# Dual space and inner/scalar product space

$V$ is vector space of finite dimension. $〈· , ·〉$ is an inner product on $V$.(Field $F$)

We set transformation $T \colon V \rightarrow V^*$ as the following: $(T(v))(w) = 〈v , w〉$.

Prove that $T$ is Isomorphism.

I don't know how to prove that it is 1 on 1 and onto. I mean, the dualic space is confusing me since I don't understand it properly.

For 1-1 : I need to assume that $〈v_1,w_1〉 = 〈v_2,w_2〉$ and show that $v_1=v_2$ and $w_1 = w_2$? I'm not sure what is my domain... $T$ is from $V$ to $V^*$ so should I show that $v_1 = v_2$ only?

For onto: I need to show that for every functional from $V^*$ there is $v$ from $V$ such that it equals? doesn't make sense because the inner product is a scalar from $F$. I feel very helpless about this, could someone help me please ?

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Let $V$ be a finite dimensional vector space over $\mathbb K$ (choose for example $\mathbb R$).

• On the dual space $V^{*}$: why?

The dual space $V^{*}$ is defined as

$$V^{*}:=\{ \varphi: V\rightarrow \mathbb K,~~ \varphi~ \text{linear}\};$$

in other words, the dual space is the space of all linear functionals, i.e. those maps from the original vector space $V$ to the ground field $\mathbb K$ which are linear.

Why is this new space interesting?

In a certain sense, introducing $V^{*}$ we used all information we had from the beginning to produce a new linear space. In fact, we started just with a pair $(V,\mathbb K)$ and we arrive at a new vector space $V^{*}$ whose elements "connect" $V$ to $\mathbb K$ respecting the non trivial structure on $V$, i.e. the linear space structure.

Moreover, if $V$ is finite dimensional, then $V^{*}$ is finite dimensional as well. Let $\{e_i\}_{i=1,\dots,n}$ be a basis of $V$ ($n$ is the dimension of $V$). The dual set

$$\{\varphi_j\}_{j=1,\dots,n}$$

of elements $\varphi_j\in V^{*}$ s.t.

$$\varphi_j(e_i):=\delta_{ij}$$

is a basis of $V^{*}$. To prove it you can check any book of linear algebra.

Note that we used only the existence of the pair $(0,1)$, with $0:=\delta_{ij}$ for $i\neq j$ and $1:=\delta_{ij}$ for $i=j$ in the ground field $\mathbb K$ to introduce our basis $\{\varphi_j\}_{j=1,\dots,n}$. Such pair $(0,1)$ always exists by definition of field: the $0$ is the zero element of the addition, while $1$ denotes the unit element of multiplicative composition.

• On $T:V\rightarrow V^{*}$

Let $T:V\rightarrow V^{*}$ be the linear map

$$T(v)(w):=\langle v,w\rangle,$$

denoting by $\langle \cdot,\cdot \rangle$ an inner product on $V$. We want to prove that $T$ is an isomorphism. We need to check that $T$ is injective and surjective.

On injectivity: as $T$ is linear, to prove injectivity is equivalent to show that

$$\operatorname{Ker}(T)=\{v\in V: T(v)=0 ~\text{in}~ V^{*}\}=\{0\}.$$

By definition, $T(v)=0$ in $V^{*}$ (here "0" denotes the zero map in $V^{*}$!) if

$$\forall w\in V\Rightarrow T(v)(w)=\langle v,w\rangle=0~~(*)$$

in $\in\mathbb K$; $(*)$ holds if and only if $v=0$. In fact, in $(*)$ we can choose $w=v$, arriving at

$$T(v)(v)=\langle v,v\rangle=0\Leftrightarrow v=0$$

by definition of inner product (check it!). In other words we have proved that $T(v)=0\Leftrightarrow v=0$, i.e. $\operatorname{Ker}(T)=\{0\}$.

On surjectivity You need to prove that

$$\forall \varphi\in V^{*}~~ \exists v\in V : T(v)=\varphi$$

in $V^{*}$, i.e.

$$\forall w\in V~~ T(v)(w)=\varphi(w)$$

in $\mathbb K$. This last statement is equivalent to

$$\forall w\in V~~ \langle v,w\rangle=\varphi(w),$$

which can be easily proven using a base on $V$ and the dual basis on $V^{*}$, as above. I leave it to you. I hope it helps.

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Thank you so much I'm very thankful and out of words... You helped me alot !!!! :) – user84636 Jul 10 '13 at 19:08
Can you explain me please how to prove the onto? I thought about Reisz but the order of the vectors is different so i don't think it's right – user84636 Jul 10 '13 at 19:48
you are welcome :) Suppose that $e_i$ is an orthonormal basis of $V$ (you can always get one using Gram-Schmidt). Then $\langle e_i,e_j\rangle=\delta_{ij}$. Let $\varphi_i$ the dual basis in $\in V^{*}$. So any $\varphi\in V^{*}$ is given by a unique linear combinations of the $\varphi_i$'s. Question: can you find a $v\in V$ s.t. $T(v)=\varphi_i$, i.e. $T(v)(e_j)=\varphi_i(e_j)$, for all $e_j$? If yes, you have solved surjectivity, as your map $T$ is linear... – Avitus Jul 11 '13 at 12:13

You are wrong about the injectivity claim. Let me translate it correctly for you.

Injectivity of $T: V \rightarrow V^*$ means that given $v \in V$, such that $T(v)=0$ implies that $v=0$.

Now what does $T(v)=0$ mean? Well, $T(v)$ is a linear functional $T(v): V \rightarrow F$, defined by $T(v)(w)=\langle v,w \rangle$, and a linear functional is zero, if and only if it maps everything to zero. So $$T(v)=0$$ really means that $$\langle v,w \rangle=0$$ for all $w \in V$. But from the properties of an inner product you can deduce that, if for some $v \in V$ we have $\langle v,w \rangle=0$ for all $w \in V$, then necessarily $v=0$ (do you see how?).

As for surjectivity, since $V$ is a finite-dimensional $F$-vector space, $V^*$ has the same dimension as $V$ (dual basis). Now what can you conclude about an injective linear map$^{\dagger}$ between two vector spaces of the same dimension.

$^{\dagger}($Of course you also have to prove that $T$ is a $F$-linear map, but that again follows from the definition of an inner product.)

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thank you my friend! – user84636 Jul 10 '13 at 19:13
you're welcome. – Nils Matthes Jul 10 '13 at 19:45

For injectivity: You need to show that whenever $T(v_1) = T(v_2)$, then $v_1 = v_2$ for any $v_1, v_2 ∈ V$. Now $T(v_1)$ and $T(v_2)$ are elements in $V^*$ and therefore linear maps $V → F$. By definition of equality, those maps they are the same if they do some thing on $V$, i.e. if $∀ w ∈ V:\, T(v_1)(w) = T(v_2)(w)$ or by definition of $T$: $∀ w ∈ V:\, 〈v_1,w〉 = 〈v_2,w〉$.

Now, use that $T$ is a linear map and show that its kernel is zero. So show: $∀ v ∈ V:\, T(v) = 0 ⇒ v = 0$. This will do to show injectivity. The fact that any inner product is a non-degenerate bilinear map might come in handy.

For surjectivity use that any non-trivial linear map $ϕ : V → F$ has a kernel of dimension $n-1$, where $n = \dim V$. Now in the inner product space, this means that its orthogonal complement $(\ker ϕ)^⊥$ has dimension 1 and is generated by some nonzero element $v$. Show that $\ker T(v) = \ker ϕ$. Conclude that there is a $λ ∈ F$ such that $T(λv) = ϕ$. (Take $λ = ϕ(v)/〈v,v〉$.)

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what is the kernel of T? – user84636 Jul 10 '13 at 10:33
@user84636 $\ker T := \{v ∈ V;\; T(v) = 0\}$, the subspace where $T$ vanishes. You didn’t know of the kernel, but of dual spaces? That’s surprising. You can show that any linear map $T$ is injective if and only if $\ker T = 0$. That’s because $T(v_1) = T(v_2) ⇔ T(v_1 - v_2) = 0$. – k.stm Jul 10 '13 at 10:35