For any $u, v, w \in V$ and scalars $\lambda, \rho$, a vector space must satisfy:
In $\mathbb{R}^n$, addition and scalar multiplication act component by component, and the zero vector is $0 = (0,\dots,0)^\top$.
Both satisfy the same eight axioms, so every result proved for a general vector space $V$ applies to both at once.
A subset $H \subset V$ is a subspace of $V$ when it satisfies three conditions.
Condition 1 is technically redundant, since it follows from condition 3 with $\lambda = 0$, but it is worth checking first: it is the fastest way to rule out a candidate.
All three must hold for $H$ to be a subspace; failing just one is enough to rule it out.
The span route is usually faster once you can rewrite the general element of $H$ as $a\,v_1 + b\,v_2 + \cdots$ for free parameters $a, b, \dots$
A linear combination of $v_1,\dots,v_n$ is a vector $\lambda_1 v_1 + \cdots + \lambda_n v_n$ for some scalars $\lambda_i$. The span collects every such combination:
A set $\{v_1,\dots,v_n\}$ is linearly independent when the zero vector can only be built from them in the trivial way, using all zero coefficients. If some other combination also gives zero, the set is dependent.
A basis is a set of vectors that generates the whole space $V$, with no unnecessary vectors in it.
Every basis of the same $V$ has the same number of vectors. That number is called the dimension of $V$.
Every matrix $A$ has three subspaces attached to it, built from its rows, its columns, and the solutions of $Ax=0$.
| Nul(A) | Col(A) | Row(A) | |
|---|---|---|---|
| Full name | Null space | Column space | Row space |
| Defined as | $\{x : Ax=0\}$ | Span of the columns | Span of the rows |
| Subspace of | $\mathbb{R}^n$ | $\mathbb{R}^m$ | $\mathbb{R}^n$ |
| Dimension | free variables | pivot columns | pivot rows |
| Read from | reduced echelon solution | original matrix, pivot columns | any echelon form, nonzero rows |
For a square $n\times n$ matrix, $A$ is invertible if and only if $\text{Nul}A = \{0\}$.
Once you pick a basis $B = \{b_1,\dots,b_n\}$, every vector $v$ can be built from $b_1,\dots,b_n$ in exactly one way. Those weights are the coordinates of $v$ in that basis.
In general, $[v]_X$ denotes the coordinates of $v$ relative to whichever basis $X$ is being used. If no basis is mentioned, we mean the standard basis $E$, where coordinates and components are the same thing: $[v]_E = v$.
PE←B converts $B$-coordinates into standard coordinates. Its columns are the vectors of $B$ written in the standard basis: $P_{E\leftarrow B} = \begin{pmatrix}b_1 & \cdots & b_n\end{pmatrix}$, and $P_{E\leftarrow B}[v]_B = v$.
PB←E goes the other way, from standard coordinates to $B$-coordinates: $P_{B\leftarrow E} = (P_{E\leftarrow B})^{-1}$, and $P_{B\leftarrow E}\,v = [v]_B$.
PC←B converts $B$-coordinates directly into $C$-coordinates. Its columns are $[b_i]_C$: $P_{C\leftarrow B} = \begin{pmatrix}[b_1]_C & \cdots & [b_n]_C\end{pmatrix}$, and $P_{C\leftarrow B}[v]_B = [v]_C$.
Composing. $P_{D\leftarrow B} = P_{D\leftarrow C}\,P_{C\leftarrow B}$.
Follow an arrow → multiply by its matrix. Reverse an arrow → invert it. For $B \rightarrow C \rightarrow D$, write the matrices right to left in the order you travel them: $P_{D\leftarrow B} = P_{D\leftarrow C}\,P_{C\leftarrow B}$.