On expressing a subspace as intersection of kernels of linearly independent functionals

dual-spacesfunctional-analysislinear algebralinear-transformations

I tried to prove the following proposition

Let $X$ be a vector space over field $K$ of dimension $n<\infty$ and $X^*$ its algebraic dual. A subset $E\subseteq X$ is a subspace of dimension $p$ iff there exist linearly independent functionals $f_1,\ldots,f_{n-p}\in X^*$ such that $E=\bigcap_{i=1}^{n-p}f_i^{-1}(0)$.

I can prove $\Rightarrow$ by 1) taking complement dual basis corresponding to basis elements not in $E$, and 2) constructing an map $f$ from $X$ to $K^{n-p}$ as $f(x)=\bigl(f_1(x), \ldots\,f_{n-p}(x)\bigr)^T$ so $E$ is the kernel of $f$. But I was not able to proceed with $\Leftarrow$. For example, I tried to find $x_1,\ldots,x_{n-p}\in X$ such that they are independent and $f_j(x_i)=\delta_{ij}$ from independence of $f_1,\ldots,f_{n-p}$, but failed. So, can you please help me prove $\Leftarrow$ (you don't have to follow my unsuccessful attempt)? Thank you.

Best Answer

First, find $x_1,\ldots,x_{n-p}\in X$ such that $f_j(x_i)=\delta_{ij}$. Applying the same construction of complementary basis$^*$ to independent $f_1,\ldots,f_{n-p}$ in dual space $X^*$, we got $g_1,\ldots,g_{n-p}$ which satisfies $g_j(f_i)=\delta_{ij}, i,j=1..n-p$. Taking $x_1,\ldots,x_{n-p}\in X$ to be the inverse images of the canonical mapping from $X$ to $X^{**}$, we have $f_j(x_i)=g_j(f_i)=\delta_{ij}$ from the definition of $g_j$'s.

Second, define $f$ from $X$ to $K^{n-p}$ as $f(x)=\bigl(f_1(x), \ldots\,f_{n-p}(x)\bigr)^T$, which is a linear transformation. We have, from the above construction, $$f(x_1)=(1,0,\ldots,0)^T\\ f(x_2)=(0,1,\ldots,0)^T\\ \dots\\ f(x_{n-p})=(0,0,\ldots,1)^T.$$ So, $f$ is surjective because $f\bigl(\sum\limits_{i=1}^{n-p}a_ix_i\bigr)=(a_1,a_2,\ldots,a_{n-p})$ for any $a_i\in K$. It can be seen that $E$ is the kernel of $f$, thus it is a subspace. From the rank-nullity theorem, ${\rm{rank\ of\ }}f+{\rm{nullity\ of\ }}f=n$ where ${\rm{rank\ of\ }}f=\dim K^{n-p}=n-p$, we get $\dim E={\rm{nullity\ of\ }}f=p$.

*: See, for example, Nathan Jacobson, "Lectures in Abstract Algebra, II. Linear Algebra", P53, or Erwin Kreyszig, "Introductory Functional Analysis with Applications", P114.