There exist the famous theorem about a basis for dual space

Let $\mathbb V$ be finite dimensional vector space over $F$ and $\mathcal{B} = \{\alpha_1, \ldots ,\alpha_n\}$ is basis for vector space $\mathbb V$ then $\mathcal{B^*} = \{f_1, \ldots ,f_n\}$ is basis for dual space $\mathbb V^*$ such that $f_i(\alpha_{j})=\delta_{ij}$

Is the converse of this Theorem true, to explain more:

Let $\mathcal{B^*} = \{f_1, \ldots ,f_n\}$ be basis for dual space $\mathbb V^*$. Does there exist $\mathcal{B} = \{\alpha_1, \ldots ,\alpha_n\}$ such that $\mathcal{B}$ would be basis for vector space and $f_i(\alpha_{j})=\delta_{ij}$?

  • 5
    $\begingroup$ Since $\Bbb{V}$ is finite-dimensional, there's a natural isomorphism between $\Bbb{V}$ and $\Bbb{V}^{**}$. You can get the converse by applying the original theorem to $\Bbb{V}^*$ and then using that isomorphism. $\endgroup$
    – Micah
    Jul 9, 2012 at 17:54

1 Answer 1


Yes, you can prove that via the bidual space $V^{**} = \mathrm{Hom}(V^*,K)$. There is a canonical homomorphism $\iota: V \to V^{**}$ defined by $\iota(v)(f) := f(v)$ for $f \in V^{*}$. It is easy to see that $\iota$ is injective. Since $\dim V^{**} = \dim V^* = \dim V$, it is actually an isomorphism. Therefore, for every $\varphi \in V^{**}$ there exists a $v \in V$ s.t. $\varphi(f) = f(v)$ for $f \in V^*$.

Now, if $\{f_1,\ldots,f_n\}$ is a basis if $V^*$ then the elements $\{\varphi_1,\ldots,\varphi_n\}$ form a basis of $V^{**}$ where $\varphi_i(f_j) := \delta_{ij}$. By the above considerations, each $\varphi_i$ corresponds to an $\alpha_i \in V$ s.t. $f_i(\alpha_j) = \varphi_j(f_i) = \delta_{ij}$. Since $\iota: V\to V^{**}$ is an isomorphism and $\{\varphi_1,\ldots,\varphi_n\}$ is a basis of $V^{**}$ we know that $\{\alpha_1,\ldots,\alpha_n\}$ is a basis of $V$. Its dual basis is then $\{f_1,\ldots,f_n\}$ because $f_i(\alpha_j) = \delta_{ij}$.


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