# Harrison Chapman

## Linear Algebra I: Homework 4

Due: Friday, September 15, 2017
1. Let $\vec v$ be the vector in $\mathbb R^4$ which points from $P(1,-1,0,3)$ to $Q(0,-2,-3,-3)$.
1. Express $\vec v$ as a column vector.

2. Find the magnitude of $\vec v$.

3. Find the angle from $\vec v$ to the vector,

Because $\vec v \cdot \vec w = \lVert \vec v \rVert \lVert \vec w \rVert\cos\theta$, we get that

2. The matrix $R_\theta$

has a nice graphical explanation. If $\vec v$ is a 2-vector in $\mathbb R^2$, the vector $R_\theta \vec v$ (that is, the product of the matrix multiplication) has the same length as $\vec v$, but has been rotated by $\theta$ degrees counterclockwise ($\theta$ can be any angle).

Let $\vec v$ be the 2-vector

Without actually computing the vector $R_{\pi/2} \vec v$, compute the following:

1. The dot product,

The vector $R_{\pi/2}\vec v$ is the same as $\vec v$ but rotated by $\pi/2$ radians. So, it is orthogonal to $\vec v$, meaning that the dot product is $0$.

2. The magnitude,

The vector $R_{\pi/2}\vec v$ is the same as $\vec v$ but rotated by $\pi/2$ radians. So, it is the same length as $\vec v$, and its magnitude $\lVert R_{\pi/2}\vec v \rVert = \lVert \vec v \rVert = \sqrt{1^2 + (-2)^2} = \sqrt{5}$.

3. Let $\vec {r_0}$ be a fixed vector in $\mathbb R^2$. For each part, describe in words the set of all vectors $\vec r$ that satisfy the stated condition. Hint: Think about nice shapes. An answer which just re-writes the math in English will not receive full credit.

1. $\lVert \vec {r} - \vec {r_0} \rVert = 1$.

If the vector $\vec {r_0}$ is drawn starting at the origin and ending at a point $C$, then the set of all $\vec r$ is the set of all vectors which, when drawn starting at he origin, end in the unit circle around $C$.

2. $\lVert \vec {r} - \vec {r_0} \rVert \ge 1$.

If the vector $\vec {r_0}$ is drawn starting at the origin and ending at a point $C$, then the set of all $\vec r$ is the set of all vectors which, when drawn starting at he origin, end in or outside of the unit circle around $C$.

4. Explain why the line of 3-vectors,

is not a vector space.

You only have to show that $L$ fails one of the vector space axioms. The easiest in this case is that it fails to have the zero vector $\vec 0$:

Vectors in $L$ look like

and always have a 1 in the third component. But the zero vector is

which has a zero (not a 1) in the third component, and so can’t be in $L$.

5. Does there exist a linear transformation $T: \mathbb R^2 \to \mathbb R^3$ such that,

and,

No. If $T$ were linear then
So $T$ can’t be linear.