1
$\begingroup$

I have a question about the proof of the following result (from Aliprantis and Border). Recall that $ker f = \{x \in X: f(x)=0\}$.

Theorem. Let $f_1,...,f_n$ be linear functionals on a vector space $X$. Then $f$ lies in the span of $f_1,..., f_n$ (that is, $f = \sum_{i=1}^n \lambda_if_i$ for some scalars $\lambda_1,...,\lambda_n$) if and only if $\bigcap_{i=1}^n ker f_i \subset ker f$.

Proof. If $f = \sum^n_{i=1} \lambda_i f_i$, then clearly $\bigcap^n_{i=1} ker f_i \subset ker f$. For the converse, assume that$\bigcap^n_{i=1}ker f_i \subset ker f$. Then $T:X\to \mathbb R^n$ via $T(x)= (f_1(x),...,f_n(x))$ is a linear operator. Since$\bigcap^n_{i=1}ker f_i \subset ker f$, if $(f_1(x),..., f_n(x)) = (f_1(y),..., f_n(y))$, then $f_i(x − y) = 0$ for each $i$ and so $f(x) = f(y)$. Thus the linear functional $\phi:T(X) \to \mathbb R$ defined by $\phi(f_1(x),...,f_n(x) = f(x)$ is well defined. Now note that $\phi$ extends to all of $\mathbb R^n$, so there exist scalars $\lambda_1,...,\lambda_n$ such that $\phi(\alpha_1,...,\alpha_n) = \sum^n_{i=1} λ_i\alpha_i$. Thus $f(x) = \sum^n_{i=1} \lambda_i f_i(x)$ for each $x \in X$, as desired.

I'm having trouble seeing why $\phi$ can be extended to all of $\mathbb R^n$. The way the proof is written, it seems like this step should be obvious, so I guess I'm missing something easy. Any hints would be appreciated.

$\endgroup$
2

1 Answer 1

Reset to default
1
$\begingroup$

In finite dimension, every subspace has a complement. Every linear map defined on a subspace can be extended to the whole space by defining the extended map as $0$ on such a complement subspace.

$\endgroup$
7
  • $\begingroup$ What do you mean by "complement"? $\endgroup$
    – aduh
    Commented Dec 14, 2019 at 18:58
  • $\begingroup$ If $U$ is a subspace of $E$, I mean a subspace $V$ such that $U \oplus V = E$. $\endgroup$
    – Gribouillis
    Commented Dec 14, 2019 at 18:59
  • $\begingroup$ It would be helpful if you could add more details or provide a reference, so I can see how the finite dimensionality is needed for this argument. $\endgroup$
    – aduh
    Commented Dec 14, 2019 at 18:59
  • 1
    $\begingroup$ The existence of a complement does not need the vector space to be finite-dimensional. If $U$ is a subspace of $E$, it has a basis $B$. This can be extended to a basis $B'$ of $E$, and then the linear span of $B' \setminus B$ is a complement to $U$. $\endgroup$
    – spin
    Commented Dec 14, 2019 at 19:04
  • 1
    $\begingroup$ @grndl: This is explained in the answer. To define a linear map it suffices to define it on a basis and then extend linearly. So just define the extension of $\phi$ by $\phi'(x) = \phi(x)$ for $x$ in $T(X)$, and $\phi'(x)$ can be whatever you want for basis elements $x$ outside of $T(X)$, for example zero. $\endgroup$
    – spin
    Commented Dec 14, 2019 at 19:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .