27/09/2014

Basis (Linear algebra)

Basis:


Definition 1.1
A basis for a vector space is a sequence of vectors that form a set that is linearly independent and that spans the space.
We denote a basis with angle brackets  \langle \vec{\beta}_1,\vec{\beta}_2,\ldots \rangle  to signify that this collection is a sequence — the order of the elements is significant. (The requirement that a basis be ordered will be needed, for instance, in Definition 1.13.)
Example 1.2
This is a basis for  \mathbb{R}^2 .
\langle \begin{pmatrix} 2 \\ 4 \end{pmatrix},\begin{pmatrix} 1 \\ 1 \end{pmatrix} \rangle
It is linearly independent
c_1\begin{pmatrix} 2 \\ 4 \end{pmatrix}+c_2\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad\implies\quad \begin{array}{*{2}{rc}r} 2c_1 &+ &1c_2 &= &0 \\ 4c_1 &+ &1c_2 &= &0 \end{array} \quad\implies\quad c_1=c_2=0
and it spans  \mathbb{R}^2 .
\begin{array}{*{2}{rc}r} 2c_1 &+ &1c_2 &= &x \\ 4c_1 &+ &1c_2 &= &y \end{array} \quad\implies\quad c_2=2x-y\text{ and } c_1=(y-x)/2
Example 1.3
This basis for  \mathbb{R}^2
\langle \begin{pmatrix} 1 \\ 1 \end{pmatrix},\begin{pmatrix} 2 \\ 4 \end{pmatrix} \rangle
differs from the prior one because the vectors are in a different order. The verification that it is a basis is just as in the prior example.
Example 1.4
The space  \mathbb{R}^2  has many bases. Another one is this.
\langle \begin{pmatrix} 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \end{pmatrix} \rangle
The verification is easy.
Definition 1.5
For any  \mathbb{R}^n ,
\mathcal{E}_n=\langle \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix}, \dots,\, \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} \rangle
is the standard (or natural) basis. We denote these vectors by  \vec{e}_1,\dots,\vec{e}_n .

(Calculus books refer to \mathbb{R}^2's standard basis vectors  \vec{\imath}  and  \vec{\jmath}  instead of \vec{e}_1 and \vec{e}_2, and they refer to  \mathbb{R}^3 's standard basis vectors  \vec{\imath} ,  \vec{\jmath} , and  \vec{k}  instead of \vec{e}_1, \vec{e}_2, and \vec{e}_3.) Note that the symbol " \vec{e}_1 " means something different in a discussion of \mathbb{R}^3  than it means in a discussion of  \mathbb{R}^2 .
Example 1.6
Consider the space  \{a\cdot\cos\theta+b\cdot\sin\theta\,\big|\, a,b\in\mathbb{R}\}  of functions of the real variable \theta.
\langle 1\cdot\cos\theta+0\cdot\sin\theta, 0\cdot\cos\theta+1\cdot\sin\theta \rangle =\langle \cos\theta, \sin\theta \rangle
Another basis is  \langle \cos\theta-\sin\theta, 2\cos\theta+3\sin\theta \rangle . Verification that these two are bases is Problem 7.
Example 1.7
A natural for the vector space of cubic polynomials  \mathcal{P}_3  is  \langle 1,x,x^2,x^3 \rangle . Two other bases for this space are  \langle x^3,3x^2,6x,6 \rangle  and  \langle 1,1+x,1+x+x^2,1+x+x^2+x^3 \rangle . Checking that these are linearly independent and span the space is easy.
Example 1.8
The trivial space \{\vec{0}\} has only one basis, the empty one  \langle \rangle .
Example 1.9
The space of finite-degree polynomials has a basis with infinitely many elements  \langle 1,x,x^2,\ldots \rangle .
Example 1.10
We have seen bases before. In the first chapter we described the solution set of homogeneous systems such as this one
\begin{array}{*{4}{rc}r} x &+ &y & & &- &w &= &0 \\ & & & &z &+ &w &= &0 \end{array}
by parametrizing.
\{\begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}y +\begin{pmatrix} 1 \\ 0 \\ -1 \\ 1 \end{pmatrix}w \,\big|\, y,w\in\mathbb{R} \}
That is, we described the vector space of solutions as the span of a two-element set. We can easily check that this two-vector set is also linearly independent. Thus the solution set is a subspace of  \mathbb{R}^4  with a two-element basis.
Example 1.11
Parameterization helps find bases for other vector spaces, not just for solution sets of homogeneous systems. To find a basis for this subspace of \mathcal{M}_{2 \! \times \! 2}
\{\begin{pmatrix} a &b \\ c &0 \end{pmatrix} \,\big|\, a+b-2c=0\}
we rewrite the condition as a=-b+2c.
\{\begin{pmatrix} -b+2c &b \\ c &0 \end{pmatrix} \,\big|\, b,c \in \mathbb{R}\} =\{b\begin{pmatrix} -1 &1 \\ 0 &0 \end{pmatrix}+ c\begin{pmatrix} 2 &0 \\ 1 &0 \end{pmatrix} \,\big|\, b,c \in \mathbb{R}\}
Thus, this is a good candidate for a basis.
\langle \begin{pmatrix} -1 &1 \\ 0 &0 \end{pmatrix}, \begin{pmatrix} 2 &0 \\ 1 &0 \end{pmatrix} \rangle
The above work shows that it spans the space. To show that it is linearly independent is routine.
Consider again Example 1.2. It involves two verifications.
In the first, to check that the set is linearly independent we looked at linear combinations of the set's members that total to the zero vector c_1\vec{\beta}_1+c_2\vec{\beta}_2=\binom{0}{0}. The resulting calculation shows that such a combination is unique, that c_1 must be 0 and c_2 must be 0.
The second verification, that the set spans the space, looks at linear combinations that total to any member of the space c_1\vec{\beta}_1+c_2\vec{\beta}_2=\binom{x}{y}. InExample 1.2 we noted only that the resulting calculation shows that such a combination exists, that for each x,y there is a c_1,c_2. However, in fact the calculation also shows that the combination is unique: c_1 must be (y-x)/2and c_2 must be 2x-y.
That is, the first calculation is a special case of the second. The next result says that this holds in general for a spanning set: the combination totaling to the zero vector is unique if and only if the combination totaling to any vector is unique.
Theorem 1.12
In any vector space, a subset is a basis if and only if each vector in the space can be expressed as a linear combination of elements of the subset in a unique way.
We consider combinations to be the same if they differ only in the order of summands or in the addition or deletion of terms of the form " 0\cdot\vec{\beta} ".
Proof
By definition, a sequence is a basis if and only if its vectors form both a spanning set and a linearly independent set. A subset is a spanning set if and only if each vector in the space is a linear combination of elements of that subset in at least one way.
Thus, to finish we need only show that a subset is linearly independent if and only if every vector in the space is a linear combination of elements from the subset in at most one way. Consider two expressions of a vector as a linear combination of the members of the basis. We can rearrange the two sums, and if necessary add some  0\vec{\beta}_i terms, so that the two sums combine the same  \vec{\beta} 's in the same order:  \vec{v}=c_1\vec{\beta}_1+c_2\vec{\beta}_2+\cdots +c_n\vec{\beta}_n  and  \vec{v}=d_1\vec{\beta}_1+d_2\vec{\beta}_2+\cdots +d_n\vec{\beta}_n . Now
c_1\vec{\beta}_1+c_2\vec{\beta}_2+\cdots +c_n\vec{\beta}_n=d_1\vec{\beta}_1+d_2\vec{\beta}_2+\cdots +d_n\vec{\beta}_n
holds if and only if
(c_1-d_1)\vec{\beta}_1+\dots+(c_n-d_n)\vec{\beta}_n=\vec{0}
holds, and so asserting that each coefficient in the lower equation is zero is the same thing as asserting that  c_i=d_i  for each  i .
Definition 1.13
In a vector space with basis B the representation of \vec{v}with respect to B is the column vector of the coefficients used to express \vec{v} as a linear combination of the basis vectors:
{\rm Rep}_{B}(\vec{v})= \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix}
where  B=\langle \vec{\beta}_1,\dots,\vec{\beta}_n \rangle  and  \vec{v}=c_1\vec{\beta}_1+c_2\vec{\beta}_2+\cdots +c_n\vec{\beta}_n . The  c 's are the coordinates of  \vec{v}  with respect to B
We will later do representations in contexts that involve more than one basis. To help with the bookkeeping, we shall often attach a subscript B to the column vector.
Example 1.14
In  \mathcal{P}_3 , with respect to the basis  B=\langle 1,2x,2x^2,2x^3 \rangle , the representation of  x+x^2  is
{\rm Rep}_{B}(x+x^2)=\begin{pmatrix} 0 \\ 1/2 \\ 1/2 \\ 0 \end{pmatrix}_B
(note that the coordinates are scalars, not vectors). With respect to a different basis  D=\langle 1+x,1-x,x+x^2,x+x^3 \rangle , the representation
{\rm Rep}_{D}(x+x^2)=\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}_D
is different.
Remark 1.15
This use of column notation and the term "coordinates" has both a down side and an up side.
The down side is that representations look like vectors from  \mathbb{R}^n , which can be confusing when the vector space we are working with is \mathbb{R}^n, especially since we sometimes omit the subscript base. We must then infer the intent from the context. For example, the phrase "in  \mathbb{R}^2 , where \vec{v}=\binom{3}{2}" refers to the plane vector that, when in canonical position, ends at  (3,2) . To find the coordinates of that vector with respect to the basis
B=\langle \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \end{pmatrix} \rangle
we solve
c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} +c_2\begin{pmatrix} 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}
to get that c_1=3 and c_2=-1/2. Then we have this.
{\rm Rep}_{B}(\vec{v})=\begin{pmatrix} 3 \\ -1/2 \end{pmatrix}
Here, although we've ommited the subscript  B  from the column, the fact that the right side is a representation is clear from the context.
The up side of the notation and the term "coordinates" is that they generalize the use that we are familiar with:~in  \mathbb{R}^n  and with respect to the standard basis  \mathcal{E}_n , the vector starting at the origin and ending at  (v_1,\dots,v_n)  has this representation.

{\rm Rep}_{\mathcal{E}_n}(\begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}) = \begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}_{\mathcal{E}_n} .
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