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Bases and Dimension of Abstract Vector Spaces→ LTR-0055/main
Bases and Dimension of Abstract Vector Spaces
When working with \(\RR ^n\) and subspaces of \(\RR ^n\) we developed several fundamental ideas including span, linear independence, bases
and dimension. We will find that these concepts generalize easily to abstract vector spaces and that analogous results hold in
these new settings.
Linear Independence
Linear Independence Let \(V\) be a vector space. Let \(\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\) be vectors of \(V\). We say that the set \(\{\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\}\) is linearly independent if the only
solution to
If, in addition to the trivial solution, a non-trivial solution (not all \(a_1, a_2,\ldots ,a_p\) are zero) exists, then we say that the set \(\{\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\}\) is linearly
dependent.
Show that \(P=\{1 + x, 3x + x^{2}, 2 + x - x^{2}\}\) is linearly independent in \(\mathbb {P}^{2}\).
The constant term, as well as the coefficients in front of \(x\) and \(x^2\), must be equal to \(0\). This gives us the following system of
equations.
\begin{equation*} \begin{array}{rlrlrcr} a & + & & + & 2c & = & 0 \\ a & + & 3b & + & c & = & 0 \\ & & b & - & c & = & 0 \\ \end{array} \end{equation*}
The only solution is \(a = b = c = 0\). We conclude that \(P\) is linearly independent in \(\mathbb {P}^2\).
Bases and Dimension
Recall that our motivation for defining a basis of a subspace of \(\RR ^n\) was to have a collection of vectors such that every vector of
the subspace can be expressed as a unique linear combination of the vectors in that collection. Definition of a basis (??)
generalizes to abstract vector spaces as follows.
Let \(V\) be a vector space. A set \(\mathcal {B}\) of vectors of \(V\) is called a basis of \(V\) provided that
1.
\(\mbox {span}(\mathcal {B})=V\)
2.
\(\mathcal {B}\) is linearly independent.
Let \(V\) be a vector space, and let \(\mathcal {B}=\{\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_n\}\) be a basis for \(V\). Then every element \(\vec {v}\) of \(V\) has a unique representation as linear combination of
the elements of \(\mathcal {B}\).
By the definition of a basis, we know that \(\vec {v}\) can be written as a linear combination of \(\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_n\). Suppose there are two such
representations. Then,
Because \(\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_n\) are linearly independent, we have \(a_i-b_i=0\) for \(1\leq i\leq n\). Consequently \(a_i=b_i\) for \(1\leq i\leq n\).
In Bases and Dimension we defined the dimension of a subspace of \(\RR ^n\) to be the number of elements in a basis. (Definition ??)
We will adopt this definition for abstract vector spaces. As before, to ensure that dimension is well-defined we need to
establish that this definition is independent of our choice of a basis. The proof of the following theorem is identical to the proof
of its counterpart in \(\RR ^n\). (Theorem ??)
Let \(V\) be a vector space. 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 \(V\). Then \(s=t\).
Now we can state the definition.
Let \(V\) be a subspace of a vector space \(W\). The dimension of \(V\) is the number, \(m\), of elements in any basis of \(V\). We
write
\[\mbox {dim}(V)=m\]
In our discussions up to this point, we have always assumed that a basis is nonempty and hence that the dimension of the
space is at least \(1\). However, the zero space \(\{\vec {0}\}\) has no basis. To accommodate for this, we will say that the zero vector space \(\{\vec {0}\}\) is
defined to have dimension \(0\):
Our insistence that \(\mbox {dim}\{\vec {0}\} = 0\) amounts to saying that the empty set of vectors is a basis of \(\{\vec {0}\}\). Thus the statement that “the dimension of a
vector space is the number of vectors in any basis” holds even for the zero space.
Recall that the vector space \(\mathbb {M}_{m,n}\) consists of all \(m\times n\) matrices. (See Example ?? of Abstract Vector Spaces). Find a
basis and the dimension of \(\mathbb {M}_{m,n}\).
Let \(\mathcal {B}\) consist of \(m\times n\) matrices with exactly one entry equal to \(1\) and all other entries
equal to \(0\). It is clear that every \(m\times n\) matrix can be written as a linear combination of elements of \(\mathcal {B}\). It is also easy to
see that the elements of \(\mathcal {B}\) are linearly independent. Thus \(\mathcal {B}\) is a basis for \(\mathbb {M}_{m,n}\). The set \(\mathcal {B}\) contains \(mn\) elements, so \(\mbox {dim}(\mathbb {M}_{m,n})=mn\).
Recall that \(\mathbb {P}^n\) is the set of all polynomials of degree \(n\) or less. (See Example ?? of Abstract Vector Spaces) Show that \(\mbox {dim}( \mathbb {P}^{n}) = n + 1\) and that \(\{1, x, x^{2}, \dots , x^{n}\}\)
is a basis of \(\mathbb {P}^{n}\).
Each polynomial \(p(x) = a_{0} + a_{1}x + \ldots + a_{n}x^{n}\) in \(\mathbb {P}^{n}\) is clearly a linear combination of \(1, x, \dots , x^{n}\), so \(\mathbb {P}^{n} = \mbox {span}\{1, x, \dots , x^{n}\}\).
Suppose \(a_{0}1 + a_{1}x + \dots + a_{n}x^{n} = 0\), then \(a_{0} = a_{1} = \ldots = a_{n} = 0\). So \(\{1, x, \dots , x^{n}\}\) is linearly independent and hence is a basis containing \(n + 1\) vectors. Thus, \(\mbox {dim}(\mathbb {P}^{n}) = n + 1\).
The set \(\mathcal {B}\) is linearly independent. (See Practice Problem ) Every element \(X\) of \(C_A\) can be written as a linear combination of
elements of \(\mathcal {B}\). Thus \(C_A=\mbox {span}\mathcal {B}\). Therefore \(\mathcal {B}\) is a basis of \(C_A\), and \(\mbox {dim}(C_A) = 2\).
In Practice Problem ?? of Abstract Vector Spaces you demonstrated that the set of all symmetric \(n\times n\) matrices is a subspace of
\(\mathbb {M}_{n,n}\).
Let \(V\) be a subspace of \(\mathbb {M}_{2,2}\) consisting of all \(2\times 2\) symmetric matrices. Find the dimension of \(V\).
A matrix \(A\) is symmetric if \(A^{T} = A\). In other words, a matrix \(A\) is symmetric when entries directly across the main diagonal are equal, so
each \(2 \times 2\) symmetric matrix has the form
Hence the set \(\mathcal {B} = \left \{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \right \}\) spans \(V\). The reader can verify that \(\mathcal {B}\) is linearly independent. Thus \(\mathcal {B}\) is a basis of \(V\), so \(\mbox {dim}(V) = 3\).
Finite-Dimensional Vector Spaces
Our definition of dimension of a vector space depends on the vector space having a basis. In this section we will establish that
any vector space spanned by finitely many vectors has a basis.
A vector space is said to be finite-dimensional if it is spanned by finitely many vectors.
Given a finite-dimensional vector space \(V\) we will find a basis for \(V\) by starting with a linearly independent subset of \(V\) and
expanding it to a basis. The following results are more general versions of Lemmas ?? and ??, and Theorem ?? of Bases
and Dimension. The proofs are identical and we will omit them.
Let \(V\) be a vector space spanned by \(n\) vectors. If a linearly
independent subset \(S\) of \(V\) contains \(m\) vectors, then \(m\leq n\).
Let \(V\) be a vector space. Let \(\{\vec {v}_1,\ldots ,\vec {v}_k\}\) be a linearly independent subset of \(V\). 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.
Let \(V\) be a finite-dimensional vector space. Any linearly independent subset of \(V\) can be expanded to a basis of \(V\).
Coordinate Vectors
Recall that in the context of \(\RR ^n\) (and subspaces of \(\RR ^n\)) the requirement that elements of a basis be linearly independent guarantees
that every element of the vector space has a unique representation in terms of the elements of the basis. (See
Theorem ?? of Introduction to Bases) We proved the same property for abstract vector spaces in Theorem
5.
Uniqueness of representation in terms of the elements of a basis allows us to associate every element of a vector space \(V\) with
a unique coordinate vector with respect to a given basis. Coordinate vectors were first introduced in Introduction to Bases.
We now give a formal definition.
Let \(V\) be a vector space, and let \(\mathcal {B}=\{\vec {v}_1, \ldots ,\vec {v}_n\}\) be a basis for \(V\). If \(\vec {v}=a_1\vec {v}_1+\ldots +a_n\vec {v}_n\), then the vector in \(\RR ^n\) whose components are the coefficients \(a_1, \ldots ,a_n\) is said to be the
coordinate vector for \(\vec {v}\) with respect to \(\mathcal {B}\). We denote the coordinate vector by \([\vec {v}]_{\mathcal {B}}\) and write:
The order of in which vectors \(\vec {v}_1, \ldots ,\vec {v}_n\) appear in \(\mathcal {B}\) of Definition 21 is important. Switching the order of these vectors would switch the
order of the coordinate vector components. For this reason, we will often use the term ordered basis to describe \(\mathcal {B}\) in the
context of coordinate vectors.
The coordinate vector for \(p(x)=4-3x^2+5x^3\) in \(\mathbb {P}^4\) with respect to the ordered basis \(\mathcal {B}_1=\{1, x, x^2, x^3, x^4\}\) is
Now let’s change the order of the elements in \(\mathcal {B}_1\). The coordinate vector for \(p(x)=4-3x^2+5x^3\) with respect to the ordered basis \(\mathcal {B}_2=\{x^4, x^3, x^2, x, 1\}\)
is
Show that the set \(\mathcal {B}=\{x, 1+x, x+x^2\}\) is a basis for \(\mathbb {P}^2\). Keep the order of elements in \(\mathcal {B}\) and find the coordinate vector for \(p(x)=4-x+3x^2\) with respect to the ordered
basis \(\mathcal {B}\).
We will begin by showing that the elements of \(\mathcal {B}\) are linearly independent. Suppose
The solution \(a=b=c=0\) is unique. We conclude that \(\mathcal {B}\) is linearly independent.
Next, we need to show that \(\mathcal {B}\) spans \(\mathbb {P}^2\). To this end, we will consider a generic element \(p(x)=\alpha +\beta x+\gamma x^2\) of \(\mathbb {P}^2\) and attempt to express it as a linear
combination of the elements of \(\mathcal {B}\).
\[ax+b(1+x)+c(x+x^2)=\alpha +\beta x+\gamma x^2\]
then
\[b+(a+b+c)x+cx^2=\alpha +\beta x+\gamma x^2\]
Setting the coefficients of like terms equal to each other gives us
(You should verify this.) This shows that every element of \(\mathbb {P}^2\) can be written as a linear combination of elements of \(\mathcal {B}\). Therefore \(\mathcal {B}\)
is a basis for \(\mathbb {P}^2\).
To find the coordinate vector for \(p(x)=4-x+3x^2\) with respect to \(\mathcal {B}\) we need to express \(p(x)\) as a linear combination of the elements of \(\mathcal {B}\). Fortunately,
we have already done all the necessary work. For \(p(x)\), \(\alpha =4\), \(\beta =-1\) and \(\gamma =3\). This gives us the coefficients of the linear combination: \(a=\beta -\alpha -\gamma =-8\), \(b=\alpha =4\), \(c=\gamma =3\). We
now write \(p(x)\) as a linear combination
\[p(x)=-8(x)+4(1+x)+3(x+x^2)\]
The coordinate vector for \(p(x)\) with respect to \(\mathcal {B}\) is
Recall that the set \(V\) of all symmetric \(2\times 2\) matrices is a subspace of \(\mathbb {M}_{2,2}\). In Example 15 we demonstrated that \(\mathcal {B} = \left \{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \right \}\) is a basis for \(V\). Let \(A=\begin{bmatrix}2&-3\\-3&1\end{bmatrix}\).
Observe that \(A\) is an element of \(V\).
1.
Find the coordinate vector with respect to the ordered basis \(\mathcal {B}\) for \(A\).
2.
Let \(\mathcal {B}'=\left \{ \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \right \}\) be another ordered basis for \(V\). Find the coordinate vector for \(A\) with respect to \(\mathcal {B}'\).
We write \(A\) as a linear combination of the elements of \(\mathcal {B}\).
Coordinate vectors will play a vital role in establishing one of the most fundamental results in linear algebra, that all
\(n\)-dimensional vector spaces have the same structure as \(\RR ^n\). In Example ?? of Isomorphic Vector Spaces, for instance, we will
show that \(\mathbb {P}^2\) is essentially the same as \(\RR ^3\).
Prove that set \(\mathcal {B}=\left \{\begin{bmatrix}1&1\\0&0\end{bmatrix},\begin{bmatrix}1&0\\0&1\end{bmatrix}\right \}\) is a linearly independent subset of \(C_A\).
Show that the following subset of \(\mathbb {P}^{2}\) is linearly independent, and is a basis of \(\mathbb {P}^{2}\).
\[\{1 + x, 1 - x, x + x^{2}\}\]
Show that the following subset of \(\mathbb {P}^{2}\) is linearly independent, and is a basis of \(\mathbb {P}^{2}\).
\[\{x^{2}, x + 1, 1 - x - x^{2}\}\]
Show that the following subset of \(\mathbb {M}_{2,2}\) is linearly independent, and is a basis of \(\mathbb {M}_{2,2}\).
Week 15