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.

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 \(N(A)\).

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