# 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. – azarel 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. – Dylan Moreland 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 – 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
• 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. – 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