Let $M$ be a non-empty subset of a normed space $X$, and let $M^a$ denote the subspace of the dual space $X'$ that consists of all those bounded linear functionals that vanish at each point of set $M$.

Now here's Prob. 14, Sec. 2.10 in Introductory Functional Analysis With Applications by Erwin Kreyszig:

If $M$ is an $m$-dimensional subspace of an $n$-dimensional normed space $X$, show that $M^a$ is an ($n-m$)-dimensional subspace of $X'$. Formulate this as a theorem about solutions of a system of linear equations.

My effort:

Since $X$ is finite-dimensional, every linear functional on $X$ is bounded; so we can write $X' = X^*$.

Let $\{e_1, \ldots, e_m \}$ be a basis for $M$; extend it to a basis $\{ e_1, \ldots, e_m, \ldots, e_n \}$ for $X$. Then each element of $X$ has a unique representation as a linear combination of the $e_j$s.

Suppose that $x \in X$ has the unique representation $$x = \sum_{j=1}^n \xi_j e_j.$$

Then, for any $f \in X'$, we have $$f(x) = \sum_{j=1}^n \xi_j f(e_j).$$ Now, for each $j= 1, \ldots, n$, let $f_j \in X'$ be deined as $$f_j(x) = \xi_j.$$ Then we can write $$f(x) = \sum_{j=1}^n f(e_j) f_j(x).$$ So $$f = \sum_{j=1}^n \alpha_j f_j, \ \mbox{ where } \ \alpha_j \colon= f(e_j) \ \mbox{ for each } \ j= 1, \ldots, n.$$ It can also be shown that the set $\{ f_1, \ldots, f_n \}$ is linearly independent and therefore a basis for $X^* = X'$.

Now suppose that $f \in M^a$. Then $$ f(e_j) = 0 \ \mbox{ for each } \ j= 1, \ldots, m.$$ So $$f(x) = \sum_{j=m+1}^n \xi_j f(e_j) = \sum_{j=m+1}^n \alpha_j f_j(x).$$ Thus each $f \in M^a$ can be written as $$f = \sum_{j=m+1}^n \alpha_j f_j.$$ Moreover, the set $\{f_{m+1}, \ldots, f_n \}$, being a subset of a linearly independent set, is also linearly independent and hence forms a basis for $M^a$. So $M^a$ has dimention $n-m$.

Is the above proof correct?

Now it is my feeling that this result yields the following result:

Let $m < n$. Then any system of $m$ independent homogeneous simultaneous linear equations in $n$ unknowns (with real or complex numbers as co-efficients) has $n-m$ linearly independent solutions.

Is my conclusion correct?

But I'm not exactly sure how to relate the above formulation to this conclusion.

  • $\begingroup$ Hi Saaqib! For future reference, there is no second "e" in Erwin. $\endgroup$ – Ben Grossmann Jun 1 '15 at 17:34
  • $\begingroup$ You may be able to shorten your proof if you assume the rank-nullity theorem (noting, as you did, that the spaces involved are finite-dimensional). $\endgroup$ – Ben Grossmann Jun 1 '15 at 17:38
  • $\begingroup$ Your proof is correct! $\endgroup$ – Ben Grossmann Jun 1 '15 at 17:40
  • $\begingroup$ Hello, @Omnomnomnom, how're you? Thanks for the correction. Can you please write out for me that shortened proof? $\endgroup$ – Saaqib Mahmood Jun 1 '15 at 18:00
  • $\begingroup$ On second thought, I think that the shorter proof is too short, and assumes something essentially equivalent to the desired result. $\endgroup$ – Ben Grossmann Jun 1 '15 at 18:09

Your proof is correct. Well done!

As for the other result, here's one way to think about it:

Take $e_j$ to be the standard basis of $\Bbb R^n$, and take $f_j$ to be the corresponding dual basis. The system of equations $$ a_{11} x_1 + \cdots + a_{1n} x_n = 0\\ a_{21} x_1 + \cdots + a_{2n} x_n = 0\\ \vdots \\ a_{m1} x_1 + \cdots + a_{mn} x_n = 0 $$ can be rewritten as $$ (a_{11}f_1 + \cdots + a_{1n}f_n) x = 0\\ (a_{21}f_1 + \cdots + a_{2n}f_n) x = 0\\ \vdots\\ (a_{m1}f_1 + \cdots + a_{mn}f_n) x = 0 $$ That is, the solution set of the system of equation is the common zero set of the linearly independent set of functionals $\sum_{j=1}^n a_{ij}f_j$ where $i = 1,\dots,m$. That is, if $M$ denotes the solution set to the homogeneous system of equations, we have $$ M^a = \text{span}\left\{\sum_{j=1}^n a_{1j}f_j,\dots,\sum_{j=1}^n a_{mj}f_j\right\} $$ By your result, we may conclude $\dim(M) = n-m$, as desired.

  • $\begingroup$ Well, @Omnomnomnom, but you haven't specified any $M$ in your answer. $\endgroup$ – Saaqib Mahmood Jun 1 '15 at 17:58
  • $\begingroup$ I'll add that in. $M$ is the solution space to the system of equations, which is what we wanted to characterize. $\endgroup$ – Ben Grossmann Jun 1 '15 at 18:07

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