Let $f_1,f_2,\ldots, f_n$ be linear functionals on a vector space $X$. Show that there exist constants $\lambda_1,\ldots,\lambda_n$ satisfying $$f=\sum_{i=1}^n\lambda_i f_i$$ if and only if $\bigcap_{i=1}^n \ker f_i \subset \ker f$.


If there exist constants $\lambda_1,\ldots,\lambda_n$ satisfying $$f=\sum_{i=1}^n\lambda_i f_i,$$ then $x\in \bigcap_{i=1}^n \ker f_i$ gives $\sum_{i=1}^n\lambda_i f_i=0$. On the other hand, define $$V=\{y\in\mathbb{R}^n:\exists x\in X \textrm{ such that } y=\left(f_1(x),\ldots,f_n(x)\right)\},$$ and $g:V \rightarrow \mathbb{R}$ by $$g\left(\left(f_1(x),\ldots,f_n(x)\right)\right)=f(x).$$ $V$ is seen to be a vector subspace of $\mathbb{R}^n$. If $\left(f_1(x),\ldots,f_n(x)\right)=\left(f_1(y),\ldots,f_n(y)\right)$ Then $f_i(x-y)=0$ so $(x-y) \in \ker f_i$ for all $i$. By assumption, $(x-y)\in \ker f$ so $f(x)=f(y)$. Hence $g$ is well defined. Clearly $g$ is linear. Denote by $g^*$ a linear extention of $g$ to all of $\mathbb{R}^n$. Then there exist $\lambda_1,\ldots,\lambda_n$ such that $g^*(z_1,\ldots,z_n)=\sum_{i=1}^n\lambda_i z_i$. $\ ^{(1)}$ In particular, $(z_1,\ldots,z_n)=(f_1(x),\ldots,f_n(x))$ give the desired result.


I am having trouble justifying $(1)$. That is, why can we say $g^*(z_1,\ldots z_n)=\sum_{i=1}^n \lambda_i z_i$?

  • $\begingroup$ Isn't $g : V \rightarrow \mathbb{R}$, not $\mathbb{R}^n$? $\endgroup$ Commented Jul 18, 2015 at 23:42
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    $\begingroup$ This can also be done by induction. In fact, the meat of the proof is really just proving that, if $\mathrm{ker}(f) \cap \mathrm{ker}(g) \subseteq \mathrm{ker}(h)$, then $h = af + bg$ for some $a, b$. This isn't too hard to prove. $\endgroup$ Commented Jul 18, 2015 at 23:45
  • $\begingroup$ @Theo_Bendit yes thank you for pointing that out... I just made the appropriate edit $\endgroup$
    – recmath
    Commented Jul 18, 2015 at 23:45
  • $\begingroup$ It might also be worth pointing out why $g$ is well-defined. $\endgroup$ Commented Jul 18, 2015 at 23:47
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    $\begingroup$ I don't know if you're still interested, but you need to find an $x \in \mathrm{ker}(f) \setminus \mathrm{ker}(g)$ and $y \in \mathrm{ker}(g) \setminus \mathrm{ker}(f)$ (also consider the case where you can't find such both these things, i.e. where the kernels are equal). Then, you can assume without loss of generality that $g(x) = f(y) = 1$. You should be able to show then that $a = h(y)$ and $b = h(x)$ make $h = af + bg$ true. You will need to come up with a decomposition of vectors in $X$ into a sum of a multiple of $x$ plus a multiple of $y$ plus a vector in the intersection of the kernels $\endgroup$ Commented Jul 20, 2015 at 5:04

1 Answer 1


In answer to your actual question, you can say (1) is true, by virtue of the fact that every linear functional on $\mathbb{R}^n$ has that form. You can prove this by considering $g^*$ on the standard basis vectors $e_i$, and let $\lambda_i = g^*(e_i)$ for all $i$. Then linearity gives you what you need.

  • $\begingroup$ oh boy..I really should've seen that..thank you! $\endgroup$
    – recmath
    Commented Jul 19, 2015 at 0:24

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