Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$V$ 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 ?

share|cite|improve this question
up vote 8 down vote accepted

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


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


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.

share|cite|improve this answer
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.)

share|cite|improve this answer
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 $λ = ϕ(v)/〈v,v〉$.)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.