Definition 6.3.1.
A basis of a subspace \(H\) of \(\mathbb{R}^n\) is a linearly independent set that spans \(H\text{.}\)
Section 6.3 Basis of a Vector Space
Subsection 6.3.1 Basis of a Subspace
Subspaces typically contain an infinite number of vectors, so it is useful to determine a small set of vectors in the subspace which span the subspace. That is, a minimal set of vectors that span the subspace. This set of vectors which span a space is called a basis.
Example 6.3.2. The standard basis vectors.
Recall the standard basis vectors. In \(\mathbb{R}^2\text{,}\) they are given by,
\begin{equation*}
\ihat = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad \jhat = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
\end{equation*}
In \(\mathbb{R}^3\text{,}\) they are given by,
\begin{equation*}
\ihat = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \quad \jhat = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \quad \khat = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
\end{equation*}
and more generally, in \(\mathbb{R}^n\text{,}\) they are given by,
\begin{equation*}
\vec{e}_1 = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \quad \vec{e}_2 = \begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix} \quad \dots \quad \vec{e}_n = \begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix}
\end{equation*}
In general, the \(i\)th standard basis vector \(\vec{e}_i\) has a 1 in its \(i\)th component and 0's everywhere else. In each case, these vectors are linearly independent, and span \(\mathbb{R}^n\text{.}\) The fact that they span \(\mathbb{R}^n\) is clear, because any vector \(\vec{x} = (x_1, \dots, x_n)\) can be written as the linear combination,
\begin{equation*}
\vec{x} = x_1 \vec{e}_1 + \dots + x_n \vec{e}_n
\end{equation*}
Also, they are linearly independent, because they form the matrix equation \(I_n \vec{x} = \vec{0}\text{,}\) which of course is equivalently \(\vec{x} = \vec{0}\text{,}\) so there is only the trivial solution.
Example 6.3.3. Alternate bases.
In fact, for any invertible matrix \(A\) (of size \(n \times n\)), its columns forms a basis of \(\mathbb{R}^n\text{.}\) This is because, if a matrix is invertible, its columns form a linearly independent set of \(n\) vectors, which spans \(\mathbb{R}^n\text{.}\)
In this way, bases for a subspace are highly non-unique.
Example 6.3.4. Basis of the null space.
Recall that the null space of \(A\) is the solution set of \(A \vec{x} = \vec{0}\text{.}\) When this solution set is written in parametric vector form, the vectors form a basis for \(N(A)\text{.}\)
Subsection 6.3.2 Basis of the Column Space
Theorem 6.3.5.
The pivot columns of a matrix \(A\) form a basis for the column space of \(A\text{.}\)
Note that the pivot columns of \(A\) form a basis, not the pivot columns of the RREF of \(A\text{.}\) In fact, often the columns of the RREF of A will not be in the column space of \(A\text{.}\)
