Serlo: EN: Dual space

{{#invoke:Mathe für Nicht-Freaks/Seite|oben}} We have already seen the vector space of linear maps ${\mathrm{Hom}}_K(V,W)$ between two $K$-vector spaces $V$ and $W$. We will now consider the case where the vector space $W$ corresponds to the field $K$.

Motivation

Consider the following example: We want to buy apples and pears. An apple costs $$2$ and a pear $$3$. If $x\in ℝ$ is the number of apples and $y\in ℝ$ is the number of pears, how much do we have to pay in total? The formula for the total price is $2x+3y$. We can express this equation as $ℝ$-linear map Vorlage:Einrücken Let's assume that the prices increase by half. To get the formula that gives the new total price, we need to multiply the old formula by $\frac{3}{2}$. The formula that gives this price would then be $\frac{3}{2}(2x+3y)=3x+\frac{9}{2}y$. The corresponding linear map is Vorlage:Einrücken We thus recognize that $Q(x,y)=\frac{3}{2}P(x,y)$. Suppose now that the price of apples increases by $$2$ and the price of pears by $$4$. We obtain the corresponding formula for the total price by adding $2x+4y$ to the original formula, i.e. $(2x+3y)+(2x+4y)=4x+7y$. This can be understood as the addition of linear maps. We define $R,S:{ℝ}^2\to ℝ$ by $R(x,y)=2x+4y$ and $S(x,y)=4x+7y$. Then $(P+R)(x,y)=P(x,y)+R(x,y)=S(x,y)$ holds true. So in this example, we simpy added linear maps from ${ℝ}^2$ to $ℝ$ and multiplied them by scalars.

The total price is indicated by linear maps from ${ℝ}^2\to ℝ$. Such a map assigns a value, namely the price, to each vector. In other words, we can say that the mapping "measures" these vectors. This is why we call linear maps from ${ℝ}^2$ to $ℝ$ linear measurement functions. We have seen above that sums and scalar multiples of such maps are again linear maps. In other words, linear combinations of linear maps are again linear maps. So also on the set of linear maps on ${ℝ}^2$, we can find a vector space structure.

What about other vector spaces? Let's look at the $ℂ$-vector space $ℂ[x{]}_{\le n}$ of complex polynomials of degree at most $n\in \mathbb{N}$. There are a number of simple measurement functions here. These can, for example, assign to a polynomial $p$ its value at a point $a\in ℂ$: Vorlage:Einrücken Alternatively, we can assign to a polynomial the value of its derivative at the point $a\in ℂ$: Vorlage:Einrücken Since the coefficients of polynomials are scalars, we can use them to define further measurement functions. For example, for $p=a_n{x}^n+\dots +a_{1}x+a_0$, consider the mappings $f,g:ℂ[x{]}_{\le n}\to ℂ$ defined by $f(p)=a_n+\dots +a_1$ and $g(p)=a_0$. Then $(f+g)(p)=f(p)+g(p)=a_n+\dots +a_1+a_0=p(1)=:{\mathrm{eval}}_1(p)$. We can also see here that sums of measurement functions are again measurement functions.

In general, we can also consider the space of linear measurement functions $V\to K$ over an arbitrary $K$-vector space $V$. We will see that, as in the previous examples, this is a vector space. This space is called the dual space of $V$.

Definition

Mathe für Nicht-Freaks: Vorlage:Definition The following theorem states that the dual space is a vector space.

Dual Basis

We now know what the dual space $V^{*}$ of a $K$-vector space $V$ is: It consists of all linear maps from $V$ to $K$. Intuitively, we can understand these maps as linear maps that measure vectors from $V$. This is why we sometimes call elements of the dual space $V^{*}$ "(linear) measurement functions" in this article.

Motivated by this intuitive notion of "measurements", we ask ourselves: Is there a subset $M\subseteq V^{*}$ of measurement functions that can be used to uniquely determine vectors? In other words, is there a subset $M$ so that we can find a measurement function $f\in M$ with $f(v)\ne f(w)$ for every choice of vectors $v,w\in V$ with $v\ne w$?

Let's first consider what this means using an example:

In sumary, our question is: Does there exist a subset $M\subseteq V^{*}$ such that $v,w\in V$ applies to all vectors: If $f(v)=f(w)$ applies to all measurements $f\in M$, then $v=w$ must be true.

We will first try to answer this question in $K^n$.

Measurement functions for unique determination of vectors

A vector $v=(v_1,\dots ,v_n)\in K^n$ is uniquely determined by its entries $v_i$. If we select measurement functions from $(K^n{)}^{*}$ in such a way that their values provide us with the entries of a vector, then we have ensured that a vector is already uniquely determined by these values. Let us therefore consider the following mappings for $i\in \{1,\dots ,n\}$ Vorlage:Einrücken You can check that the maps $f_i$ are linear. In addition, $f_i(v)=v_i$ holds for every $i$. The map $f_i$ therefore provides the $i$-th entry of vectors in $K^n$. A vector $v\in K^n$ is already uniquely determined by the values of $f_i$: Suppose we have vectors $v=(v_1,\dots ,v_n)$ and $w=(w_1,\dots ,w_n)$ in $K^n$ with equal function values among the $f_i$, i.e., with $f_i(v)=f_i(w)$ for all $i$. Then $v_i=f_i(v)=f_i(w)=w_i$ applies for all $i$ and therefore $v=w$. Thus, if $v,w\in K^n$ with $f_i(v)=f_i(w)$ for all $i$, then $v=w$ follows.

It is also intuitively clear that we cannot omit any of the measurement functions $f_i$ in order to uniquely determine a vector by its measurement values. For example, if we omit $f_j$, $j\in \{1,\dots ,n\}$, then for Vorlage:Einrücken we may have $f_i(v)=0=f_i(w)$ for all measurement functions with $i\ne j$, but nevertheless $v\ne w$. The measurement functions $f_i$ with $i\ne j$ therefore no longer uniquely determine a vector.

So the $f_i$ with $i=1,\dots n$ form a set of measurement functions that uniquely determine vectors from $K^n$. Further, they are minimal because we cannot omit any of the functions.

Can we generalize this to a general vector space $V$? In $K^n$ we have used the fact that a vector $v=(v_1,\dots ,v_n)\in K^n$ is uniquely determined by its entries $v_i$. Now, the $v_i$ are precisely the coordinates of $v$ with respect to the standard basis $\{e_1,\dots ,e_n\}\subseteq K^n$: Vorlage:Einrücken In a general vector space $V$, we do not have a standard basis. However, as soon as we have chosen any basis $B$, we can speak of the coordinates of a vector with respect to $B$ in the same way as in $K^n$. Just as in $K^n$ with the standard basis, in $V$ with the selected basis $B$, a vector $v\in V$ is uniquely determined by its coordinates with respect to $B$. As soon as we have chosen a basis, we can try to proceed in the same way as in $K^n$.

In the following, we assume that $V$ is finite-dimensional, i.e. $\dim V=n<\mathrm{\infty }$. Let $B=\{b_1,\dots ,b_n\}$ be a basis of $V$. Then every vector $v\in V$ is of the form Vorlage:Einrücken with uniquely determined coordinates $a_1,\dots ,a_n\in K$. Analogous to $K^n$, we now define the linear measurement functions for $i\in \{1,\dots ,n\}$ in $V^{*}$ Vorlage:Einrücken One of the measurement functions $f_i$ therefore determines the $i$-th coordinate of vectors with respect to the basis $B$. Thus, Vorlage:Einrücken for every vector $v\in V$. Mathe für Nicht-Freaks: Vorlage:Warnung Since vectors in $V$ are already uniquely determined by their coordinates, they are also already uniquely determined by the values of $f_i$. In other words, for all $v,w\in V$ we have Vorlage:Einrücken For the same reason as with $K^n$, none of the $f_i$ can be omitted: If the $j$-th measurement function $f_j$, $i\in \{1,\dots ,n\}$, is missing, then any two vectors for which only the $j$-th coordinate with respect to $B$ differs, can no longer be distinguished. Mathe für Nicht-Freaks: Vorlage:Frage

The measurement functions form a basis

Let $V$ be a vector space with a fixed basis $B=\{b_1,\dots ,b_n\}$ and let the $f_i$ be defined as above. If you want to determine vectors uniquely using the values of $f_i$, you cannot do without any of the $f_i$. The reason for this is that the result of a measurement $f_j(v)$ (the $j$-th coordinate of $v$ with respect to $B$) cannot be deduced from the other measurements. That means, we cannot represent any of the measurement functions $f_j$ as a linear combination of the other $f_i$ ($i\ne j$). In other words, the measurement functions $f_i$ are linearly independent.

On the other hand, the values of $f_i$ already tell us everything there is to know about a vector $v\in V$: Its coordinates with respect to the selected basis $B$. Can all other measurement functions from $V^{*}$ therefore be combined from $f_1,\dots ,f_n$? Any measurement function $g:V\to K$ from $V^{*}$ is already uniquely determined by its values on the basis vectors $b_1,\dots ,b_n$ according to the principle of linear continuation. For $i\in \{1,\dots ,n\}$, let ${\lambda }_i=g(b_i)\in K$ be these values. Furthermore, $f_i(b_i)=1$ and $f_i(b_j)=0$ apply for $j\ne i$ and all $i\in \{1,\dots ,n\}$. By inserting the $b_i$ we obtain that Vorlage:Einrücken assume the same values on the basis vectors. According to the principle of linear continuation, the two linear maps are therefore identical. Thus, every $g\in V^{*}$ can be written as a linear combination of $f_i$. In other word, the measurement functions $f_i$ form a generating system of $V^{*}$.

Hence, $\{f_1,\dots ,f_n\}\subseteq V^{*}$ is a basis of the dual space and we can prove the following theorem:

We call the uniquely determined basis $B^{*}$ the dual basis with respect to $B$ and denote its basis vectors by $b_i^{*}=f_i$.

What happens in the infinite dimension?

Above, we only considered the case $\dim V<\mathrm{\infty }$. Can we proceed analogously if $V$ is infinite dimensional? To define the measurement functions $f_i$, we must first choose a basis of $V$. Let $B=\{b_i\mid i\in I\}\subseteq V$ be a basis of $V$, where $I$ is an (infinite) index set. The principle of linear continuation also applies in infinite dimensions: For given values ${\lambda }_i\in K$, $i\in I$, there is exactly one linear map $f:V\to K$ with $f(b_i)={\lambda }_i$ for all $i\in I$. Just as in the finite-dimensional case, we can therefore define the map $f_i:V\to K$ for $i\in I$ using the rule Vorlage:Einrücken We can then show that $\{f_i\mid i\in I\}$ is also a linearly independent subset of $V^{*}$ in infinite dimensions. The proof is analogous to the proof of linear independence in the theorem on the dual basis.

However, in infinitely many dimensions, $\{f_i\mid i\in I\}$ cannot be a generating system of $V^{*}$: One can consider the function Vorlage:Einrücken which assumes the value 1 on all basis vectors. This function cannot be represented as a finite linear combination of $f_i$.

So in infinitely many dimensions, the "dual basis" $\{f_i\mid i\in I\}$ is not a basis of the dual space.

Exercises

In the last task, we required $K\ne {𝔽}_2$ because we needed an element in the proof that is neither $0$ nor $1$. The field ${𝔽}_2$ only consists of the elements $0$ and $1$. This means that if we want to construct a linear map $f:V\to K$ that has an $n-1$-dimensional subspace $U$ as its kernel, then we must define it as Vorlage:Einrücken This map is linear amd it is the only way to have a linear map with kernel $U$. Thus, for $K={𝔽}_2$ we arrive at a different result in the last sub-exercise: The map is then unique.

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