Week 1 Activity

Section 1.7 Week 1 Activity

Subsection 1.7.1 Higher-dimensional Space

Activity 1.7.1.

In 2-dimensional space, an infinitely long line divides the space into two pieces: one cannot pass from one side of the line to the other without crossing the line. In 3-dimensional space, however, an infinitely long line does not do this; I can go around the line, and there isn’t a division of space into two halves.

In 3-dimensional space, an infinitely extended plane also divides space into two pieces, and it is impossible to pass from one side to the other without going across the plane. In 4-dimensional space, does an infinitely extended plane still divide it into two pieces, or can I go around it, like I go around the line in 3-dimensional space?

Solution.

This is a tricky thing to try to visualize (being that 4-dimensional space is not visualizable). However, I can argue by analogy. How do I get around the line in 3-dimensions? I move to the side, then move past the line, instead of running into it. I can do this because I have two degrees of freedom in directions other than the direction of the line.

The same thing works for a plane in 4-dimensional space. While in 3-dimensional space, I only have one dimension of movement apart from the plane. To get to the other side, moving in this one degree of freedom, I have to cross the plane. In 4-dimensional space, I have two other degrees of freedom. Like getting around the line in 3-dimensional space, I can move to the side, then move past the plane to reach “the other side.” There really isn’t another side, much like a line in 3-dimensional space doesn’t really create two sides of space.

Activity 1.7.2.

In 2-dimensional space, a square or rectangle is enough to trap or contain something that lives in the plane. In 3-dimensional space, a cube (or other box) is enough to trap or contain something. What do I need to contain something in 4-dimensional space? In 5-dimensional space?

Solution.

To contain something in 4-dimensional space, I need a four dimensional box. Since I have four independent and free directions of movement, I have to close of movement in all four of these directions. This box will have a front/back, a top/bottom, a left/right, and then an additional pair of sides which block in the fourth direction of movement. In five dimensional space, I need a five-dimensional box: it will have a front/back, top/bottom, left/right, and two other independent boundary pairs.

Activity 1.7.3.

In 2-dimensional space, the slices of a rectangle are lines. In 3-dimensional space, the slices of a box are rectangles. What are the slices of a 4-dimensional box? A 5-dimensional box?

Solution.

By a “slice,” it seems I am referring to a piece of the object of one smaller dimension. Therefore, the slices of this four-dimensional box must be three-dimensional boxes. In this way, I can think of a four-dimensional box as an infinite stack of three-dimensional boxes, all stacked in some new direction of movement. (The same way I can think of a cube as an infinite stack of squares moving in a direction perpendicular to the squares themselves). Likewise, the slices of a five-dimensional box are themselves four-dimensional boxes.

Subsection 1.7.2 Vector Arithmetic

Activity 1.7.4.

Consider these vectors.

\begin{align*} u \amp = \left( \begin{matrix} 3 \\ 0 \end{matrix} \right) \amp v \amp = \left( \begin{matrix} -2 \\ -1 \end{matrix} \right) \amp w \amp = \left( \begin{matrix} -3 \\ 2 \end{matrix} \right) \end{align*}

Calculate and draw the following vectors.

\begin{align*} \amp -u \\ \amp -v \\ \amp -w \\ \amp u+v\\ \amp u+w\\ \amp v+w\\ \amp u-v\\ \amp v-u\\ \amp u-w\\ \amp w-u\\ \amp v-w\\ \amp w-v \end{align*}

In addition, calcluate these lengths.

\begin{align*} \amp |u|\\ \amp |v|\\ \amp |w|\\ \amp |u-v|\\ \amp |u-w|\\ \amp |v-w| \end{align*}

Solution.

I’ll do the arithmetic one by one. Then, at the end, all the vectors will be drawn in Figure 1.7.1.

\begin{align*} \amp -u = - \left( \begin{matrix} 3 \\ 0 \end{matrix} \right) = \left( \begin{matrix} -3 \\ 0 \end{matrix} \right) \\ \amp -v = - \left( \begin{matrix} -2 \\ -1 \end{matrix} \right) = \left( \begin{matrix} 2 \\ 1 \end{matrix} \right) \\ \amp -w = - \left( \begin{matrix} -3 \\ 2 \end{matrix} \right) = \left( \begin{matrix} 3 \\ -2 \end{matrix} \right) \\ \amp u+v = \left( \begin{matrix} 3 \\ 0 \end{matrix} \right) + \left( \begin{matrix} -2 \\ -1 \end{matrix} \right) = \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \\ \amp u+w = \left( \begin{matrix} 3 \\ 0 \end{matrix} \right) + \left( \begin{matrix} -3 \\ 2 \end{matrix} \right) = \left( \begin{matrix} 0 \\ 2 \end{matrix} \right) \\ \amp v+w = \left( \begin{matrix} -2 \\ - 1 \end{matrix} \right) + \left( \begin{matrix} -3 \\ 2 \end{matrix} \right) = \left( \begin{matrix} -5 \\ 1 \end{matrix} \right) \\ \amp u-v = \left( \begin{matrix} 3 \\ 0 \end{matrix} \right) - \left( \begin{matrix} -2 \\ -1 \end{matrix} \right) = \left( \begin{matrix} 5 \\ 1 \end{matrix} \right) \\ \amp v-u = \left( \begin{matrix} -2 \\ -1 \end{matrix} \right) - \left( \begin{matrix} 3 \\ 0 \end{matrix} \right) = \left( \begin{matrix} -5 \\ -1 \end{matrix} \right) \\ \amp u-w = \left( \begin{matrix} 3 \\ 0 \end{matrix} \right) - \left( \begin{matrix} -3 \\ 2 \end{matrix} \right) = \left( \begin{matrix} 6 \\ -2 \end{matrix} \right) \\ \amp w-u = \left( \begin{matrix} -3 \\ 2 \end{matrix} \right) - \left( \begin{matrix} 3 \\ 0 \end{matrix} \right) = \left( \begin{matrix} -6 \\ 2 \end{matrix} \right)\\ \amp v-w = \left( \begin{matrix} -2 \\ -1 \end{matrix} \right) - \left( \begin{matrix} -3 \\ 2 \end{matrix} \right) = \left( \begin{matrix} 1 \\ -3 \end{matrix} \right)\\ \amp w-v = \left( \begin{matrix} -3 \\ 2 \end{matrix} \right) - \left( \begin{matrix} -2 \\ -1 \end{matrix} \right) = \left( \begin{matrix} -1 \\ 3 \end{matrix} \right) \end{align*}

Figure 1.7.1. Vector Arithmetic

In addition, calcluate these lengths.

\begin{align*} \amp |u| = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 \\ \amp |v| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{5} \\ \amp |w| = \sqrt{(-3)^2 + 2^2} = \sqrt{13} \\ \amp |u-v| = \sqrt{5^2 + 1^2} = \sqrt{26} \\ \amp |u-w| = \sqrt{6^2 + (-2)^2} = \sqrt{40} = 2 \sqrt{10} \\ \amp |v-w| = \sqrt{1^2 + (-3)^2} = \sqrt{10} \end{align*}

Activity 1.7.5.

Consider these vectors.

\begin{align*} u \amp = \left( \begin{matrix} 6 \\ 0 \\ -1 \\ -1 \\ 1 \end{matrix} \right) \amp v \amp = \left( \begin{matrix} 3 \\ -3 \\ 1 \\ -3 \\ 1 \end{matrix} \right) \amp w \amp = \left( \begin{matrix} 0 \\ -1 \\ 0 \\ 3 \\ 5 \end{matrix} \right) \end{align*}

Calculate the following vectors and length. (Since this is in \(\RR^5\text{,}\) drawing these is no longer an option.)

\begin{align*} \amp -u \\ \amp -v \\ \amp -w \\ \amp u+v\\ \amp u+w\\ \amp v+w\\ \amp u-v\\ \amp v-u\\ \amp u-w\\ \amp w-u\\ \amp v-w\\ \amp w-v\\ \amp |u|\\ \amp |v|\\ \amp |w|\\ \amp |u-v|\\ \amp |u-w|\\ \amp |v-w| \end{align*}

Solution.

I’ll do the calculation one by one.

\begin{align*} \amp -u = - \left( \begin{matrix} 6 \\ 0 \\ -1 \\ -1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} -6 \\ 0 \\ 1 \\ 1 \\ -1 \end{matrix} \right) \\ \amp -v = - \left( \begin{matrix} 3 \\ -3 \\ 1 \\ -3 \\ 1 \end{matrix} \right) \left( \begin{matrix} -3 \\ 3 \\ -1 \\ 3 \\ -1 \end{matrix} \right)\\ \amp -w = - \left( \begin{matrix} 0 \\ -1 \\ 0 \\ 3 \\ 5 \end{matrix} \right) = \left( \begin{matrix} 0 \\ 1 \\ 0 \\ -3 \\ -5 \end{matrix} \right) \\ \amp u+v = \left( \begin{matrix} 6 \\ 0 \\ -1 \\ -1 \\ 1 \end{matrix} \right) + \left( \begin{matrix} 3 \\ -3 \\ 1 \\ -3 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 9 \\ -3 \\ 0 \\ -4 \\ 2 \end{matrix} \right) \\ \amp u+w = \left( \begin{matrix} 6 \\ 0 \\ -1 \\ -1 \\ 1 \end{matrix} \right) + \left( \begin{matrix} 0 \\ -1 \\ 0 \\ 3 \\ 5 \end{matrix} \right) = \left( \begin{matrix} 6 \\ -1 \\ -1 \\ 2 \\ 6 \end{matrix} \right) \\ \amp v+w = \left( \begin{matrix} 3 \\ -3 \\ 1 \\ -3 \\ 1 \end{matrix} \right) + \left( \begin{matrix} 0 \\ -1 \\ 0 \\ 3 \\ 5 \end{matrix} \right) = \left( \begin{matrix} 3 \\ -4 \\ 1 \\ 0 \\ 6 \end{matrix} \right) \\ \amp u-v = \left( \begin{matrix} 6 \\ 0 \\ -1 \\ -1 \\ 1 \end{matrix} \right) - \left( \begin{matrix} 3 \\ -3 \\ 1 \\ -3 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 3 \\ 3 \\ -2 \\ 2 \\ 0 \end{matrix} \right) \\ \amp v-u = \left( \begin{matrix} 3 \\ -3 \\ 1 \\ -3 \\ 1 \end{matrix} \right) - \left( \begin{matrix} 6 \\ 0 \\ -1 \\ -1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} -3 \\ -3 \\ 2 \\ -2 \\ 0 \end{matrix} \right) \\ \amp u-w = \left( \begin{matrix} 6 \\ 0 \\ -1 \\ -1 \\ 1 \end{matrix} \right) - \left( \begin{matrix} 0 \\ -1 \\ 0 \\ 3 \\ 5 \end{matrix} \right) = \left( \begin{matrix} 6 \\ 1 \\ -1 \\ -4 \\ -4 \end{matrix} \right) \\ \amp w-u = \left( \begin{matrix} 0 \\ -1 \\ 0 \\ 3 \\ 5 \end{matrix} \right) - \left( \begin{matrix} 6 \\ 0 \\ -1 \\ -1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} -6 \\ -1 \\ 1 \\ 4 \\ 4 \end{matrix} \right)\\ \amp v-w = \left( \begin{matrix} 3 \\ -3 \\ 1 \\ -3 \\ 1 \end{matrix} \right) - \left( \begin{matrix} 0 \\ -1 \\ 0 \\ 3 \\ 5 \end{matrix} \right) = \left( \begin{matrix} 3 \\ -2 \\ 1 \\ -6 \\ -4 \end{matrix} \right)\\ \amp w-v = \left( \begin{matrix} 0 \\ -1 \\ 0 \\ 3 \\ 5 \end{matrix} \right) - \left( \begin{matrix} 3 \\ -3 \\ 1 \\ -3 \\ 1 \end{matrix} \right) = \left( \begin{matrix} -3 \\ 2 \\ -1 \\ 6 \\ 4 \end{matrix} \right) \\ \amp |u| = \sqrt{6^2 + 0^2 + (-1)^2 + (-1)^2 + 1^2} = \sqrt{39} \\ \amp |v| = \sqrt{3^2 + (-3)^2 + 1^2 + (-3)^2 + 1^2} = \sqrt{29} \\ \amp |w| = \sqrt{0^2 + (-1)^2 + 0^2 + 3^2 + 5^2} = \sqrt{35} \\ \amp |u-v| = \sqrt{3^2 + 3^2 + (-2)^2 + 2^2 + 0^2} = \sqrt{26} \\ \amp |u-w| = \sqrt{6^2 + 1^2 + (-1)^2 + (-4)^2 + (-4)^2} = \sqrt{70} \\ \amp |v-w| = \sqrt{3^2 + (-2)^2 + 1^2 + (-6)^2 + (-4)^2} = \sqrt{66} \end{align*}

Subsection 1.7.3 Proof Questions

Activity 1.7.6.

If \(a \in \RR\) and \(u \in \RR^3\text{,}\) prove that \(|au| = |a||u|\text{.}\) (Use the definition of scalar multiplication and the definition of length of a vector. Start with the left side and try to turn it into the right or vice-versa. Do not reduce both equations to equivalent expressions -- that’s not a valid proof technique. You may try some examples out if you wish, but your final solution should be general. Finally, note that the notation here is a bit subtle: \(|au|\) and \(|u|\) are the lengths of vectors, defined by Pythagorean calculations, but \(|a|\) is just the absolute value of a real number.)

Solution.

I will start with the left side and try to manipulate it into the right side. I’ll write \(u\) in components and use the definition of the scalar multiplication to start.

\begin{equation*} |au| = \left| a \left( \begin{matrix} u_1 \\ u_2 \\ u_3 \end{matrix} \right) \right| = \left| \left( \begin{matrix} au_1 \\ au_2 \\ au_3 \end{matrix} \right) \right| \end{equation*}

This is the definition of scalar multiplication: multiply the scalar by each component of the vector. Now I use the square root definition of the length of a vector.

\begin{equation*} \left| \left( \begin{matrix} au_1 \\ au_2 \\ au_3 \end{matrix} \right) \right| = \sqrt{(au_1)^2 + (au_2)^2 + (au_3)^2} \end{equation*}

Now I need to do some algebra with the square root. I can distribute the exponent, using the exponent rule \((xy)^a = x^ay^a\text{.}\)

\begin{equation*} \sqrt{(au_1)^2 + (au_2)^2 + (au_3)^2} = \sqrt{a^2 u_1^2 + a^2 u_2^2 + a_2u_3^2} \end{equation*}

Now I can factor out the \(a^2\text{.}\) Once it is factored out, I can also remove it from the square root.

\begin{equation*} \sqrt{a^2 u_1^2 + a^2 u_2^2 + a_2u_3^2} = \sqrt{a^2(u_1^2 + u_2^2 + u_3^2)} = |a| \sqrt{u_1^2 + u_2^2 + u_3^2} \end{equation*}

When I take \(a^2\) out of the square root, since the square root also returns a positive number, I don’t get \(a\) back. If \(a\) was negative to start, then I will have turned \(a\) from negative to positive. Therefore, I get the absolute value of \(a\) back. Finally, I can recognize the remaining term as the definition of the length of a vector.

\begin{equation*} |a| \sqrt{u_1^2 + u_2^2 + u_3^2} = |a| |u| \end{equation*}

Now I have started with the left side and produced the right. This proves that \(|au| = |a||u|\text{,}\) as desired.

Activity 1.7.7.

If \(u\) and \(v\) are vectors in \(\RR^3\text{,}\) prove that \(|u-v| = |v-u|\text{.}\) Then adapt your argument to work for vectors in \(\RR^n\text{.}\)

Solution.

I will write \(u\) and \(v\) in components, take their difference and then apply the length operation. I will start with the left side and try to produce the right.

\begin{equation*} |u-v| = \left| \left( \begin{matrix} u_1 \\ u_2 \\ u_3 \end{matrix} \right) - \left( \begin{matrix} v_1 \\ v_2 \\ v_3 \end{matrix} \right) \right| \end{equation*}

The difference is done componentwise.

\begin{equation*} \left| \left( \begin{matrix} u_1 \\ u_2 \\ u_3 \end{matrix} \right) - \left( \begin{matrix} v_1 \\ v_2 \\ v_3 \end{matrix} \right) \right| = \left| \left( \begin{matrix} u_1 - v_1 \\ u_2 - v_2 \\ u_3 - v_3 \end{matrix} \right) \right| \end{equation*}

Then I use the square root definition of the length.

\begin{equation*} \left| \left( \begin{matrix} u_1 - v_1 \\ u_2 - v_2 \\ u_3 - v_3 \end{matrix} \right) \right| = \sqrt{(u_1-v_1)^2 + (u_2-v_2)^2 + (u_3-v_3)^2} \end{equation*}

Now, each of the three terms is square. Since they are square, any positive or negative will become positive since squares are always positive. Therefore, I could multiply by \((-1)\) inside the square root to no effect. This will interchange the order of the difference in each term.

\begin{equation*} \sqrt{(u_1-v_1)^2 + (u_2-v_2)^2 + (u_3-v_3)^2} = \sqrt{(v_1-u_1)^2 + (v_2-u_2)^2 + (v_3-u_3)^2} \end{equation*}

Then I repeat the steps from before in reverse. This is the definition of the length of a vector with the terms in the squares as components.

\begin{equation*} \sqrt{(v_1-u_1)^2 + (v_2-u_2)^2 + (v_3-u_3)^2} = \left| \left( \begin{matrix} v_1 - u_1 \\ v_2 - u_2 \\ v_3 - u_3 \end{matrix} \right) \right| \end{equation*}

Since each term is a difference, I can write this as the difference of two vectors.

\begin{equation*} \left| \left( \begin{matrix} v_1 - u_1 \\ v_2 - u_2 \\ v_3 - u_3 \end{matrix} \right) \right| = \left| \left( \begin{matrix} v_1 \\ v_2 \\ v_3 \end{matrix} \right) - \left( \begin{matrix} u_1 \\ u_2 \\ u_3 \end{matrix} \right) \right| \end{equation*}

Finally, these are just the component of the vectors \(u\) and \(v\text{.}\)

\begin{equation*} = \left| \left( \begin{matrix} v_1 \\ v_2 \\ v_3 \end{matrix} \right) - \left( \begin{matrix} u_1 \\ u_2 \\ u_3 \end{matrix} \right) \right| = |v-u| \end{equation*}

I have started from the left and produced the right in a string of equality, so I have proven the desired result.

If I wanted to adapt this to a more general proof for \(\RR^n\text{,}\) the steps of the proof would be identical, but I would have vectors with \(n\) components instead of just three components. I would use ellipses to indicate the list of components. The proof would start like this.

\begin{equation*} |u-v| = \left| \left( \begin{matrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{matrix} \right) - \left( \begin{matrix} v_1 \\ v_2 \\ \\ \vdots \\ v_n \end{matrix} \right) \right| \end{equation*}

The difference is done componentwise.

\begin{equation*} \left| \left( \begin{matrix} u_1 \\ u_2 \\\vdots \\ u_n \end{matrix} \right) - \left( \begin{matrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{matrix} \right) \right| = \left| \left( \begin{matrix} u_1 - v_1 \\ u_2 - v_2 \\ \vdots \\ u_n - v_n \end{matrix} \right) \right| \end{equation*}

I won’t repeat all the steps, but they would proceed exactly in the same sequence as before, resulting in \(|v-u|\) eventually at the end.

Activity 1.7.8.

Prove that the zero vector is the only vector in \(\RR^n\) of length zero. (To prove that it is the only vector with this property, assume that \(u \in \RR^n\) is a vector with length zero and prove that \(u\) must be the zero vector. This will show that the zero vector is the only vector with the desired property).

Solution.

Let \(u\) be an arbitrary vector in \(\RR^n\) of length 0. I will start by writing \(u\) in components and use the definition of length. Note I need ellipses to show all the components of \(u\text{,}\) since I am working in an arbitrary dimension \(n\text{.}\)

\begin{equation*} |u| = \left| \left( \begin{matrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{matrix} \right) \right| \end{equation*}

Then I use the definition of length. By assumption, I set this length equal to zero.

\begin{equation*} |u| = \sqrt{(u_1)^2 + (u_2)^2 + \ldots + (u_n)^2} = 0 \end{equation*}

I can square both sides of the equation.

\begin{equation*} \left( \sqrt{(u_1)^2 + (u_2)^2 + \ldots + (u_n)^2} \right)^2 = 0^2 \end{equation*}

The square root and squaring cancel each other out.

\begin{equation*} (u_1)^2 + (u_2)^2 + \ldots + (u_n)^2 = 0 \end{equation*}

On the left, I have the sum of a bunch of non-negative numbers (since the square of any real number must be zero or postive). On the right I have zero. The only way that a bunch of non-negative numbers add up to zero is if they are all zero to begin with. Therefore, I see that for each index \(i\text{,}\) the component \(u_i\) must be zero. But that means that

\begin{equation*} \left( \begin{matrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{matrix} \right) = \left( \begin{matrix} 0 \\ 0 \\ \vdots \\ 0 \end{matrix} \right) \end{equation*}

This is precisely the definition of the zero vector: the vector where all components are zero. Therefore, \(u\) must be the zero vector. To conclude, I assumed that \(u\) had length zero and thus concluded that it must be the zero vector. This proved that the zero vector is the only vector with length zero.

Subsection 1.7.4 Conceptual Review Questions

How is it reasonable to talk about higher-dimensional geometric spaces?
What is cartesian or euclidean space? What are coordinates?
What is a vector?
What is a scalar?
How is vector arithmetic similar to and different from number arithmetic.
What is the length of a vector?

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