List 1

1. Find a representation of $C_3$ over $\mathbb{R}$ which is irreducible but not one-dimensional.

Solution. Define $M=\left[\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right]$ and $\rho :C_3\to \mathrm{GL}({\mathbb{R}}^2),a^i\mapsto M^i$ where $C_3=\{1,a,a^2\}$. If $\rho$ is reducible, then there exists a vector $v$ such that $Mv=\lambda v$ for some $\lambda \in \mathbb{R}$. However, since the eigenvalues of $M$ are $\{e^{2\pi i/3},e^{4\pi i/3}\}$, $\lambda$ does not exist. Therefore, $\rho$ is a irreducible representation with dimension $2$. \end{proof}

2. Decompose the regular representation $\mathbf{k}\left[S_3\right]$ of $S_3$ (left action of $S_3$ on $\mathbf{k}\left[S_3\right]$ ).

Solution. Firstly, assume that $\mathrm{char}k\nmid 6$. Let $V$ be the $S_3$-module corresponding its regular representation. Then by Maschke’s theorem, there exists $V_1,\cdots ,V_n$ such that $V=V_1\oplus \cdots \oplus V_n$. Since $S_3$ is non-abelian, it has at least $1$ irreducible module which is not one-dimensional. Note that $6=1^2+1^2+2^2$, so $V=V_1\oplus V_2\oplus V_3$ with $\dim V_1=\dim V_2=1$ and $\dim V_3=\dim V_4=2$.

Note that the trivial representation and the sign representation are two non-isomorphic one-dimensional representations, and the standard representation of $S_3$ is also irreducible. It follows that $V_1$, $V_2$ are modules corresponding to trivial representation and sign representation, respectively. And $V_3\cong V_4$ are isomorphic, which correspond to the standard representation.

• $k\mid 6$? \end{proof}

3. Let $G$ be the dihedral group, the group of symmetries of a regular $n$-gon.

• (i) If $n$ is odd, construct two distinct one-dimensional representations of $G$. In $n$ is even, construct four distinct one-dimensional representations of $G$.
• (ii) If $n$ is odd, construct $\frac{n-1}{2}$ distinct two-dimensional irreducible representations of $G$. In $n$ is even, construct $\frac{n-2}{2}$ distinct two-dimensional irreducible representations of $G$.

Solution. We divide the problem into $2$ parts: $n$ is odd and $n$ is even.

(i) When $n$ is odd, define ${\sigma }_1:G\to \{1\}$ and

$${\sigma }_2:G\to \{1,-1\},a^i\mapsto 1,a^{i}b\mapsto -1$$

where $i\in \{0,\cdots ,n-1\}$. Then ${\sigma }_1$ and ${\sigma }_2$ are two distinct one-dimensional representations of $G$. Now consider two-dimensional representations $\rho$. Note that $\rho (a)$ and $\rho (a^{-1})$ are similar, they have the same eigenvalues. Let $\omega =e^{2\pi i/n}$. Define ${\rho }_i(a)=\left[\begin{array}{cc}{\omega }^i& \\ & {\omega }^{n-i}\end{array}\right]$ and ${\rho }_i(b)=\left[\begin{array}{cc}& 1\\ 1& \end{array}\right]$. Then ${\rho }_i:G\to \mathrm{GL}({\mathbb{C}}^2)$, $i\in \{1,\cdots ,\frac{n-1}{2}\}$ are all two-dimensional irreducible representation.

(ii) When $n$ is even, define ${\sigma }_1,{\sigma }_2$ as in (i). Define ${\sigma }_3$ by ${\sigma }_3(a)=-1,{\sigma }_3(b)=1$. Define ${\sigma }_4$ by ${\sigma }_4(a)=-1,{\sigma }_4(b)=-1$. It is easy to verify that ${\sigma }_3,{\sigma }_4$ are representations of $G$. Thus, all one-dimensional representations of $G$ are ${\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_4$. Similarly, let ${\rho }_i$ be the two-dimensional representation defined above. Then $\{{\rho }_i:i=1,\cdots ,\frac{n-2}{2}\}$ is a set of distinct two-dimensional irreducible representations. \end{proof}

4. Show that the representations constructed in the previous exercise are all irreducible representations of the dihedral group.

\begin{proof} Since $2n=1^2\cdot 2+2^2\cdot \frac{n-1}{2}$ and $2n=1^2\cdot 4+2^2\cdot \frac{n-2}{2}$, the representations constructed in the previous exercise are all irreducible representations of the dihedral group. \end{proof}

5. Let $V$ and $W$ be two vector spaces. Denote by $\mathrm{Hom}(V,W)$ the $k$-vector space of linear maps $T:V\to W$. Let $\pi$ and $\rho$ be representations of $G$ in $V$ and $W$ respectively. Show that the mapping

$${\mathrm{Hom}}_{\pi ,\rho }(g):\mathrm{Hom}(V,W)\to \mathrm{Hom}(V,W),T\to \rho (g)T\pi \left(g^{-1}\right)$$

defines structure of a $G$ module on $\mathrm{Hom}(V,W)$.

\begin{proof} If $g=1$, then ${\mathrm{Hom}}_{\pi ,\rho }(1)(T)=\rho (1)T\pi (1)=T$. To show ${\mathrm{Hom}}_{\pi ,\rho }(g_1{g}_2)={\mathrm{Hom}}_{\pi ,\rho }(g_1){\mathrm{Hom}}_{\pi ,\rho }(g_1)$, we only need to show that $\rho (g_1)\rho (g_2)T{\pi }^{-1}(g_2){\pi }^{-1}(g_1)=\rho (g_1{g}_2)T{\pi }^{-1}(g_1{g}_2)$ for all $T\in {\mathrm{Hom}}_{\pi ,\rho }$. Since $\rho$ and $\pi$ are representations of $G$, we have that $\rho (g_1)\rho (g_2)=\rho (g_1{g}_2)$ and ${\pi }^{-1}(g_1{g}_2)=(\pi (g_1)\pi (g_2){)}^{-1}={\pi }^{-1}(g_2){\pi }^{-1}{g}_1$. Therefore, the mapping defines structure of a $G$-module on $\mathrm{Hom}(V,W)$ \end{proof}

6. For $V$ a $\mathbf{k}$ vector space denote by $V^{\ast }=\mathrm{Hom}(V,\mathbf{k})$ the dual space of $V$. If $\rho$ is a representation of $G$ in $V$ then using the notation of previous exercise we call representation ${\rho }^{\ast }={\mathrm{Hom}}_{\rho ,\text{ triv }}$ of $G$ in $\mathrm{Hom}(V,\mathbf{k})$ the dual representation of $\rho$. Verify that $\left({\rho }^{\ast }(g)f\right)(v)=f\left(\rho \left(g^{-1}\right)v\right),v\in V,f\in V^{\ast },g\in G$.

\begin{proof} Trivial. \end{proof}

7. Show that there is a $3$-dimensional real representation $\rho :Q_8\to {\mathrm{GL}}_{\mathbb{R}}\left({\mathbb{R}}^3\right)$ of the quaternion group $Q_8$ for which

$$\rho (i)\left(x{e}_1+y{e}_2+z{e}_3\right)=x{e}_1-y{e}_2-z{e}_3,\rho (j)\left(x{e}_1+y{e}_2+z{e}_3\right)=-x{e}_1+y{e}_2-z{e}_3$$

Show that this representation is not irreducible and determine $\ker \rho$.

\begin{proof} Since $\rho (i)=\mathrm{diag}(1,-1,-1)$ and $\rho (j)=\mathrm{diag}(-1,1,-1)$, we have that $\rho (k)=\mathrm{diag}(-1,-1,1)$. Then $⟨e_i⟩$ are $\rho (G)$-invariant subspaces for $i\in \{1,2,3\}$ and so the representation is not irreducible. Since $\rho (i^2)=\rho (j^2)=\rho (k^2)=\rho (-1)=\mathrm{id}$ and $\rho (i),\rho (j),\rho (k)$ are not identity, we have that $\ker \rho =\{1,-1\}$. \end{proof}

8. If $p$ is a prime and $G$ is a non-trivial finite $p$-group, show that there is a non-trivial $1$-dimensional representation of $G$. More general show this holds for a finite solvable group.

\begin{proof} Since finite $p$-group is finite, we only need to show a finite solvable group $G$ has a non-trivial one-dimensional representation. Let $M$ be a maximal normal subgroup of $G$. If $G$ is solvable, then $G/M\cong {\mathbb{Z}}_q$ for some prime $q$ and $G={\cup }_{i=0}^{q-1}M{g}_0^i$ for some $g_0\in G$ with $(g_0{)}^q\in M$. Define $\rho :G\to \mathbb{C}$ as $g\mapsto (e^{2\pi i/q}{)}^i$ if $g\in M{g}_0^i$. Then $\rho$ is a non-trivial 1-dimensional representation of $G$. \end{proof}

9. Let $p$ be a prime. We now consider the two maps ${\rho }_1:\mathbb{Z}\to {\mathrm{GL}}_2(\mathbb{C})$ and ${\rho }_2:C_p\to$ ${\mathrm{GL}}_2\left({\mathbb{F}}_p\right)$ given by

$${\rho }_1(n)=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right],{\rho }_2\left(g^n\right)=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$$

Show that these define representations of $\mathbb{Z}$ (over $\mathbb{C})$ and $C_p$ (over ${\mathbb{F}}_p$). Are they irreducible? Are they indecomposable?

\begin{proof} They are not irreducible but indecomposable. \end{proof}

10. Let $V$ be a finite dimensional complex vector space, and let $\mathrm{GL}(V)$ be the group of invertible linear transformations of $V$. Show that $S^{n}V$ and ${\Lambda }^{m}V$ with $m\le \dim (V)$ are representations of $GL(V)$. Show that they are irreducible.

\begin{proof} The definition of $S^n$ (symmetric power) and ${\Lambda }^m$ (exterior power) are written as follows.

It is easy to verify they are representations of $\mathrm{GL}(V)$. It suffices to show they are irreducible. Note that ${\Lambda }^{m}V$ is irreducible, as $\mathrm{GL}(V)$ is transitive on elements of ${\Lambda }^{m}V$ (each element determine a set of basis). Similarly, for a set of basis $\{v_1,\cdots ,v_n\}$, the basis of $S^{n}V$ is $L:=\{v_{i_1}\otimes \cdots \otimes v_{i_n}:1⩽i_1⩽\cdots ⩽i_n⩽n\}$. And $S^{n}V$ is a cyclic module as for any $v\in L$ there exists $g\in \mathrm{GL}(V)$ such that $v_1\otimes \cdots \otimes v_n^g=v$. \end{proof}

11. Let $X$ be a set and let $\mathcal{F}(X)$ be the set of $\mathbf{k}$-valued functions on $X$. Consider a left action $l:G\times X\to X$ on $X$. For every $\phi \in \mathcal{F}(X),g\in G$ and $x\in X$ set ${\sigma }_g\phi (x)=\phi \left(g^{-1}\cdot x\right)$. Prove that $\sigma$ is a representation of $G$ in $\mathcal{F}(X)$. Define a right action $r:X\times G\to X$ by $x\cdot g=g^{-1}\cdot x$, and consider the representation $\rho$ of $G$ in $\mathcal{F}(X)$ associated with this action. Check that $\rho$ and $\sigma$ are equivalent representations.

\begin{proof} It is easy to check that $\sigma$ is a representation. Now we show that $\rho$ and $\sigma$ are equivalent representations. Note that $\rho$ is defined as ${\rho }_g\phi (x)=\phi (x\cdot g)$. Since $x\cdot g=g^{-1}\cdot x$, we have that $\rho$ and $\sigma$ are equivalent where the identity map $T:\mathcal{F}(X)\to \mathcal{F}(X)$ satisfying $T\circ {\rho }_g={\sigma }_g\circ T$ for all $g\in G$. \end{proof}

12. Let $\mathbf{k}$ be a field of characteristic different from $2$. Let $G$ be a finite group of even order and let $G^{\prime }\le G$ be a subgroup of order $2$ with generator $\gamma$. Consider the regular representation of $G$, which comes from the natural left action of $G$ on $\mathbf{k}[G]$. Consider the action of the generator $\gamma$ by multiplication of basis vectors on the right. Denote the $+1$ and $-1$ eigenspaces for this action by $\mathbf{k}[G{]}^{+},\mathbf{k}[G{]}^{-}$respectively. Show that the subspaces $\mathbf{k}[G{]}^{\pm }$are $G$-subspaces and that

$$\dim \mathbf{k}[G{]}^{+}=\dim \mathbf{k}[G{]}^{-}=\frac{|G|}{2}$$

Deduce that $\mathbf{k}[G]=\mathbf{k}[G{]}^{+}\oplus \mathbf{k}[G{]}^{-}$as $G$-representations.

\begin{proof} Assume that $|G|=2n$ and $G=\{g_1,\cdots ,g_{2n}\}$, then $V=\mathbf{k}[G]=\mathrm{span}\{g_1,\cdots ,g_{2n}\}$. It is easy to verify that $\gamma$ is a linear transformation and ${\gamma }^2=\mathrm{id}$, then $\gamma$ has two eigenvalues $\pm 1$. Since the matrix of $\gamma$ is trace $0$, we have that $\dim \mathbf{k}[G{]}^{+}=\dim \mathbf{k}[G{]}^{-}=\frac{|G|}{2}$. Furthermore, they are $G$-subspaces as $\gamma$ acts on $V$ by multiplication on the right. \end{proof}

13. Let $A$ be an associative algebra, and let $V$ be a representation of $A$. By ${\mathrm{End}}_A(V)$ one denotes the algebra of all homomorphism of representations $V\to V$. Show that ${\mathrm{End}}_A(A)=A^{op}$, the algebra $A$ with opposite multiplication.

\begin{proof}

I am not sure whether we can assume $A$ has a multiplicative identity. But I do assume it anyway.

For any $f\in {\mathrm{End}}_A(A)$, we have $f(r)=f(r\cdot 1)=rf(1)$ and hence it is a right multiplication by some $a=f(1)\in A$. Furthermore, if ${\phi }_a:x\mapsto xa$ for every $a\in A$, then ${\phi }_a{\phi }_b(x)={\phi }_{ba}(x)$ and so $a{\cdot }_{{\mathrm{End}}_A(A)}b=ba$. Thus ${\mathrm{End}}_A(A)=A^{op}$. \end{proof}

14. Let $V$ be an irreducible finite dimensional representation of algebra $A$, and let $v_1,\dots ,v_n\in V$ be any linearly independent vectors. Then for any $w_1,\dots ,w_n\in V$ there exists an element $a\in A$ such that $a{v}_i=w_i$.

\begin{proof} Assume that $\dim V=n$. Define $\phi :\mathrm{End}(V)\to nV,x\mapsto (x{v}_1,\cdots ,x{v}_n)$ where $v_1,\cdots ,v_n$ is a set of basis of $V$. Since $\phi (ax)=a\phi (x)$ for all $a\in A$ and $\phi$ is a bijection, then $\mathrm{End}(V)$ and $nV$ are isomorphic $A$-modules. Let $\rho :A\to \mathrm{End}(V)$ be the corresponding representation. Then $\rho (A)⩽\mathrm{End}(V)$ is a sub-module of $\mathrm{End}(V)\simeq nV$. By ^ey40pv, we have that $\rho (A)\simeq rV$ for some $r<n$. Suppose $\psi :\rho (A)\to rV$ is an isomorphisms between $A$-modules, then by Proposition 2.2 there exists a matrix $M_{r\times n}$ such that $\psi (\rho (a))M=\phi (\rho (a))$ for all $a\in A$.

Note that there exists a non-zero vector $(w_1,\cdots ,w_n{)}^T$ such that $M(w_1,\cdots ,w_n{)}^T=\vec{0}$. It follows that for any $a\in A$, $\phi (\rho (a))(w_1,\cdots ,w_n{)}^T=0$. Take $a=1$, then $(v_1,\cdots ,v_n)(w_1,\cdots ,w_n{)}^T=0$, which contradicts with $v_1,\cdots ,v_n$ linearly independent. Thus $\rho (A)\simeq nV$ and so $\rho$ is surjective. For any $w_1,\cdots ,w_n\in V$, there exists $x\in \mathrm{End}(V)$ such that $x{v}_i=w_i$. Then there is an $a\in A$ such that $\rho (a)=x$ and $a{v}_i=w_i$. Now we finish the proof. \end{proof}

• 15. Let $A$ be an algebra over a field $\mathbf{k}$. The center $Z(A)$ of $A$ is the set of all elements $z\in A$ which commute with all elements of $A$. Show that if $V$ is an irreducible finite dimensional representation of $A$, then any element $z\in Z(A)$ acts in $V$ by multiplication by some scalar ${\xi }_V(z)$. Show that ${\chi }_V(z):Z\to \mathbf{k}$ is a homomorphism.

16. Let $d\in \mathbb{N}$. Show that the only irreducible representation of ${\mathrm{Mat}}_d(k)$ is $k^d$, and every finite dimensional representation of ${\mathrm{Mat}}_d(k)$ is a direct sum of copies of $k^d$.

\begin{proof} For every $(i,j)\in \{1,2,\dots ,d{\}}^2$, let $E_{ij}\in {\mathrm{Mat}}_d(k)$ be the matrix with $1$ in the $i$ th row of the $j$ th column and $0$‘s everywhere else. Let $V$ be a finite dimensional representation of ${\mathrm{Mat}}_d(k)$.

For any $v\in V$, since $(E_{11}+\cdots +E_{dd})v=v$, we have that $V=E_{11}V+\cdots +E_{dd}V$. Assume that $E_{ii}{v}_i=E_{jj}{v}_j$ for some $i\ne j$. It follows that $0=E_{jj}{E}_{ii}{v}_i=E_{jj}{E}_{jj}{v}_j=E_{jj}{v}_i$ and so $E_{ii}v\cap E_{jj}v=\{0\}$ for any $i\ne j$. Therefore, $V=E_{11}V\oplus E_{22}V\oplus \dots \oplus E_{dd}V$. In addition, it is easy to verify that ${\Phi }_i:E_{11}V\to E_{ii}V,v\mapsto E_{i1}v$ is an isomorphism for every $i\in \{1,2,\dots ,d\}$ by ${\Phi }_i$ invertible.

For every $v\in E_{11}V$, denote $S(v)=⟨E_{11}v,E_{21}v,\dots ,E_{d1}v⟩$. Then $S(v)$ is a sub-representation of $V$ isomorphic to $k^d$. Let $\{v_1,\cdots ,v_k\}$ be a basis of $E_{11}V$, then $V=S\left(v_1\right)\oplus S\left(v_2\right)\oplus \dots \oplus S\left(v_k\right)$ is a direct sum of copies of $k^d$. \end{proof}

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Exercise 1. If $T:V\to W$ is a morphism of two $G$-modules, then $\ker T$ is a submodule of $V$ and $\mathrm{Im}T$ is a submodule of $W$.

\begin{proof} Let $\sigma :G\to \mathrm{GL}(V)$ and $\rho :G\to \mathrm{GL}(W)$ be representations of $G$ in $V$ and $W$, respectively.

i) For any $v\in \ker T$ and any $g\in G$, we have that $T(\sigma (g)v)=\rho (g)(Tv)=\rho (g)(0)=0$. Thus $\sigma (g)v\in \ker T$ for all $g\in G$. Therefore, $\ker T$ is a submodule of $V$.

ii) For any $w\in \mathrm{Im}T$, there exists $u\in V$ such that $w=Tu$. Then for any $g\in G$, $\rho (g)w=\rho (g)(Tu)=T(\sigma (g)u)$. It yields that $\rho (g)w\in \mathrm{Im}T$ for all $g\in G$. Therefore, $\mathrm{Im}T$ is a submodule of $W$. \end{proof}

Exercise 2. Suppose ${\rho }_1$, ${\rho }_2$ are representations of $G$ in $V$ and $W$, respectively. Define ${\rho }_1\oplus {\rho }_2$ as $({\rho }_1\oplus {\rho }_2)(g)=({\rho }_1(g),{\rho }_2(g))\in \mathrm{GL}(V\oplus W)$. Show that ${\rho }_1\oplus {\rho }_2$ is a representation in $V\oplus W$.

\begin{proof} Assume that $\dim V=n$ and $\dim W=m$, then $\dim (V\oplus W)=n+m$. Let ${\mathcal{B}}_1$, ${\mathcal{B}}_2$ be bases of $V$ and $W$, respectively. Then there is a basis $\mathcal{B}$ of $V\oplus W$ such that $[({\rho }_1(g),{\rho }_2(g)){]}_{\mathcal{B}}=\mathrm{diag}([{\rho }_1(g){]}_{{\mathcal{B}}_1},[{\rho }_2(g){]}_{{\mathcal{B}}_2})$. For any $g_1,g_2\in G$,

$$\begin{array}{rl}[({\rho }_1\oplus {\rho }_2)(g_1{g}_2){]}_{\mathcal{B}}& =[({\rho }_1(g_1{g}_2),{\rho }_2(g_1{g}_2)){]}_{\mathcal{B}}\\ & =\mathrm{diag}([{\rho }_1(g_1){\rho }_1(g_2){]}_{{\mathcal{B}}_1},[{\rho }_2(g_1){\rho }_2(g_2){]}_{{\mathcal{B}}_2})\\ & =\mathrm{diag}([{\rho }_1(g_1){]}_{{\mathcal{B}}_1},[{\rho }_2(g_1){]}_{{\mathcal{B}}_2})\mathrm{diag}([{\rho }_1(g_2){]}_{{\mathcal{B}}_1},[{\rho }_2(g_2){]}_{{\mathcal{B}}_2})\\ & =[({\rho }_1(g_1),{\rho }_2(g_1)){]}_{\mathcal{B}}[({\rho }_1(g_2),{\rho }_2(g_2)){]}_{\mathcal{B}}\end{array}$$

and so $({\rho }_1\oplus {\rho }_2)(g_1{g}_2)=({\rho }_1\oplus {\rho }_2)(g_1)({\rho }_1\oplus {\rho }_2)(g_2)$. Also, it is easy to verify $({\rho }_1\oplus {\rho }_2)(e)=({\mathrm{Id}}_V,{\mathrm{Id}}_W)=\mathrm{Id}$. Therefore, ${\rho }_1\oplus {\rho }_2$ is a representation in $V\oplus W$. \end{proof}

Exercise 3. Suppose $V$ is a $G$-module. Let $V^{\ast }={\mathrm{Hom}}_k(V,k)$, and let ${\rho }^{\ast }$ is a representation of $G$ on $V^{\ast }$. For any $f\in {\mathrm{Hom}}_k(V,k)$, define ${\rho }^{\ast }(g)f=f\circ \rho (g^{-1})$. Prove that:

• ${\rho }^{\ast }$ is a representation of $G$ in $V^{\ast }$.
• If $\dim V<\infty$, then $\dim V^{\ast }=\dim V$ and $(V^{\ast }{)}^{\ast }\simeq V$.

\begin{proof} i) For any $g_1,g_2\in G$ and any $f\in V^{\ast }$,

$$\begin{array}{rl}{\rho }^{\ast }(g_1{g}_2)f& =f\circ \rho (g_2^{-1}{g}_1^{-1})=f\circ (\rho (g_2^{-1})\rho (g_1^{-1}))\\ & =f\circ \rho (g_2^{-1})\circ \rho (g_1^{-1})={\rho }^{\ast }(g_2)f\circ \rho (g_1^{-1})\\ & ={\rho }^{\ast }(g_1){\rho }^{\ast }(g_2)f.\end{array}$$

In addition, we have that ${\rho }^{\ast }(e)f=f\circ {\mathrm{Id}}_V=f$ and so ${\rho }^{\ast }(e)={\mathrm{Id}}_{V^{\ast }}$. Therefore, ${\rho }^{\ast }$ is a representation of $G$ in $V^{\ast }$.

ii) Since $\dim V<\infty$, we assume that $\dim V=n$ and $\{e_1,\cdots ,e_n\}$ is a basis of $V$. For $i\in \{1,\cdots ,n\}$, define $f_i\in V^{\ast }$ as $f_i(e_j)={\delta }_{ij}$, where $\delta$ is the Kronecker delta. If there exist $k_1,\cdots ,k_n$ such that ${\sum }_{i=1}^n{k}_i{f}_i=0$, then $0={\sum }_{i=1}^n{k}_i{f}_i(e_j)=k_j$ for all $j$. Thus $k_i$ are all zero and so $f_1,\cdots ,f_n$ are linear independent. Note that for any $f\in V^{\ast }$ with $f(e_i)={\alpha }_i$, there is $f={\sum }_{i=1}^n{\alpha }_i{f}_i$. So $\{f_1,\cdots ,f_n\}$ is a basis of $V^{\ast }$. Therefore, $\dim V^{\ast }=\dim V$.

Define $\widetilde{e_i}\in (V^{\ast }{)}^{\ast }$ as $\widetilde{e_i}(f):=f(e_i)$. It is easy to verify that $T:V\to (V^{\ast }{)}^{\ast },e_i\mapsto \widetilde{e_i}$ is a linear transformation. Since $T$ is bijective, then $T$ is an isomorphism and so $V\simeq (V^{\ast }{)}^{\ast }$. \end{proof}