Erwin Kreyszig's Introductory Functional Analysis With Applications, Section 2.9, Problem 12

Question

If $f_1, \ldots, f_p$ are linear functionals on an $n$-dimensional vector space $X$, where $p<n$, then how to show that there is a vector $x \ne 0$ in $X$ such that $f_1(x) = 0, \ldots, f_p(x)=0$? What consequences does this result have with respect to linear equations?

My feeling is that one of the consequences of this result with respect to linear equations is the following fact:

Any system of $p$ homogeneous linear equations in $n$ unknowns, where $p < n$, must have a non-trivial solution.

Am I right?

And if so, then what other consequences does our result entail?

Answer 1

Find an $x=\sum_{i=1}^n\alpha_ie_i\not=0$ s.t. $f_1(x)=...=f_p(x)=0$.

This results in the following system of $p$ equalities in $n$ unknowns:

$$f_1(x)=\sum_{i=1}^n\alpha_i f_1(e_i)=0$$ $$\vdots$$ $$f_p(x)=\sum_{i=1}^n\alpha_i f_p(e_i)=0$$

Such that we know from linear algebra that this system must have a non trivial solution (since $p<n$), and so there must be an $x\not=0$ as desired.

Answer 2

Suppose that there does not exist such $x$. Then the mapping $\phi : X\to \mathbb R^p$ defined by $$\phi (x)=(f_1(x),\cdots,f_p (x))$$ is injective. Therefore, $dim X\leq dim \mathbb R^p$ or $n\leq p$. A contradiction.