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I was wondering if some one could please shed some light on why or how a dual space itself becomes a vector space over the field. Finite-Dimensional Vector Spaces by Paul Halmos states:

. . . to every vector space V we make correspond the dual space $V^*$ consisting of all the linear functionals on $V$. . . . 

p. 21, notation edited

The book goes on to present the defining property of a linear functional and the definition of the linear operations for linear functionals.

Also, for the sake of completion, a linear functional is defined by the text as

a scalar-valued function $y$ defined for every vector $x$, with the property that (identically in the vectors $x_{1}$ and $x_{2}$ and the scalar $\alpha _{1}$ and $\alpha _{2}$)

$$y( \alpha _{1}x_{1}+\alpha _{2}x_{2}) =\alpha _{1}\,y\left( x_{1}\right) +\alpha _{2}\,y\left( x_{2}\right)$$

p. 20

Based on these definitions, isn't $V^*$ composed of scalar-valued functions with the above property? I fail to see any vectors present in $V^*$. Yet the book later assumes that we must know that and starts defining a dual space $V^{**}$ of a dual space $V^*$ of a vector space $V$.

Any help would be much appreciated.

Comic Book Guy
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    A vector is an element of a vector space. Thus, if $V'$ has a vector space structure all its elements are vectors. – azarel Feb 20 '12 at 18:53
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    Halmos surely gives the axioms for a vector space at some point. Check that the "linear operations for linear functionals" make $V'$ into a vector space. – Dylan Moreland Feb 20 '12 at 19:04

1 Answers1

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Let's go back further:

Let $\mathbf{V}$ and $\mathbf{W}$ be any two vector spaces over the same field $\mathbf{F}$. Let $\mathcal{L}(\mathbf{V},\mathbf{W})$ be the set of linear transformations $T\colon \mathbf{V}\to\mathbf{W}$.

We will make $\mathcal{L}(\mathbf{V},\mathbf{W})$ into a vector space over $\mathbf{F}$. In order to do this, we need to define an "addition of linear transformations" and a "scalar multiplication of elements of $\mathbf{F}$ by linear transformations" (that is, our "vectors" will be linear transformations from $\mathbf{V}$ to $\mathbf{W}$; remember that a vector space is just a set with a "vector addition" and a "scalar multiplication" that satisfy certain properties, and we call the elements of the set "vectors"; they don't have to be "tuples" in the usual sense).

So, given two linear transformations $T,U\colon \mathbf{V}\to\mathbf{W}$, we need to define a new linear transformation that is called the "sum of $T$ and $U$". I'm going to write this as $T\oplus U$, to distinguish the "sum of linear transformations" from the sum of vectors. Since we want $T\oplus U$ to be a linear transformation (which is a special kind of function) from $\mathbf{V}$ to $\mathbf{W}$, in order to specify it we need to say what the value of $T\oplus U$ is at every $\mathbf{v}\in \mathbf{V}$. My definition is: $$(T\oplus U)(\mathbf{v}) = T(\mathbf{v}) + U(\mathbf{v}),$$ where the sum on the right is taking place in $\mathbf{W}$. This makes sense, because $T$ and $U$ are already functions from $\mathbf{V}$ to $\mathbf{W}$, so $T(\mathbf{v})$ and $U(\mathbf{v})$ are vectors in $\mathbf{W}$, which we can add.

Is $T\oplus U$ a linear transformation from $\mathbf{V}$ to $\mathbf{W}$? First, it is a function from $\mathbf{V}$ to $\mathbf{W}$. Now, to check that it is a linear transformation, we need to check that for all $\mathbf{v}_1,\mathbf{v}_2\in\mathbf{V}$ and all $\alpha\in \mathbf{F}$, we have $$(T\oplus U)(\mathbf{v}_1+\mathbf{v}_2) = (T\oplus U)(\mathbf{v}_1)+(T\oplus U)(\mathbf{v}_2)\quad\text{and}\quad (T\oplus U)(\alpha\mathbf{v}_1) = \alpha((T\oplus U)(\mathbf{v}_1)).$$ Indeed, since $T$ and $U$ are themselves linear transformations, we have: $$\begin{align*} (T\oplus U)(\mathbf{v}_1+\mathbf{v}_2) &= T(\mathbf{v}_1+\mathbf{v}_2) + U(\mathbf{v}_1+\mathbf{v}_2) &\text{(by definition of }T\oplus U\text{)}\\ &= T(\mathbf{v}_1)+T(\mathbf{v}_2) + U(\mathbf{v}_1)+U(\mathbf{v}_2) &\text{(by linearity of }T\text{ and }U\text{)}\\ &= T(\mathbf{v}_1)+U(\mathbf{v}_1) + T(\mathbf{v}_2)+U(\mathbf{v}_2)\\ &= (T\oplus U)(\mathbf{v}_1) + (T\oplus U)(\mathbf{v}_2) &\text{(by definition of }T\oplus U\text{)}\\ (T\oplus U)(\alpha\mathbf{v}_1) &= T(\alpha\mathbf{v}_1) + U(\alpha\mathbf{v}_1) &\text{(by definition of }T\oplus U\text{)}\\ &= \alpha T(\mathbf{v}_1) + \alpha U(\mathbf{v}_1) &\text{(by linearity of }T\text{ and }U\text{)}\\ &= \alpha(T(\mathbf{v}_1) + U(\mathbf{v}_1))\\ &= \alpha((T\oplus U)(\mathbf{v}_1)) &\text{(by definition of }T\oplus U\text{)} \end{align*}$$ so $T\oplus U$ is indeed an element of $\mathcal{L}(\mathbf{V},\mathbf{W})$.

I'll let you verify that $(S\oplus T)\oplus U = S\oplus (T\oplus U)$ for all $S,T,U\in\mathcal{L}(\mathbf{V},\mathbf{W})$ (since this is an equality of functions, you need to check that they have the same value at every $\mathbf{v}\in \mathbf{V}$). That $T\oplus U=U\oplus T$ for all $T,U\in\mathcal{L}(\mathbf{V},\mathbf{W})$; that if $\mathbf{0}$ is the linear transformation that sends every $\mathbf{v}\in\mathbf{V}$ to $\mathbf{0}\in\mathbf{W}$, then $T\oplus\mathbf{0}=T$ for all $T$; and that given $T\in\mathcal{L}(\mathbf{V},\mathbf{W})$, and we define $-T$ to be the function $(-T)(\mathbf{v}) = -(T(\mathbf{v}))$, then $T\oplus (-T) = \mathbf{0}$.

Now we define a scalar multiplication, which I will denote by $\odot$ (again, to avoid confusion with the scalar multiplication from $\mathbf{V}$ and $\mathbf{W}$. Given $T\colon \mathbf{V}\to\mathbf{W}$ and $\alpha\in\mathbf{F}$, define $(\alpha\odot T)$ to be the function $$(\alpha\odot T)(\mathbf{v}) = \alpha T(\mathbf{v}).$$ I will let you verify that this definition works, in that $\alpha\odot T$ is a linear transformation when $T$ is a linear transformation; and that it satisfies the necessary properties:

  • $\alpha\odot(\beta\odot T) = (\alpha\beta)\odot T$;
  • $1\odot T = T$;
  • $(\alpha + \beta)\odot T = (\alpha\odot T)\oplus (\beta\odot T)$;
  • $\alpha\odot(T\oplus U) = (\alpha\odot T)\oplus (\alpha\odot U)$.

So $(\mathcal{L}(\mathbf{V},\mathbf{W}),\oplus,\odot)$ is a vector space over $\mathbf{F}$ whenever $\mathbf{V}$ and $\mathbf{W}$ are vector spaces over $\mathbf{F}$.


So now, dual spaces: Note that $\mathbf{F}$ is always a vector space over itself, by defining vector addition to be the same as the addition of $\mathbf{F}$, and scalar multiplication to be the same as multiplication in $\mathbf{F}$.

So if $\mathbf{V}$ is any vector space over $\mathbf{F}$, then we can consider $\mathcal{L}(\mathbf{V},\mathbf{F})$: this makes sense, because both $\mathbf{V}$ and $\mathbf{F}$ are vector spaces over $\mathbf{F}$; and this is itself a vector space over $\mathbf{F}$ with vector addition $\oplus$ and scalar multiplication $\odot$ as defined above.

This vector space, $\mathcal{L}(\mathbf{V},\mathbf{F})$, is called the dual space of $\mathbf{V}$. We write $\mathbf{V}^*$ instead of $\mathcal{L}(\mathbf{V},\mathbf{F})$, and the elements of $\mathbf{V}^*$ are called "functionals".

By abuse of notation, we usually write $+$ instead of $\oplus$ (just like we use the same symbol for the addition of $\mathbf{V}$ and the addition of $\mathbf{W}$), and $\cdot$ or just juxtaposition instead of $\odot$.

The equation you have, $$y(\alpha_1 x_1 + \alpha_2x_2) = \alpha_1y(x_1) + \alpha_2y(x_2)$$ is just telling you that the function $y$ is a linear transformation from $\mathbf{V}$ to $\mathbf{F}$.

It is traditional to use boldface lower case letters like $\mathbf{f}$, $\mathbf{g}$, $\mathbf{h}$ to represent functionals. This to remind us that even though they are vectors in the vector space $\mathbf{V}^*$, they are "really" functions (when they are at home).


In fact, you could go back even further. If $\mathbf{W}$ is a vector space over $\mathbf{F}$, and $X$ is any set, then we can look at $$\mathcal{F}(X,\mathbf{W}) = \{f\colon X\to\mathbf{W}\mid f\text{ is a function}\}.$$ Then $\mathcal{F}(X,\mathbf{W})$ is a vector space, with addition $(f\oplus g)(x) = f(x)+g(x)$ and scalar multiplication $(\alpha\odot f)(x) = \alpha f(x)$. The case of $\mathcal{L}(\mathbf{V},\mathbf{W})$ corresponds to looking at a subspace of $\mathcal{F}(\mathbf{V},\mathbf{W})$ consisting of linear transformations.

This is a standard construction in abstract algebra. Whenever $A$ is an algebra (in the sense of General Algebra; a group, semigroup, ring, vector space, lattice, etc), and $X$ is a set, the collection of all function $f\colon X\to A$ becomes an algebra of the same type under "pointwise operations". In fact, this is nothing more than a "direct power" (a direct product in which every factor is the same) indexed by $X$.

Arturo Magidin
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    You should really write a book on linear algebra; I'm confident it would become an instant best-seller – ItsNotObvious Feb 20 '12 at 19:28
  • Wow, Thanks so much for that insight. I think i understand finally what vector spaces are and how generic they are and hence why so many other fields rely on vector spaces. – Comic Book Guy Feb 20 '12 at 19:40
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    A really great answer. Thanks - this really helped me also. –  Jan 29 '13 at 11:59
  • This answer helped me the most out of the other resources I found when I wanted to learn what the dual space is. I feel like I still can't visualize it though. I came from [here](http://math.stackexchange.com/a/1279872/8931), and I was thinking maybe this great answer could be made even better if you could touch upon how this relates/applies to matrix transposition. For me personally I'm still missing the bit where you are allowed to refer to functions as vector spaces in the general sense. – Steven Lu May 14 '15 at 14:44
  • For instance, what little intuition I picked up so far is telling me that the dual of $\mathbb{R}^n$ vector space is represented by the row vectors of the same dimension. – Steven Lu May 14 '15 at 14:49
  • This answer is *amazing* it clarified so much! However I still have one question: I see that we *can* make maps to a field be vector spaces, but in what concrete situation did they see the need to do this? – Pineapple Fish Jun 02 '20 at 15:44
  • @BenjaminThoburn: You need it to be able to do linear algebra! The field is the one-dimensional vector space over itself. Note that a “normal” linear transformation, say from $\mathbb{R}^3$ to $\mathbb{R}^2$, can be thought of as being made up of *two* linear function $\mathbb{R}^3\to \mathbb{R}$: what happens in the first coordinate, and what happens in the second coordinate. Without this way to think about it, there’s a whole bunch of stuff you wouldn’t be able to do, for lack of a 1-dimensional vector space. – Arturo Magidin Jun 02 '20 at 16:12