Compute a basis for the subspace spanned by a given set of Euclidean vectors, and determine the dimension of the subspace.
Subsection2.6.1Class Activities
Observation2.6.1.
Recall from section Section 2.3 that a subspace of a vector space is the result of spanning a set of vectors from that vector space.
Recall also that a linearly dependent set contains “redundant” vectors. For example, only two of the three vectors in Figure 14 are needed to span the planar subspace.
Activity2.6.2.
Consider the subspace of \(\IR^4\) given by \(W=\vspan\left\{
\left[\begin{array}{c}2\\3\\0\\1\end{array}\right],
\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right],
\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right],
\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right]
\right\}
\text{.}\)
(a)
Mark the column of \(\RREF\left[\begin{array}{cccc}
2&2&2&1\\
3&0&-3&5\\
0&1&2&-1\\
1&-1&-3&0
\end{array}\right]\) that shows that \(W\)’s spanning set is linearly dependent.
(b)
What would be the result of removing the vector that gave us this column?
The set still spans \(W\text{,}\) and remains linearly dependent.
The set still spans \(W\text{,}\) but is now also linearly independent.
The set no longer spans \(W\text{,}\) and remains linearly dependent.
The set no longer spans \(W\text{,}\) but is now linearly independent.
Definition2.6.3.
Let \(W\) be a subspace of a vector space. A basis for \(W\) is a linearly independent set of vectors that spans \(W\) (but not necessarily the entire vector space).
Observation2.6.4.
So given a set \(S=\{\vec v_1,\dots,\vec v_m\}\text{,}\) to compute a basis for the subspace \(\vspan S\text{,}\) simply remove the vectors corresponding to the non-pivot columns of \(\RREF[\vec v_1\,\dots\,\vec v_m]\text{.}\) For example, since
the subspace \(W=\vspan\setList{
\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}2\\4\\6\end{array}\right],
\left[\begin{array}{c}0\\-2\\-2\end{array}\right],
\left[\begin{array}{c}1\\2\\1\end{array}\right]
}\) has \(\setList{
\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}0\\-2\\-2\end{array}\right]
}\) as a basis.
If \(\{\vec{b}_1, \vec{b}_2,\ldots, \vec{b}_m\}\) and \(\{\vec{c}_1,\vec{c}_2,\ldots,\vec{c}_n\}\) are each a basis for a vector space \(V\text{,}\) then \(m=n.\)
If \(\{\vec{v}_1,\vec{v}_2\ldots, \vec{v}_n\}\) is linearly independent, then so is \(\{\vec{v}_1,\vec{v}_1 + \vec{v}_2, \ldots, \vec{v}_1 + \vec{v}_2 + \cdots + \vec{v}_n\}\text{.}\)
Let \(V\) be a vector space of dimension \(n\text{,}\) and \(\vec{v} \in V\text{.}\) Then there exists a basis for \(V\) which contains \(\vec{v}\text{.}\)
Exploration2.6.12.
Suppose we have the set of all function \(f:S \rightarrow \mathbb{R}\text{.}\) We claim that this is a vector space under the usual operation of function addition and scalar multiplication. What is the dimension of this space for each choice of \(S\) below:
\(\displaystyle S = \{1\}\)
\(\displaystyle S = \{1,2\}\)
\(\displaystyle S = \{1,2,\ldots ,n\}\)
\(\displaystyle S = \mathbb{R}\)
Exploration2.6.13.
Suppose you have the vector space \(V = \left\{\left(\begin{array}{c}x\\y\\z\end{array}\right)\in \mathbb{R}^3: x + y + z = 1\right\}\) with the operations \(\left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) \oplus \left(\begin{array}{c}x_2\\y_2\\z_2\end{array}\right) = \left(\begin{array}{c}x_1 + x_2 - 1\\y_1 + y_2\\z_1+z_2\end{array}\right) \mbox{ and } \alpha\odot\left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) = \left(\begin{array}{c}\alpha x_1 - \alpha +1\\\alpha y_1\\\alpha z_1\end{array}\right).\) Find a basis for \(V\) and determine it’s dimension.