Vector space of real numbers

Unit 9 Vector space of real numbers

Key unit competence: Determine the magnitude of vectors, the angle between two vectors and the dot product of two vectors.

Key ideas

• ${\mathbb{R}}^2=\{(x,y):x,y\in \mathbb{R}\}$ with addition and multiplication by a scalar is a vector space: it satisfies closure, associativity, commutativity, zero vector, opposites, and distributive laws.
• A subspace $W$ of ${\mathbb{R}}^2$ is a non-empty subset closed under addition and scalar multiplication (it contains $\vec{0}$). The subspaces of ${\mathbb{R}}^2$ are $\{\vec{0}\}$, lines through the origin, and ${\mathbb{R}}^2$.
• A linear combination of $\vec{u},\vec{v}$ is $a\vec{u}+b\vec{v}$. Vectors are linearly independent if $a\vec{u}+b\vec{v}=\vec{0}\Rightarrow a=b=0$; in ${\mathbb{R}}^2$, $\left(\begin{array}{c}a\\ b\end{array}\right)$ and $\left(\begin{array}{c}c\\ d\end{array}\right)$ are independent iff $ad-bc\ne 0$.
• A basis is a linearly independent set that spans the space; $\dim {\mathbb{R}}^2=2$. Standard basis $\{\vec{i},\vec{j}\}$.
• Dot product: $\vec{u}\cdot \vec{v}=x_1{x}_2+y_1{y}_2=|\vec{u}||\vec{v}|\cos \theta$. Magnitude $|\vec{u}|=\sqrt{x^2+y^2}=\sqrt{\vec{u}\cdot \vec{u}}$.
• Angle: $\cos \theta =\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$. Orthogonal when $\vec{u}\cdot \vec{v}=0$. Unit vector $\frac{\vec{u}}{|\vec{u}|}$.

Formulas to know

Common misconceptions

• "Any two non-zero vectors form a basis of ${\mathbb{R}}^2$." No: they must be linearly independent (not parallel).
• "The line $y=x+1$ is a subspace of ${\mathbb{R}}^2$." No: it does not contain $(0,0)$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Show that $\{(2,1),(1,3)\}$ is a basis of ${\mathbb{R}}^2$ and write $(4,7)$ in this basis.

   Show answer

   $2\cdot 3-1\cdot 1=5\ne 0$: independent, so a basis. $2a+b=4$, $a+3b=7$: $a=1$, $b=2$.

   Common mistake: Checking only that the vectors are non-zero.

2. Find $|\vec{u}-\vec{v}|$ for $\vec{u}=(4,1)$, $\vec{v}=(1,5)$.

   Show answer

   $\vec{u}-\vec{v}=(3,-4)$, magnitude 5.

   Common mistake: Subtracting the magnitudes ($\sqrt{17}-\sqrt{26}$).

3. Find the angle between $\vec{u}=(2,-1)$ and $\vec{v}=(1,3)$.

   Show answer

   $\vec{u}\cdot \vec{v}=-1$; $|\vec{u}|=\sqrt{5}$, $|\vec{v}|=\sqrt{10}$; $\cos \theta =-\frac{1}{\sqrt{50}}$, $\theta \approx 98.1^{\circ }$.

   Common mistake: Using $|\vec{u}{|}^2$ instead of $|\vec{u}|$.

4. Explain why $\{(x,y):x\ge 0,y\ge 0\}$ is not a subspace.

   Show answer

   It is not closed under multiplication by negative scalars: $-1\cdot (1,1)=(-1,-1)$ is not in the set.

   Common mistake: Saying it does not contain the zero vector (it does).

5. Prove that $W=\{(x,y):y=-x\}$ is a subspace of ${\mathbb{R}}^2$.

   Show answer

   $(0,0)\in W$. If $(a,-a),(b,-b)\in W$, their sum $(a+b,-(a+b))\in W$; $k(a,-a)=(ka,-ka)\in W$.

   Common mistake: Checking closure under addition only.

6. Given $|\vec{u}|=2$, $|\vec{v}|=5$ and $\vec{u}\cdot \vec{v}=-5$, find the angle between them.

   Show answer

   $\cos \theta =\frac{-5}{10}=-\frac{1}{2}$, $\theta =120^{\circ }$.

   Common mistake: Taking the positive value $60^{\circ }$.

7. Find a unit vector perpendicular to $(1,2)$.

   Show answer

   $(-2,1)$ is perpendicular; unit: $\left(-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)$.

   Common mistake: Forgetting to divide by the magnitude.

8. Show that $(1,1)$ and $(1,-1)$ are orthogonal and use them as a basis to write $(3,5)$.

   Show answer

   Dot product $1-1=0$. $a+b=3$, $a-b=5$: $a=4$, $b=-1$: $(3,5)=4(1,1)-(1,-1)$.

   Common mistake: Sign errors in solving.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $|\vec{u}|=3$, $|\vec{v}|=4$ and the angle between them is $60^{\circ }$. Find $\vec{u}\cdot \vec{v}$.	a. $3.4$
   2. Find the angle between $(1,1)$ and $(1,0)$.	b. $45^{\circ }$
   3. Find the angle between $(3,4)$ and $(4,3)$.	c. about $16.3^{\circ }$
   4. Find the scalar projection of $\vec{u}=(2,3)$ on $\vec{v}=(4,3)$.	d. $6$

   Show answer

   1 → d ($6$) · 2 → b ($45^{\circ }$) · 3 → c (about $16.3^{\circ }$) · 4 → a ($3.4$)

   • Step 1. $\vec{u}\cdot \vec{v}=|\vec{u}||\vec{v}|\cos 60^{\circ }$.
     Step 2. $3\times 4\times \frac{1}{2}=6$.
   • Step 1. Dot product 1; magnitudes $\sqrt{2}$ and 1.
     Step 2. $\cos \theta =\frac{1}{\sqrt{2}}$, so $\theta =45^{\circ }$.
   • Step 1. Dot product $12+12=24$; both magnitudes are 5.
     Step 2. $\cos \theta =\frac{24}{25}$, $\theta \approx 16.3^{\circ }$.
   • Step 1. $\vec{u}\cdot \vec{v}=8+9=17$ and $|\vec{v}|=5$.
     Step 2. Projection $=\frac{17}{5}=3.4$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\left(-\frac{3}{5},\frac{4}{5}\right)$  ·  $k=6$  ·  $120^{\circ }$  ·  $k=-\frac{3}{2}$
   1. Find the angle between $(1,0)$ and $(-1,\sqrt{3})$. →
   2. Find $k$ so that $(k,3)$ and $(2,-4)$ are orthogonal. →
   3. Find the unit vector in the direction of $(-6,8)$. →
   Show answer

   1: $120^{\circ }$ · 2: $k=6$ · 3: $\left(-\frac{3}{5},\frac{4}{5}\right)$

   • Step 1. Dot product $-1$; magnitudes 1 and 2.
     Step 2. $\cos \theta =-\frac{1}{2}$, so $\theta =120^{\circ }$.
   • Step 1. $2k-12=0$.
     Step 2. $k=6$.
   • Step 1. $|(-6,8)|=\sqrt{36+64}=10$.
     Step 2. Divide each component by 10.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Which set is a subspace of ${\mathbb{R}}^2$?	a. $90^{\circ }$
   2. Find the angle between $(2,1)$ and $(-1,2)$.	b. $\{(-4,1)\}$
   3. Find the coordinates of $(5,8)$ in the basis $\{(1,2),(3,5)\}$.	c. $\{(x,y):3x-y=0\}$
   4. Give a basis of $W=\{(x,y):x+4y=0\}$.	d. $(-1,2)$

   Show answer

   1 → c ($\{(x,y):3x-y=0\}$) · 2 → a ($90^{\circ }$) · 3 → d ($(-1,2)$) · 4 → b ($\{(-4,1)\}$)

   • Step 1. $3x-y=0$ is a line through the origin: it contains $\vec{0}$ and is closed.
     Step 2. The others miss $\vec{0}$ or are not closed under addition or scalar multiplication.
   • Step 1. Dot product: $-2+2=0$.
     Step 2. A zero dot product means the vectors are perpendicular.
   • Step 1. $a+3b=5$ and $2a+5b=8$.
     Step 2. From the first: $a=5-3b$; then $10-6b+5b=8$, so $b=2$, $a=-1$.
     Step 3. Check: $-(1,2)+2(3,5)=(5,8)$. ✓
   • Step 1. $x=-4y$, so $(x,y)=y(-4,1)$.
     Step 2. One vector spans $W$: dimension 1.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $3$  ·  $\frac{3}{5}$  ·  $\sqrt{37}$  ·  $a=\frac{3}{2}$, $b=\frac{5}{2}$
   1. With $|\vec{u}|=3$, $|\vec{v}|=4$ and $\vec{u}\cdot \vec{v}=6$, find $|\vec{u}+\vec{v}|$. →
   2. Write $(4,-1)$ as $a(1,1)+b(1,-1)$. →
   3. Find the length of the projection of $\vec{u}=(3,4)$ on the direction of $\vec{v}=(1,0)$. →
   Show answer

   1: $\sqrt{37}$ · 2: $a=\frac{3}{2}$, $b=\frac{5}{2}$ · 3: $3$

   • Step 1. $|\vec{u}+\vec{v}{|}^2=|\vec{u}{|}^2+2\vec{u}\cdot \vec{v}+|\vec{v}{|}^2$.
     Step 2. $9+12+16=37$.
   • Step 1. $a+b=4$ and $a-b=-1$.
     Step 2. Add: $2a=3$, $a=\frac{3}{2}$; then $b=\frac{5}{2}$.
   • Step 1. $\frac{\vec{u}\cdot \vec{v}}{|\vec{v}|}=\frac{3}{1}$.
     Step 2. $=3$.

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