1 Span and Basis
1.1 Span
The span of a collection of vectors \(\vec{v}_{1}, \dots, \vec{v}_{n}\in \mathcal{V}\) is defined as \[ \text{span}_{F}(\vec{v}_{1}, \dots, \vec{v}_{n}):=\{ \alpha_{1}\vec{v}_{1}+\dots+\alpha_{n}\vec{v}_{n}; \alpha_{1}, \dots, \alpha_{n}\in F \}, \] i.e. the set of all possible linear combinations.
More generally, if \(S \subset \mathcal{V}\) is any subset of vectors, then the span of \(S\) is the smallest subspace of \(\mathcal{V}\) containing \(S\). The span of a set of vectors is always a subspace, and if \(W=\text{span}_{F}(\vec{v}_{1}, \dots, \vec{v}_{n})\) then we say \(W\) is spanned by vectors \(\vec{v}_{1}, \dots, \vec{v}_{n}\in \mathcal{V}\).
For vectors \(\vec{v}_{1}, \dots, \vec{v}_{n}\in\mathcal{V}\), if one of them \(\vec{v}_{i}\) can be written as a linear combination of the others, then we say the whole sequence is linearly dependent. Otherwise, we call the sequence linear independent.
1.2 Basis
Let \(\mathcal{B}:=(\vec{v}_{1}, \dots, \vec{v}_{n})\) be a finite sequence of vectors \(\vec{v}_{i}\in\mathcal{V}\) . We say \(\mathcal{B}\) forms a basis when: 1. \(\mathcal{B}\) spans \(\mathcal{V}\); 2. \(\mathcal{B}\) is linearly independent.
A sequence \(\mathcal{B}:=(\vec{v}_{1}, \dots, \vec{v}_{n})\) of vectors in \(\mathcal{V}\) forms a basis if and only if every vector \(\vec{v}\in\mathcal{V}\) can be written uniquely as a linear combination of the vectors in \(\mathcal{B}\).
If a vector space \((\mathcal{V},F)\) has a basis \(\mathcal{B}:=(\vec{v}_{1}, \dots, \vec{v}_{n})\) with \(n\) elements, then we say \(V\) has dimension \(n\), and we write \(\text{dim}_{F}(\mathcal{V})=n\). Clearly, if \(\mathcal{V}\) has two bases \(\mathcal{B}_{1}\) and \(\mathcal{B}_{2}\) then they must be of the same length. If a vector space \((\mathcal{V}, F)\) has finite basis \(\mathcal{B}\) then we say \((\mathcal{V}, F)\) is finite-dimensional, otherwise we say it is infinite-dimensional.
Let \((\mathcal{V},F)\) be a finite-dimensional vector space, and assume \(S\) is a set of vectors that spans \(\mathcal{V}\). Then there is a sequence of vectors in \(S\) that forms a basis of \(\mathcal{V}\). We construct basis iteratively by selecting vectors not currently in the span of the collection until the collection spans \(\mathcal{V}\).
If \(\vec{v}_{1}, \dots, \vec{v}_{r}\) is a linearly independent sequence in a finite-dimensional vector space \(\mathcal{V}\), then it can be extended to a basis of \(\mathcal{V}\), in other words we can find \(\vec{v}_{r+1},\dots,\vec{v}_{n}\) such that \(\vec{v}_{1}, \dots, \vec{v}_{n}\) is a basis of \(\mathcal{V}\).
Let \(\vec{v}_{1}, \dots, \vec{v}_{n}\) be \(n\) vectors in an \(n\)-dimensional vector space \(\mathcal{V}\). These vectors form a basis of \(\mathcal{V}\) if either: 1. \(\vec{v}_{1}, \dots, \vec{v}_{n}\) are linearly independent; 2. \(\vec{v}_{1}, \dots, \vec{v}_{n}\) span \(\mathcal{V}\).
For basis \(\mathcal{B}=(\vec{v}_{1}, \dots, \vec{v}_{n})\) of vector space \((\mathcal{V}, F)\) we can write any vector \(\vec{v}\in\mathcal{V}\) as \[ \vec{v}=\alpha_{1}\vec{v}_{1}+\dots+\alpha_{n}\vec{v}_{n}, \] where the scalars \(\alpha_{i}\in F\) are called the coordinates of \(\vec{v}\) with respect to basis \(\mathcal{B}\). The coordinate vector / matrix of \(\vec{v}\) with respect to \(\mathcal{B}\) is \[ [\vec{v}]_{\mathcal{B}}=\begin{bmatrix} \alpha_{1} & \cdots & \alpha_{n} \end{bmatrix}^T. \] Some common vector spaces, for example the Euclidean spaces, have standard coordinate systems which are defined with respect to the standard bases. For \(\mathbb{R}^2\) for example these are \[ \vec{v}_{1}:=\begin{bmatrix} 1 & 0 \end{bmatrix}^T\quad\&\quad\vec{v}_{2}=\begin{bmatrix} 0 & 1 \end{bmatrix}^T. \]
In the following definition, the symbol \([\vec{x}]\) denotes the coordinates of \(\vec{x}\) in the standard basis \(F^n\). Let \(S:=(\vec{x}_{1}, \dots, \vec{x}_{n})\) be a sequence of vectors in \(F^m\). The matrix of column vectors of \(S\) is the \(M_{m \times n}(F)\) matrix with columns \([\vec{x}_{i}]\). The matrix of row vectors of \(S\) is the \(M_{n \times m}(F)\) matrix with rows \([\vec{x}_{i}]^T\).
For \(A \in M_{n \times m}(F)\) the row space of \(A\) is the subspace in \(F^m\) spanned by its rows. The row rank of \(A\) is the dimension of the row space. The column space of \(A\) is the subspace in \(F^n\) spanned by its columns. The column rank of \(A\) is the dimension of the column space. Row operations do not change the row space of \(A\) and column operations do not change the column space of \(A\).
For \(A,B \in M_{n \times m}(F)\) the following statements are equivalent:
- \(A\) and \(B\) are row-equivalent to each other;
- The row space of \(A\) is equal to the row space of \(B\).