# Why is a dual space a vector space?

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.

• A vector is an element of a vector space. Thus, if $V'$ has a vector space structure all its elements are vectors. Feb 20 '12 at 18:53
• 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. Feb 20 '12 at 19:04

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 $$X$$.

• You should really write a book on linear algebra; I'm confident it would become an instant best-seller 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. Feb 20 '12 at 19:40
• A really great answer. Thanks - this really helped me also.
– user60088
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, 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. 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. May 14 '15 at 14:49