This is a matrix.

\[\begin{bmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \end{bmatrix}\]

It has 2 rows and 3 columns, so it is a \(2 \times 3\) matrix.

Matrix addition and multiplication-by-a-scalar is done componentwise. Matrix multiplication is done trickily; it’s associative, distributive, commutative with scalars, linear, anticommutative-under-transposition.

Vectors are like columns of matrices, or matrices with one column. Except row vectors are rows of matrices.

The three **elementary row operations** are:

- exchanging two rows
- multiplying a row by a nonzero scalar
- adding a multiple of a row to another row

They can each be represented by multiplying by a simple matrix to the left. Also, they’re uniquely reversible.

A matrix is in **echelon form** if the nonzero elements are like an inverted staircase. This is vague because echelon form is not unique and usually you can just row-reduce as well. A matrix is in **row-reduced echelon form** if the nonzero elements are like an inverted staircase and every column where the staircase steps down has one 1 and the rest 0s. Row-reduced echelon form is useful for proving stuff. In both cases, the first nonzero entry in a nonzero row is called a *pivot entry* and the column it belongs to is called a *pivot column*. Other columns are *nonpivot* or *free*.

A set of vectors, or anything else that behaves kind of like vectors, is a *vector space* if it’s closed under addition and multiplication by scalars. Also, it should probably be nonempty, so it includes \(\vec{\textbf{0}}\).

The *nullspace* of a matrix \(A\) is the space of column vectors \(\vec{\textbf{X}}\) so that \(AX = \vec{\textbf{0}}\).

Vector spaces have unique dimension, i.e. all spanning bases have the same size. Proof: Take one spanning base and express it in terms of another. Adjust the base elements one by one to match each other.

The dimension of the nullspace is the *nullity*.