Basis

Think of a basis as a minimal set of building blocks for a space. In the plane $\mathbb{R}^2$, the two arrows $(1,0)$ (pointing right) and $(0,1)$ (pointing up) let you reach any point: the point $(3,-2)$ is just $3$ steps right and $2$ steps down, i.e. $(3,-2) = 3\,(1,0) + (-2)\,(0,1).$ Two properties make this work, and a basis is exactly the set that has both:

• Nothing is wasted (independence). None of the building blocks is redundant — you cannot build one of them out of the others. If you tried to use $(1,0)$ and $(2,0)$, the second adds no new direction, so you could never leave the $x$-axis.
• Nothing is missing (spanning). The building blocks are enough to reach every vector in the space. Using only $(1,0)$ you can reach the $x$-axis but never $(0,1)$.

When both hold, every vector has exactly one recipe in terms of the basis. Those unique numbers $(3,-2)$ above are the coordinates of the vector. A basis is what turns an abstract space into a grid you can do arithmetic on.

Let $V$ be a vector space (for us, usually $V=\mathbb{R}^n$). A set of vectors $B = \{v_1, v_2, \dots, v_k\}$ in $V$ is a basis of $V$ if:

1. $B$ is linearly independent: the only scalars $c_1,\dots,c_k$ with $c_1 v_1 + c_2 v_2 + \cdots + c_k v_k = \mathbf{0}$ are $c_1 = c_2 = \cdots = c_k = 0$. (Here $\mathbf{0}$ is the zero vector and the $c_i$ are real numbers.)
2. $B$ spans $V$: every vector $w \in V$ can be written as some combination $w = c_1 v_1 + c_2 v_2 + \cdots + c_k v_k.$

Put together, these guarantee the key payoff: unique representation. Every $w \in V$ equals $c_1 v_1 + \cdots + c_k v_k$ for one and only one choice of scalars. Spanning gives at least one way; independence forbids a second way. A quick consequence: the zero vector can never belong to a basis, since any set containing $\mathbf{0}$ is automatically dependent (take $1\cdot\mathbf{0}=\mathbf 0$).

The most familiar basis of $\mathbb{R}^n$ is the standard basis $\{e_1, e_2, \dots, e_n\}$, where $e_i$ is the vector with a $1$ in position $i$ and $0$ everywhere else. For $\mathbb{R}^3$: $e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},\quad e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix},\quad e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.$ These are clearly independent and clearly span, so they form a basis.

The deep fact is that the count never changes: every basis of $\mathbb{R}^n$ has exactly $n$ vectors. This number $n$ is the dimension of the space. Two useful shortcuts follow for $\mathbb{R}^n$:

• Fewer than $n$ vectors can never span $\mathbb{R}^n$ (too few to reach everything).
• More than $n$ vectors are always dependent (too many to avoid redundancy).

So only a set of exactly $n$ vectors has any chance of being a basis of $\mathbb{R}^n$. And when you have exactly $n$ of them, the two conditions collapse into one: for $n$ vectors in $\mathbb{R}^n$, linearly independent $\iff$ spanning. Checking either one is enough.

Given a candidate set in $\mathbb{R}^n$, run this checklist:

Step 1 — Count. How many vectors? If it is not exactly $n$, stop: it is not a basis (too few to span, or too many to be independent).

Step 2 — Test the $n$ vectors. Place the $n$ vectors as the columns of an $n\times n$ matrix $A$. Then the set is a basis of $\mathbb{R}^n$ if and only if $A$ is invertible, which you can detect by $\det(A) \neq 0.$ A nonzero determinant means the columns are independent (and therefore spanning), so they form a basis. A zero determinant means they are dependent, so they do not.

Example of the quick test. Is $\{(2,1),(4,2)\}$ a basis of $\mathbb{R}^2$? Form $A=\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$. Then $\det(A) = 2\cdot 2 - 4\cdot 1 = 0$, so it is not a basis — indeed $(4,2) = 2(2,1)$. This determinant test is the workhorse for every basis question in $\mathbb{R}^n$.