# linear functionals on a real vector space

Let $X$ be a real vector space. Let $\varphi_1,\dots, \varphi_n$ and $\psi$ be linear maps from $X$ to $\mathbb{R}$. The following are equivalent:

1. There exist $\lambda_1, \dots, \lambda_n \in \mathbb{R}$ such that $\psi = \sum_{i=1}^n \lambda_i \varphi_i$.

2. $\bigcap_{i=1}^n \ker{\varphi_i} \subset \ker{\psi}$.

It is clear that 1) implies 2). But not at all that 2) implies 1) (at least to me). Here is what I tried:

The statement is easy if $n=1$: Indeed if $\varphi_1 = 0$, set $\lambda_1 = 0$, and otherwise pick $x_1 \notin \ker \varphi_1$ and set $\lambda_1 = \psi(x_1)/\varphi(x_1)$. Because then, since an arbritrary $x \in X$ can be written as $x = \alpha_1 x_1 + v$ with $\alpha_1 \in \mathbb{R}$ and $v \in \ker \varphi_1$, we get $$\lambda_1 \varphi_1(x) = \frac{\psi(x_1)}{\varphi(x_1)} \varphi_1(\alpha_1 x_1 + v) = \psi(\alpha_1 x_1) = \psi(\alpha_1 x_1 + v) = \psi(x).$$

Now let $n\geq 2$ and suppose the implication $2) \rightarrow 1)$ holds for $n-1$. If, for some index $i \in \lbrace 1, \dots ,n \rbrace$, the set $S_i:=\bigcap_{j\neq i} \ker\varphi_j \setminus \ker \varphi_i$ was empty, then $\bigcap_{j=1}^n \ker \varphi_j = \bigcap_{j\neq i} \ker\varphi_j$ and by induction we could express $\psi$ as a linear combination of the $\varphi_j$'s with $j\neq i$. Thus, suppose $S_i \neq \emptyset$ for all $i$ and pick $x_i \in S_i$. Now by plugging the $x_i's$ into the equation for $\psi$ we're after, we find $\lambda_i = \psi(x_i)/\varphi_i(x_i)$. So it should be possible to prove that we have $\psi = \sum_{i=1}^n {\frac{\psi(x_i)}{\varphi_i(x_i)}}\varphi_i$, right?

But how? I couldn't do it. Is there a mistake in my reasonning? I used the direct sum decompositions $X = \mathbb{R}x_i \oplus \ker\varphi_i$ to write an arbitrary $x \in X$ in $n$ different ways and to get rid of the denomintors $\varphi_i(x_i)$ but then I am left with $\psi(\alpha_1 x_1 + \dots + \alpha_n x_n)$ which should equal $\psi(x)$.

This is fine. Write $x\in X$ as $\sum a_ix_i+x'$ with $\varphi_i(x')=0$ for every $i$. Then $\psi(x)=\sum a_i\psi(x_i)$ and $(\sum \psi(x_i)/\varphi_i(x_i)\varphi_i)(x)=\sum a_i \psi(x_i)/\varphi_i(x_i)\varphi_i(x_i)=\sum a_i\psi(x_i)$, as desired.
• Yeah, sure! Or just use that the $x_i$ are linearly independent since they're all in the kernels of different sets of functionals, so you could choose a basis including them. – Kevin Carlson Oct 29 '15 at 22:28