# Proving that $\phi (T) = T^*$ is an isomorphism between vector spaces

I am tasked with the following:

I am thus tasked with proving:

$$1)$$ $$\phi(T)$$ is linear, so that it respects closure under scalar multiplication and addition.

$$2)$$ $$\phi(T)$$ is a bijection.

I only need justification as to whether or not my explanation as to whether $$\phi(T)$$ is a bijection is indeed valid or not. First, I will note that linear functionals are isomorphisms, so are bijective themselves and respect closure under addition and scalar multiplication. I work to first prove that $$\phi$$ is injective.

## Injectivity

Suppose $$\phi$$ is injective.

$$\implies \phi(T_n) = \phi(T_m) \iff T_n = T_m$$

$$\implies T^*_n = T^*_m \iff T_n = T_m$$ $$\implies f \ (T_n) = f \ (T_m) \iff T_n = T_m$$ $$\implies f(T_n - T_m) = 0 \iff T_n = T_m$$

Let's suppose $$T_n \ne T_m$$.

$$\implies f(T_n-T_m) = 0 \implies T_n - T_m \in \text{Ker} \ f$$

Which is a contradiction, as $$f$$ is surjective.

## Surjectivity

Assume $$\exists \ T \in \mathcal L (V,W)$$ such that $$\phi(T) = 0$$.

$$\implies T^* = 0$$. $$\implies f(T) = 0$$ $$\implies T \in \text{Ker} \ f$$

Which is a contradiction. Hence, $$\phi$$ is surjective, making it a bijection and $$\phi$$ an isomorphism.

Did I make any mistakes here?

• You prove injectivity of $\;\phi\;$ ...by assuming $\;\phi\;$ is injective?! – DonAntonio Nov 16 '18 at 20:25
• Ack! I hope not. I thought I proved by contradiction eventually with suppose $T_n \ne T_m$ – sangstar Nov 16 '18 at 20:26
• Also observe that in (1)-(2) you're supposed, I guess, to prove $\;\phi\;$ is linear and injective, not $\;\phi(T)\;$ , which is merely a linear map $\;W^*\to V^*\;$ ... – DonAntonio Nov 16 '18 at 20:28

## 2 Answers

Fill in details:

By definition, if $$\;T:V\to W\;$$ is a linear map, then $$\;T^*:W^*\to V^*\;$$ is defined as

$$\;T^*(g)v:=g(Tv)\;,\;\;v\in V\;,\;\;g\in W^*\;$$ , thus

$$\;\phi\;$$ is linear, because

$$\phi(T+S):=(T+S)^*=T^*+S^*$$

and the last equality follows from

$$(T+S)^*(g)v:=g(T+S)v=g(Tv+Sv)=g(Tv)+g(Sv)=:T^*(g)v+S^*(g)v$$

And also

$$T\in\ker \phi\implies \phi(T):=T^*=0^*\implies\forall\,g\in W^*\,,\,\,T^*(g)v=g(Tv)=0\,,\,\,\forall v\in V\implies$$

$$Tv\in\ker g\,,\,\,\forall g\in W^*\implies Tv=0\,\,\forall\,v\in V\implies T\equiv0.$$

DonAntonio already touched linearity and injectivity questions. For surjectivity, you seem to be proving injectivity instead. However, to correctly prove surjectivity, you are going to need to use a dimension-counting argument. This is because, if $$W$$ is infinite-dimensional and $$V\neq 0$$, then $$\dim_\mathbb{K}\big(\mathcal{L}(V,W)\big)<\dim_\mathbb{K}\big(\mathcal{L}(W^*,V^*)\big)\,,$$ where $$\mathbb{K}$$ is the ground field. However, the dual map $$\phi:\mathcal{L}(V,W)\to\mathcal{L}(W^*,V^*)$$ is still an injective linear map, regardless of the dimensions of $$V$$ and $$W$$. The proofs of linearity and injectivity are essentially unchanged.

Since $$V$$ and $$W$$ in the problem statement are both finite-dimensional, $$\dim_\mathbb{K}\big(\mathcal{L}(V,W)\big)=\dim_\mathbb{K}(V)\,\dim_\mathbb{K}(W)=\dim_\mathbb{K}(W^*)\,\dim_\mathbb{K}(V^*)=\dim_\mathbb{K}\big(\mathcal{L}(W^*,V^*)\big)\,.$$ Thus, any injective linear map from $$\mathcal{L}(V,W)$$ to $$\mathcal{L}(W^*,V^*)$$ is automatically surjective, whence bijective.

Interestingly, if $$W$$ is finite-dimensional and $$V$$ is infinite-dimensional, the map $$\phi$$ is still an isomorphism. We are left to show that $$\phi$$ is surjective. To show this, let $$S:W^*\to V^*$$ be a linear map. Let $$n:=\dim_\mathbb{K}(W)$$. Pick a basis $$\{w_1,w_2,\ldots,w_n\}$$ of $$W$$, along with the dual basis $$\{f_1,f_2,\ldots,f_n\}$$ of $$W^*$$ (i.e., $$f_i(w_j)=\delta_{i,j}$$ for $$i,j=1,2,\ldots,n$$, where $$\delta$$ is the Kronecker delta). For each $$w\in W$$, write $$w^{**}\in W^{**}$$ for its double dual. Ergo, we see that $$S$$ takes the form $$S=\sum_{i=1}^n\,e_i\otimes w_i^{**}$$ for some $$e_1,e_2,\ldots,e_n\in V^*$$ (namely, $$e_i:=S(f_i)$$ for $$i=1,2,\ldots,n$$). Define $$T:=\sum_{i=1}^n\,w_i\otimes e_i\,.$$ Then, for all $$j=1,2,\ldots,n$$ and $$v\in V$$, we have $$\big(T^*(f_j)\big)(v)=f_j\big(T(v)\big)=f_j\left(\sum_{i=1}^n\,e_i(v)\,w_i\right)=\sum_{i=1}^n\,e_i(v)\,f_j(w_i)=\sum_{i=1}^n\,e_i(v)\,\delta_{i,j}=e_j(v)\,.$$ However, as $$e_j=S(f_j)$$, we get $$\big(S(f_j)\big)(v)=e_j(v)$$ for all $$j=1,2,\ldots,n$$ and $$v\in V$$. This proves that $$S(f_j)=T^*(f_j)$$ for $$j=1,2,\ldots,n$$. Because $$f_1,f_2,\ldots,f_n$$ span $$W^*$$, we get $$S=T^*=\phi(T)$$. Therefore, $$\phi$$ is surjective whenever $$W$$ is finite-dimensional. (Consequently, the dual map $$\phi:\mathcal{L}(V,W)\to\mathcal{L}(W^*,V^*)$$ is an isomorphism if and only if $$W$$ is finite-dimensional or $$V=0$$.)

• Surjectivity is automatic once we have injectivity since all the involved linear spaces are finite dimensional...and of the same (finite) dimension, of course. – DonAntonio Nov 16 '18 at 21:49
• That is what I wrote. I merely wanted to discuss what happens when $V$ or $W$ is not finite-dimensional. – Batominovski Nov 16 '18 at 21:50
• Why does involved linear spaces being finite dimensional imply subjectivity when injective? – sangstar Nov 17 '18 at 16:51
• Because the dimension of the image of $\phi$ equals the dimension of the domain, but the dimension of the domain also equals the dimension of the codomain. Hence, the image is a subspace of the codomain with the same finite dimension. – Batominovski Nov 17 '18 at 17:09