# Spaces of linear maps and dual space

Yesterday I learned about dual spaces when reading about spaces of linear maps. The concept of a linear map and why linear maps form a vector space is clear to me. But there are some details about the dual space and its basis that I could not fully understand.

The text I am reading states the following:

Furthermore, for fixed vector spaces $$U$$ and $$V$$ over $$K$$, the operations of addition and scalar multiplication on the set $$\operatorname{Hom}_K(U,V)$$ of all linear maps from $$U$$ to $$V$$ makes $$\operatorname{Hom}_K(U,V)$$ into a vector space over $$K$$.

Given a vector space $$U$$ over a field $$K$$, the vector space $$U^{*} = \operatorname{Hom}_K(U,K)$$ plays a special role. It is often called the dual space or the space of covectors of $$U$$. One can think of coordinates as elements of $$U^{*}$$. Indeed, suppose that $$U$$ is finite-dimensional and let $$e_{1},...,e_{n}$$ be a basis of $$U$$. Every $$x \in U$$ can be uniquely written as $$x=\alpha_{1}e_{1}+...+\alpha_{n}e_{n}, \alpha_{i} \in K.$$ The scalars $$\alpha_{1},...\alpha_{n}$$ depend on $$x$$ as well as on the choice of basis, so for each $$i$$ one can write the coordinate function $$e^{i}: U \to K, e^{i}(x)=\alpha_{i}.$$ It is routine to check that each $$e^{i}$$ is a linear map, and indeed the functions $$e^{1},...,e^{n}$$ form a basis of the dual space $$U^{*}$$.

Now I have two questions:

1) The text states that $$\operatorname{Hom}_K(U,V)$$ is a vector space for fixed $$U$$ and $$V$$. This is perfectly clear to me, but is it correct that the dual space is $$U^{*}=\operatorname{Hom}_K(U,K)$$, i.e. it consists of all linear maps from $$U$$ to the field $$K$$? At first I thought this was a typo, but from what I've read on wikipedia and from other sources the notation seems to be correct. It also seems to make no sense to say the coordinate functions form a basis for $$\operatorname{Hom}_K(U,V)$$.

2) It is easy to see that the coordinate functions $$e^{1},...,e^{n}$$ are linear maps and I have also tried to check that the claim that they form a basis for $$U^{*}$$. However, I am unsure if my proof is correct and I think this is mainly because of my confusion stated in the first question.

My proof goes as follows:

We need to prove that $$e^{1},...,e^{n}$$ are linearly independent and span $$U^{*}$$.

First note that linear maps are uniquely determined by their action on a basis. Now let $$0_{UK}:U \to K, 0_{UK}(u)=0$$ $$\forall u$$ be the zero map. To prove linear independence we need to show that

$$(*)$$ $$b_{1}e^{1}+...+b_{n}e^{n}=0_{UK} \implies b_{i}=0$$ $$\forall i$$

or in other words the only linear combination of $$e^{1},...,e^{n}$$ that gives $$0_{UK}$$ is the trivial linear combination. Now assume $$b_{i}=0$$ for some $$i$$, then(*) clearly fails if $$x$$ has a non-zero ith coordinate and so $$b_{1}e^{1}+...+b_{n}e^{n}$$ is not the zero map.

To prove that $$e^{1},...,e^{n}$$ span $$U^{*}$$ we need to prove that any vector in $$U^{*}$$ (every linear map) can be written as a linear combination of $$e^{1},...,e^{n}$$, i.e. $$T(u)=k_{1}e^{1}+...k_{n}e^{n}$$ for any vector $$u \in U$$. To see this note that \begin{align*} T(u)&=T(\alpha_{1}e_{1}+...\alpha_{n}e_{n}) \\ &=\alpha_{1}T(e_{1})+...\alpha_{n}T(e_{n}) \\ &=e^{1}T(e_{1})+...e^{n}T(e_{n}) \end{align*} where $$T(e_{i})$$ are scalars by definition of $$T$$.

Is my proof correct or have I missed anything?

Thanks very much for any hints and comments.

• What do you denote (1) which has to hold for any $x$? – Bernard Jan 2 '20 at 9:42
• Sorry, this is a typo. I changed (1) to $(*)$ later. Actually I wanted to remove this part since it became clear to me when I wrote it down. Thanks. – DerivativesGuy Jan 2 '20 at 9:56

## 2 Answers

• The linear independence can be shown more directly, suppose that $$b_1e^1 + b_2e^2 + \cdots + b_ne^n = \textit0 \quad \textrm{for some } b_1,\dots,b_n \in K$$ where $$\textit0 : U \to K$$ is the zero map. Now, for all $$i$$, $$1\leq i\leq n$$, \begin{align} b_i = b_ie^i(e_i) &= \sum_{j=1}^n b_j e^j(e_i) \\ &= \Big( \sum_{j=1}^n b_j e^j \Big)(e_i) = \textit0(e_i) = 0 \end{align} so, $$b_1 = b_2 = \cdots = b_n = 0$$.

• Also, to check that $$e^1,\dots,e^n$$ spans $$U^*$$, let $$f$$ be arbitrary in $$U^*$$, and observe that for any $$x\in U$$ written as $$x = \alpha_1e_1 + \cdots + \alpha_n e_n$$, we have \begin{align} f(x) &= \sum_{j=1}^n \alpha_j f(e_j) \\ &= \sum_{j=1}^n f(e_j) \alpha_j \\ &= \sum_{j=1}^n f(e_j) e^j(x) = \Big( \sum_{j=1}^n f(e_j)e^j \Big) (x) \end{align} so, $$f = f(e_1)e^1 + f(e_2)e^2 + \cdots + f(e_n)e^n$$

Your proof seems a bit confusing. You would make it clearer with the initial observation that any linear map from a vector space $$U$$ to a vector space $$V$$ is entirely determined by its values at the vectors $$e_i$$ of a basis of $$U$$. The rest should follow almost immediately.

• Thanks for your comment. You are right, I have implicitly taken this for granted since it was stated earlier in the text I am reading. I will add it to the proof. – DerivativesGuy Jan 2 '20 at 10:35
• And then all verifications have to be made only for the vectors of the basis. – Bernard Jan 2 '20 at 10:37
• Yes, this is clear to me. I think the main confusion for me was what I have stated in question 1). If my understanding is correct, then my proof should be fine. – DerivativesGuy Jan 2 '20 at 10:42