In Linear Algebra, *basis* is a set of linearly independent vectors $\mathbf v_1, ..., \mathbf v_n$

Vectors $\mathbf v_1, ..., \mathbf v_l$ *span a (sub)space* $\iff$ this space consists of all possible linear combinations of these vectors

- columns of a matrix $A$ span it's column space $C(A)$
- are such $\mathbf v_i$ independent? - depends

Basis of a vector space is a sequence of vectors $\mathbf v_1, \mathbf v_2, ..., \mathbf v_d$ that

- are linearly independent and
- span the entire space

Standard Basis:

- the identity $I_d$,
- e.g. for $\mathbb R^3$, $\mathbf e_1 = \begin{bmatrix}

1 \\ 0 \\ 0 \end{bmatrix}$, $\mathbf e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, $\mathbf e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

Non-Example:

- $\begin{bmatrix}

1 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 2 \\ 5 \end{bmatrix}$, linearly independent, but don't span $\mathbb R^3$

- $\begin{bmatrix}

1 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 2 \\ 5 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 3 \\ 7 \end{bmatrix}$, the 3rd vector is a linear combination of first 2

- the first case is 2 vectors on a plane, and 2nd is 3 vectors on a plane

Another example:

- $\begin{bmatrix}

1 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 2 \\ 5 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 1 \\ 8 \end{bmatrix}$

- if we put the vectors as columns of a matrix, then the rank should be equal to the number of vectors
- $\begin{bmatrix}

1 & 2 & 3\\ 1 & 2 & 1\\ 2 & 5 & 8 \\ \end{bmatrix}$, rank is 3

Also,

- take any invertible $n \times n$ matrix, take the columns from it and get a basis

Every space has some basis, and each basis of this space has the same number of vectors. The number of vectors in the basis is the *dimension* of this space.