Print this problem set out (there are X problems!) and answer the problems on the given sheet.
  1. Consider the following definition:
    If $a_1,\ldots,a_n$ are elements of some ring $R$, and $x_1,\ldots,x_n$ are objects on which addition is defined and multiplication by elements of $R$ is defined, we call $$ a_1 \cdot x_1 + a_2\cdot x_2 + \cdots + a_n\cdot x_n $$ a linear combination of the $x_i$'s. We refer to the $a_i$'s as the coefficients in the linear combination.
    So, for example, the dot product of two vectors in $R^n$ is a linear combination, since $ [a_1\ \ldots\ a_n] \cdot [x_1\ \ldots\ x_n] = a_1 \cdot x_1 + a_2\cdot x_2 + \cdots + a_n\cdot x_n $. The $a_i$'s and $x_i$'s are all just elements of of the ring $R$. Of course, it's much more exciting to have linear combinations of vectors!

    1. What is the following linear combination of column vecotors as a single column vector? [Remember a column vector is literally just a vector where we've written the components vertically instead of horizontally!] $$ 2\cdot \begin{bmatrix} -1\\ 0\\ 2 \end{bmatrix} + -1\cdot \begin{bmatrix} 0\\ 3\\ 3 \end{bmatrix} + \frac{1}{3}\cdot \begin{bmatrix} 6\\ 3\\ -9 \end{bmatrix} $$
    2. Find real number values for $x_1$, $x_2$ and $x_3$, not all zero(!), that makes the following equation involving a linear combination of vectors true.
      Note: I'm expecting you to play around a bit with different numbers to find these values. We will eventually learn how to solve something like this systematically, but for now, think of it like a puzzle. $$ x_1\cdot \begin{bmatrix} -1\\ 0\\ 2 \end{bmatrix} + x_2\cdot \begin{bmatrix} 0\\ 3\\ 3 \end{bmatrix} + x_3\cdot \begin{bmatrix} 6\\ 3\\ -9 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} $$
  2. If we have a vector $\boldsymbol{w}$ in vector space $R^n$, we can think of the dot product $\boldsymbol{w}\cdot\boldsymbol{x}$ as defining a function that takes a vector $\boldsymbol{x}$ as input, and outputs a scalar (i.e. an element of $R$). If we have multiple "$\boldsymbol{w}$" vectors, we could think of each of them defining a function, and collect their outputs on the same input $\boldsymbol{x}$ into a new vector. This would give us a function that takes a vector as input and produces a vector as output. Here's an example: $$ f(\boldsymbol{x}) = \begin{bmatrix} \boldsymbol{w_1}\cdot \boldsymbol{x}\\ \boldsymbol{w_2}\cdot \boldsymbol{x}\\ \boldsymbol{w_3}\cdot \boldsymbol{x} \end{bmatrix} , \text{ where } \begin{array}{l} \boldsymbol{w_1} = [-1\ 0\ \ \ 6\ ]\\ \boldsymbol{w_2} = [\ \ \ 0\ 3\ \ \ 3\ ]\\ \boldsymbol{w_3} = [\ \ \ 2\ 3\ -9] \end{array} $$ What is $f(\boldsymbol{x})$ when $\boldsymbol{x} = [4\ -1\ 2]$?

  3. Let $\boldsymbol{u}$ be a vector in $\mathbb{R}^n$, i.e. the vector space with dimension $n$ and over the real numbers.
    1. Why does $\sqrt{\boldsymbol{u}\cdot \boldsymbol{u}}$ make sense, but $\sqrt{\boldsymbol{u}}$ does not?
    2. Prove that $\boldsymbol{u}\cdot\boldsymbol{u} \geq 0$
    3. Let's restrict ourselves to $n=2$ and ring $\mathbb{R}$. Then we can view any vector $\boldsymbol{u} = [u_1\ u_2]$ as being the point $(x,y) = (u_1,u_2)$ in the Cartesian plane.
      Prove the following: given two vectors $\boldsymbol{u}$ and $\boldsymbol{v}$ in $\mathbb{R}^2$, the distance between the point defined by $\boldsymbol{u}$ and point defined by $\boldsymbol{v}$ is equal to $\sqrt{(\boldsymbol{u} - \boldsymbol{v}) \cdot (\boldsymbol{u} - \boldsymbol{v})}$.

      Example: If $\boldsymbol{u} = [4\ 2]$ and $\boldsymbol{v} = [1\ -2]$. They define points $(4,2)$ and $(1,-2)$. The distance between these points is $\sqrt{(4 - 1)^2 + (2 - - 2)^2} = 5$, and $\sqrt{(\boldsymbol{u} - \boldsymbol{v}) \cdot (\boldsymbol{u} - \boldsymbol{v})}=5$ as well.

    Discussion: For vectors over the real numbers, $\sqrt{\boldsymbol{v}\cdot \boldsymbol{v}}$ is the distance from the origin to the point $\boldsymbol{v}$ defines. You are most used to points in 2-space and 3-space, but all the concepts of geometry make sense for larger values of $n$ as well, and this interpretation of distance holds no matter what the dimension. This gives us one notion of "size" for vectors over $\mathbb{R}$: the size of a vector is its distance from the origin.
    The Euclidean norm of a vector $\boldsymbol{v}$ over the ring $\mathbb{R}$, denoted $||\boldsymbol{v}||_2$, is defined as $||\boldsymbol{v}||_2 = \sqrt{\boldsymbol{v}\cdot \boldsymbol{v}}$.