## Motivation

If you have taken high school physics, you might have seen that a vector is a quantity with magnitude and direction. Even though such a definition of a vector is sufficient for problems in high school physics, we will need a more general (as a result more abstract) notion of a vector in order to deal with mathematics and applications beyond high school physics.

Like the definition of a field, the definition of a vector space can seem overwhelmingly technical at first sight. The properties of a vector space are extracted from the many common properties displayed by different structures. As a result, vector spaces allow us to talk about all these different structures under the same abstraction.

To motivate the definition of a vector space, we highlight some central ideas with an examples.

## Solutions to $$Ax = 0$$

Consider the system $\begin{array}{rcrcrcrccc} u & - & 3v & & & + &2 z & = & 0 \\ & & & & y & - & z & = & 0. \end{array}$ This system can be written as $$A x = 0$$ with $$A = \begin{bmatrix} 1 & -3 & 0 & 2 \\ 0 & 0 & 1 & -1\end{bmatrix}$$ and $$x = \begin{bmatrix} u \\ v \\ y \\ z \end{bmatrix}$$.

Note that $$A$$ is already in reduced row-echelon form. We see that $$v$$ and $$z$$ are free variables. Hence, all the solutions are given by $\begin{bmatrix} u \\ v \\ y \\ z \end{bmatrix} = s \begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t \begin{bmatrix} -2 \\ 0 \\ 1 \\ 1 \end{bmatrix}$ for all $$s, t \in \mathbb{R}$$.

Note that the system can also be written as $\begin{array}{rcrcrcrccc} 2z & & & - & 3v & + & u & = & 0 \\ -z & + & y & & & & & = & 0. \end{array}$ In matrix form, it is $$A' x' = 0$$ with $$A' = \begin{bmatrix} 2 & 0 & -3 & 1 \\ -1 & 1 & 0 & 0 \end{bmatrix}$$ and $$x' = \begin{bmatrix} z \\ y \\ v \\ u \end{bmatrix}$$.

Row-reducing $$A'$$ gives the matrix $$\begin{bmatrix} 1 & 0 & -\frac{3}{2} & \frac{1}{2} \\ 0 & 1 & -\frac{3}{2} & \frac{1}{2}\end{bmatrix}$$. The third and fourth columns of this matrix are nonpivot columns. But the variables correspond to these columns are the variables in the third and fourth entries of $$x'$$; that is, $$v$$ and $$u$$. Hence, all the solutions are given by $\begin{bmatrix} u \\ v \\ y \\ z \end{bmatrix} = s' \begin{bmatrix} 0 \\ 1 \\ \frac{3}{2} \\ \frac{3}{2} \end{bmatrix}+ t' \begin{bmatrix} 1 \\ 0 \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}$ for all $$s', t' \in \mathbb{R}$$.

So we have two different-looking descriptions of all the solutions to $\begin{array}{rcrcrcrccc} u & - & 3v & & & + &2 z & = & 0 \\ & & & & y & - & z & = & 0. \end{array}$ Namely, $\begin{bmatrix} u \\ v \\ y \\ z \end{bmatrix} = s \begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t \begin{bmatrix} -2 \\ 0 \\ 1 \\ 1 \end{bmatrix}~~~~\text{for all } s,t \in \mathbb{R}$ and $\begin{bmatrix} u \\ v \\ y \\ z \end{bmatrix} = s' \begin{bmatrix} 0 \\ 1 \\ \frac{3}{2} \\ \frac{3}{2} \end{bmatrix}+ t' \begin{bmatrix} 1 \\ 0 \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}\ ~~~~\text{for all } s', t' \in \mathbb{R}.$

We can easily see a similarity between $$\begin{bmatrix} -2 \\ 0 \\ 1 \\ 1 \end{bmatrix}$$ in the first description and $$\begin{bmatrix} 1 \\ 0 \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}$$ in the second description; the former can be obtained by multiplying the latter by $$-2$$. However, there is no immediate connection between $$\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}$$ and $$\begin{bmatrix} 0 \\ 1 \\ \frac{3}{2} \\ \frac{3}{2} \end{bmatrix}$$. How are they related given that both descriptions are for the same solution set?

As $$\begin{bmatrix} u \\ v \\ y \\ z\end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ \frac{3}{2} \\ \frac{3}{2} \end{bmatrix}$$ is a solution, there must exist real numbers $$s$$ and $$t$$ such that $$\begin{bmatrix} 0 \\ 1 \\ \frac{3}{2} \\ \frac{3}{2} \end{bmatrix} = s \begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t \begin{bmatrix} -2 \\ 0 \\ 1 \\ 1 \end{bmatrix}$$. Note that setting $$s = 1$$ and $$t = \frac{3}{2}$$ works.

Similarly, as $$\begin{bmatrix} u \\ v \\ y \\ z\end{bmatrix} =\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}$$ is a solution, there must exist real numbers $$s'$$ and $$t'$$ such that $$\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix} = s' \begin{bmatrix} 0 \\ 1 \\ \frac{3}{2} \\ \frac{3}{2} \end{bmatrix}+ t' \begin{bmatrix} 1 \\ 0 \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}.$$ Setting $$s' = 1$$ and $$t' = 3$$ works.

We are now ready to make some general remarks.

Observe that both descriptions are in the form $$\alpha t + \alpha' t'$$ where $$\alpha$$ and $$\alpha'$$ are scalars and $$t$$ and $$t'$$ are tuples of the same size. We call $$\alpha t + \alpha' t'$$ a linear combination of $$t$$ and $$t'$$. (That both descriptions above are linear combinations of the same number of tuples is not a coincidence. We will see more on this phenomenon when we discuss bases and dimensions of vector spaces.)

More generally, if $$t_1,\ldots,t_k$$ are $$n$$-tuples and $$\alpha_1,\ldots,\alpha_k$$ are scalars, then $\alpha_1 t_1 + \cdots + \alpha_k t_k$ is called a linear combination of $$t_1,\ldots,t_k$$. It can be seen that if $$A \in \mathbb{F}^{m\times n}$$ for some positive integers $$m$$ and $$n$$, the solutions to $$Ax = 0$$ can be expressed as linear combinations of $$n$$-tuples in $$\mathbb{F}^n$$.

Let us make a few more observations on the solutions to $$Ax = 0$$ in general.

Recall that for a given $$A\in \mathbb{F}^{m \times n}$$, $$N(A)$$ denotes the set of solutions to the system $$Ax = 0$$. (As usual, $$0$$ denotes the $$m$$-tuple of 0's.)

Take any two $$u, v \in N(A)$$. Then $$u + v$$ also gives a solution to $$Ax = 0$$. To see this, note that $$A(u+v) = Au + Av = 0 + 0 = 0$$. Hence, $$u + v \in N(A)$$. In other words, adding two elements of $$N(A)$$ results in an element of $$N(A)$$.

Now, take $$\alpha \in \mathbb{F}$$ and $$v \in N(A)$$. Then $$\alpha v$$ also gives a solution to $$Ax = 0$$. To see this, note that $$A(\alpha v) = \alpha (Av) = \alpha 0 = 0$$. Hence, $$\alpha v \in N(A)$$. In other words, taking a scalar multiple of an element of $$N(A)$$ results in an element of $$N(A)$$.

## Exercise

Write $$\begin{bmatrix} 3 \\ 0 \end{bmatrix}$$ as a linear combination of $$\begin{bmatrix} 1 \\ 2 \end{bmatrix}$$ and $$\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$.