𝜆𝑣=𝑤$, contradicting the assumption that $𝑤$ is not a multiple of $𝑣$. Suppose ${𝑣}_1=0$. Then ${𝑣}_2\ne 0$, because $𝑣$ is a nonzero vector, so ${𝑤}_1=\frac{{𝑤}_2}{{𝑣}_2}{𝑣}_1$. Take $𝜆=\frac{{𝑤}_2}{{𝑣}_2}$. Then both $𝜆{𝑣}_2={𝑤}_2$ and $𝜆{𝑣}_1={𝑤}_1$ hold. Hence $𝜆𝑣=𝑤$, contradicting the assumption that $𝑤$ is not a multiple of $𝑣$. Thus $𝐷\ne 0$.

Let $(𝑥,𝑦)\in {𝐑}^2$. Then we have a linear system of equations comprising $𝑥=𝑎{𝑣}_1+𝑏{𝑤}_1$ and $𝑦=𝑎{𝑣}_2+𝑏{𝑤}_2$. Solving the system we have $𝑎=\frac{𝑥{𝑤}_2-𝑦{𝑤}_1}{{𝑣}_1{𝑤}_2-{𝑣}_2{𝑤}_1}$ and $𝑏=\frac{𝑦{𝑣}_1-𝑥{𝑣}_2}{{𝑣}_1{𝑤}_2-{𝑣}_2{𝑤}_1}$. ${𝑣}_1{𝑤}_2-{𝑣}_2{𝑤}_1\ne 0$ ($𝐷\ne 0$), so there exist $𝑎,𝑏$ such that $(𝑥,𝑦)=𝑎𝑣+𝑏𝑤$. So $(𝑥,𝑦)\in 𝑊$ and $(𝑥,𝑦)$ was arbitrary, hence ${𝐑}^2\subseteq 𝑊$, and because $𝑊\subseteq 𝑈$, so ${𝐑}^2\subseteq 𝑈$. $𝑈\subseteq {𝐑}^2$, since $𝑈$ is a subset of ${𝐑}^2$, and ${𝐑}^2\subseteq 𝑈$, thus $𝑈={𝐑}^2$.

Proof. $\{0\}$ is a subspace of ${𝐑}^3$: $0\in \{0\}$; $0+0\in \{0\}$; let $𝜆\in 𝐑$, so $𝜆0\in \{0\}$.

We define a line in ${𝐑}^3$ containing the origin as $𝐿=\{𝑎𝑣:𝑎\in 𝐑\}$ for some nonzero $𝑣\in {𝐑}^3$. Then $𝐿$ is a subspace of ${𝐑}^3$: take $𝑎=0$, so $0𝑣=0\in 𝐿$; let $𝑎𝑣,𝑏𝑣\in 𝐿$, so $𝑎𝑣+𝑏𝑣=(𝑎+𝑏)𝑣\in 𝐿$ where $𝑎+𝑏\in 𝐑$; let $𝜆\in 𝐑$, so $𝜆(𝑎𝑣)=(𝜆𝑎)𝑣\in 𝐿$ where $𝜆𝑎\in 𝐑$.

We define a plane in ${𝐑}^3$ containing the origin as $𝑃=\{𝑎𝑣+𝑏𝑢:𝑎,𝑏\in 𝐑\}$ for some vectors $𝑣,𝑢\in {𝐑}^3$ where neither of them is a multiple of the other. Then $𝑃$ is a subspace of ${𝐑}^3$: take $𝑎=𝑏=0$, so $0𝑣+0𝑢=0\in 𝑃$; let $𝑎𝑣+𝑏𝑢\in 𝑃$ and $𝑐𝑣+𝑑𝑢\in 𝑃$, so $(𝑎𝑣+𝑏𝑢)+(𝑐𝑣+𝑑𝑢)=(𝑎+𝑐)𝑣+(𝑏+𝑑)𝑢\in 𝑃$ where $𝑎+𝑐,𝑏+𝑑\in 𝐑$; let $𝜆\in 𝐑$, so $𝜆(𝑎𝑣+𝑏𝑢)=𝜆(𝑎𝑣)+𝜆(𝑏𝑢)=(𝜆𝑎)𝑣+(𝜆𝑏)𝑢\in 𝑃$ where $𝜆𝑎,𝜆𝑏\in 𝐑$.

${𝐑}^3$ is a subspace of ${𝐑}^3$: $0\in {𝐑}^3$; let $𝑢,𝑣\in {𝐑}^3$, so $𝑢+𝑣\in {𝐑}^3$; let $𝜆\in 𝐑$ and $𝑢\in {𝐑}^3$, so $𝜆𝑢\in {𝐑}^3$.

Let $𝑈$ be a subspace of ${𝐑}^3$. A subspace of a finite-dimensional ${𝐑}^3$ is finite-dimensional with $dim𝑈\le dim{𝐑}^3$, hence $dim𝑈\in \{0,1,2,3\}$.

Case 1. Suppose $dim𝑈=0$. Then by definition $𝑈$ has a basis of length $0$, hence $span()=𝑈$. By definition of a span of an empty list, $𝑈=\{0\}$.

Case 2. Suppose $dim𝑈=1$. Then by definition $𝑈$ has a basis of length $1$, hence $span(𝑣)=𝑈$ where $(𝑣)$ is linearly independent, so $𝑣$ is some nonzero vector (otherwise it is linearly dependent). By definition of a span, $𝑈=\{𝑎𝑣:𝑎\in 𝐑\}$, i.e., a line containing the origin.

Case 3. Suppose $dim𝑈=2$. Then by definition $𝑈$ has a basis of length $2$, hence $span(𝑣,𝑢)=𝑈$ where $(𝑣,𝑢)$ is linearly independent. Suppose $𝑣=𝜆𝑢$ for some $𝜆\in 𝐑$. Then $𝑣-𝜆𝑢=0$, contradicting the linear independence of $(𝑣,𝑢)$ (coefficient of $𝑣$ is $1$). Symmetrically, suppose $𝑢=𝜆𝑣$. Then $𝑢-𝜆𝑣=-𝜆𝑣+𝑢=0$, again, contradicting the linear independence of $(𝑣,𝑢)$ (coefficient of $𝑢$ is $1$). So $𝑣,𝑢$ are vectors where neither is a multiple of the other. Then by definition of a span, $𝑈=\{𝑎𝑣+𝑏𝑢:𝑎,𝑏\in 𝐑\}$, i.e., a plane containing the origin.

Case 4. Suppose $dim𝑈=3$. $dim{𝐑}^3=3$. We have $dim𝑈=dim{𝐑}^3=3$, so by 2.39 (subspace of full dimension equals the whole space) $𝑈={𝐑}^3$.

Proof. (a) $dim𝑈\le dim{\mathrm{𝒫︀}}_4(𝐅)$ because ${\mathrm{𝒫︀}}_4(𝐅)$ is finite-dimensional and $𝑈$ is a subspace of ${\mathrm{𝒫︀}}_4(𝐅)$ (by 2.37). Take $𝑝(𝑥)={𝑥}^4$ such that $𝑝\in {\mathrm{𝒫︀}}_4(𝐅)$. Then $𝑝\notin 𝑈$ since $𝑝(6)=6^4=1296\ne 0$. So $dim𝑈\ne dim{\mathrm{𝒫︀}}_4(𝐅)=5$, because $dim𝑈=5$ implies $𝑈={\mathrm{𝒫︀}}_4(𝐅)$ (by 2.39) which is not the case. Take polynomials $𝑥-6$, ${(𝑥-6)}^2$, ${(𝑥-6)}^3$, and ${(𝑥-6)}^4$ where each one is in $𝑈$ ($6-6=0$; ${(6-6)}^2=0$; ${(6-6)}^3=0$; ${(6-6)}^4=0$). Suppose $𝑎,𝑏,𝑐,𝑑\in 𝐅$ and $𝑎(𝑥-6)+{𝑏(𝑥-6)}^2+{𝑐(𝑥-6)}^3+{𝑑(𝑥-6)}^4=0$ We can see that the left side has a $𝑑{𝑥}^4$ term. Because the right side has no ${𝑥}^4$ term, this implies that $𝑑=0$. Because $𝑑=0$, we see that the left side has a $𝑐{𝑥}^3$ term, which implies that $𝑐=0$. Because $𝑐=𝑑=0$, we can conclude that $𝑏=0$. Because $𝑏=𝑐=𝑑=0$, we can also conclude that $𝑎=0$. Hence the list $𝑥-6,{(𝑥-6)}^2,{(𝑥-6)}^3,{(𝑥-6)}^4$ is a linearly independent list in $𝑈$. Thus $4\le dim𝑈<5$, so $dim𝑈=4$.