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Recall that a basis of a subspace \(V\) of \(\RR ^n\) is a subset of \(V\) that is linearly independent and spans \(V\). A basis allows us to uniquely
express every element of \(V\) as a linear combination of the elements of the basis. Several questions may come to mind at this
time. Does every subspace of \(\RR ^n\) have a basis? We know that bases are not unique. If there is more than one basis, what, if
anything, do they have in common?
If you answered that \(V\) is a line in \(\RR ^3\), you are correct. While the two vectors span the line, it is not necessary to have both of them
in the spanning set to describe the line.
What is the minimum number of vectors needed to span a line?
Answer: \(\answer {1}\).
Observe also that the vectors in the given spanning set are not linearly independent, so they do not form a basis for \(V\). How
many vectors would a basis for \(V\) have?
Geometrically, \(W\) is a plane in \(\RR ^3\). Note that the vectors in the spanning set are linearly independent. Can we remove one of the
vectors and have the remaining vector span the plane?
What is the minimum number of vectors needed to span a plane?
Answer: \(\answer {2}\).
How many vectors would a basis for a plane have?
Answer: \(\answer {2}\).
Our observations in Exploration hint at the idea of dimension. We know that a line is a one-dimensional object, a plane is a
two-dimensional object, and the space we reside in is three-dimensional.
Based on our observations in Exploration , it makes sense for us to define dimension of a vector space (or a subspace) as the
minimum number of vectors required to span the space (subspace). We can accomplish this by defining dimension as the
number of elements in a basis.
We have to proceed carefully because we don’t want the dimension to depend on our choice of a basis. So, before we state
our definition, we need to make sure that every basis for a given vector space (or subspace) has the same number of
elements.
Suppose \(\mathcal {B}=\{\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_t\}\) and \(\mathcal {C}=\{\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_s\}\) be two bases of \(\RR ^n\) (or a subspace \(V\) of \(\RR ^n\)). Then \(s=t\).
Suppose \(s\neq t\). Without loss of generality, assume that \(s>t\). Because \(\mathcal {B}\) spans \(V\), every \(\vec {w}_i\) of \(\mathcal {C}\) can be written as a linear combination of
elements of \(\mathcal {B}\):
Recall our assumption that \(s>t\). By Theorem ??, we know that the system has infinitely many solutions. This shows that equation
(??) has a nontrivial solution (in fact, infinitely many of them). But this shows that \(\{\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_s\}\) is linearly dependent and contradicts our
assumption that \(\mathcal {C}\) is a basis of \(V\). We conclude that \(s=t\).
Let \(V\) be a subspace of \(\RR ^n\). The dimension of \(V\) is the number, \(m\), of elements in any basis of \(V\). We write
\[\mbox {dim}(V)=m\]
We know that vectors \(\vec {e}_1, \ldots ,\vec {e}_n\) form a basis of \(\RR ^n\). Therefore \(\mbox {dim}(\RR ^n)=n\).
It is easy to verify that \(\{\vec {0}\}\) is a subspace of \(\RR ^n\). Because \(\{\vec {0}\}\) is linearly dependent, it does not have a basis. We define \(\text {dim}\{\vec {0}\}=0\).
The following section will guarantee that dimension is defined for every subspace of \(\RR ^n\).
Every Subspace of \(\RR ^n\) has a Basis
If a linearly independent subset of \(\RR ^n\) contains \(m\) vectors, then \(m\leq n\).
See Practice Problem .
Let \(\{\vec {v}_1,\ldots ,\vec {v}_k\}\) be a linearly independent subset of \(\RR ^n\). If \(\vec {u}\) is not in \(\mbox {span}(\vec {v}_1,\ldots ,\vec {v}_k)\), then \(\{\vec {u},\vec {v}_1,\ldots ,\vec {v}_k\}\) is linearly independent.
We need to show that \(a=a_1=\ldots =a_k=0\). Suppose \(a\neq 0\), then \(\vec {u}=\frac {-a_1}{a}\vec {v}_1+\ldots +\frac {-a_k}{a}\vec {v}_k\). But this contradicts the assumption that \(\vec {u}\) is not in the span of \(\vec {v}_1,\ldots ,\vec {v}_k\). So, \(a=0\). But \(a_1=\ldots =a_k=0\) because \(\vec {v}_1,\ldots ,\vec {v}_k\) are
linearly independent.
This means that (??) has only the trivial solution, and \(\{\vec {u},\vec {v}_1,\ldots ,\vec {v}_k\}\) is linearly independent.
Let \(V\) be a subspace of \(\RR ^n\). Any linearly independent subset of \(V\) can be expanded to a basis of \(V\).
Suppose that \(X=\{\vec {v}_1,\ldots ,\vec {v}_k\}\) is a linearly independent subset of \(V\). If \(\mbox {span}(X) = V\) then \(X\) is already a basis of \(V\). If \(\mbox {span}(X) \neq V\), choose \(\vec {u}_1\) in \(V\) such that \(\vec {u}_1\) is not in \(\mbox {span}(X)\). The set \(\{\vec {u}_1, \vec {v}_1,\ldots ,\vec {v}_k\}\) is
linearly independent by Lemma 8.
If \(\mbox {span}(\vec {u}_1, \vec {v}_1,\ldots ,\vec {v}_k) = V\) we are done; otherwise choose \(\vec {u}_{2} \in V\) such that \(\vec {u}_{2}\) is not in \(\mbox {span}(\vec {u}_1, \vec {v}_1,\ldots ,\vec {v}_k)\). Then \(\{\vec {u}_1,\vec {u}_2, \vec {v}_1,\ldots ,\vec {v}_k\}\) is linearly independent, and the process continues. We claim
that a basis of \(V\) will be reached eventually. If no basis of \(V\) is ever reached, the process creates arbitrarily large independent
sets in \(\RR ^n\). But this is impossible by Lemma 6.
