# Linear functionals form a basis for the dual space if and only if they perform an isomorphism

Actually I understand what dual space is basically about, but still I wasn't able to prove the following: Prove that linear functionals $l_1, ..., l_n$ form a basis for the dual space $V^*$ if and only if the linear transformation $$L(\vec x) = (l_1(\vec x), \cdots , l_n(x))^T :V \rightarrow \mathbb R^n$$ is an isomorphism.

Many thanks! It may sounds like a stupid question but for sure it will be a great step for me!

## 1 Answer

I have a very hands-on approach which involves turning things into matrices. First off, we might as well assume $V$ has dimension $n$; otherwise the equivalence holds in a trivial way. Further, we might as well assume that $V=\mathbb{R}^n$ (solving the problem for $V=\mathbb{R}^n$ should be enough since results can be carried through the linear isomorphism $V\cong \mathbb{R}^n$). Let $\vec{e}_1,\ldots,\vec{e}_n$ be the standard basis.

Now our linear functionals $l_1,\ldots,l_n:\mathbb{R}^n\rightarrow \mathbb{R}$ are really row vectors $\vec{l}_1^T,\ldots, \vec{l}_n^T$. (This is because there is for every functional $l:\mathbb{R}^n\rightarrow\mathbb{R}$ a vector $\vec{l}$, namely the one with $i^\text{th}$ entry $l(\vec{e}_i)$, such that $l(\vec{v})=\vec{l}^T\vec{v}$). The linear map $L:\mathbb{R}^n\rightarrow\mathbb{R}^n$ that you define in your question is represented in the standard basis by the matrix whose rows are $\vec{l}_1^T,\ldots,\vec{l}_n^T$. The proof is now crystal clear: This matrix is invertible if and only if the functionals $l_1,\ldots,l_n$ were linearly independent to begin with.