I am currently reading "Finite dimensional vectors spaces" by Halmos. Could anyone check if my proof of a specific exercise (page 27 exercise 5) is rigorous enough and correct? The topics are annihilator and linear functionals.
Prove that if $m < n$, and if $y_{1}, \dots, y_{m}$ are linear functionals on an n-dimensional vector space $\mathbb{V}$, then there exists a non-zero vector $x$ in $\mathbb{V}$ such that $y_{j}(x) = 0$ for $j = 1, \dots, m$. What does this result say about the solutions of linear equations?
The basic idea here is to use the isomorphism between $\mathbb{V}$ and $\mathbb{V}''$. If $\mathbb{V}$ is a finite-dimensional vector space, then corresponding to every linear functional $z_{0}$ on $\mathbb{V}'$ there is a vector $x_{0}$ in $\mathbb{V}$ such that $z_{0}(y) = y(x_{0})$ for every y in $\mathbb{V}'$. Now we define $S := span\{y_{1}, \dots, y_{m}\} \subset \mathbb{V}'$. $S$ is a subspace of $\mathbb{V}'$. Since $y_{1}, \dots, y_{m}$ are not necessarily linearly independent it follows that $\dim S \leq m < n = \dim \mathbb{V}'$. We describe the annihilator of $S$ with $S^0$ which is a $n-m$-dimensional subspace of $\mathbb{V}''$. Since $m < n$ we know that $S^0$ has at least dimension 1 and this means that $S^0$ contains at least one linear functional which isn't the 0-functional. Using the isomorphism between $\mathbb{V}$ and $\mathbb{V}''$ we pick a vector $z_{0} \in \mathbb{V}''$. The theorem above ensures that there exists a vector $x_{0}$ such that $z_{0}(y) = y(x_{0})$ for every $y \in \mathbb{V}'$. From this it follows that there exists a vector $x_{0}$ such that $z_{0}(y) = y_{j}(x_{0})$ for $j = 1, \dots, m$. Since a linear functional is always surjective we can deduce that $z_{0}(y) = y_{j}(x_{0}) = 0$ for $j = 1, \dots, m$. We can make a more specific statement in $x_{0} \in \cap_{j=1}^m \ker y_{j}$.
The result tells us that given $m$ linear equations with the right side of the equations being 0, we can find a non-trivial solution.
Also the next exercise is quite similar but I am not sure how to start it. Maybe someone has a hint for me?
Suppose that $m < n$, and that $y_{1}, \dots, y_{m}$ are linear functionals on an n-dimensional vector space $\mathbb{V}$. Under what conditions on the scalars $\alpha_{1}, \dots, \alpha_{m}$ is it true that there exists a vector $x$ in $\mathbb{V}$ such that $y_{j}(x) = \alpha_{j}$ for $j = 1, \dots, m$. What does this result say about the solutions of linear equations?
Thanks for any help!