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Two Hamel bases of the same vector space have the same cardinality, so we define the dimension of a vector space as the cardinality of one of its Hamel basis.

Let $E$ be an infinite-dimensional Banach space. Let $G$ be a closed subspace of $E$. A subset $L$ of $E$ is said to be a (topological) complement of $G$ if $L$ is a closed subspace of $E$ such that $G\cap L = \{0\}$ and $G+L=E$.

I would like to confirm that

If $G$ has finite codimension, then $G$ has a complement.

Could you have a check on my below understanding?

Let $n := \operatorname{codim} G < \infty$. By rank-nullity theorem, $n + \dim G = \dim E$. Let $(e_i)_{i\in I}$ be a Hamel basis of $G$. Then there is a linearly independent set $(e'_i)_{i=1}^n$ such that $(e_i)_{i\in I} \cup (e'_i)_{i=1}^n$ is a Hamel basis of $E$. Let $L := \operatorname{span} (\{e'_1, \ldots, e'_n\})$. Then $L$ is the required complement of $G$.

Analyst
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    Looks fine to me. You should add that a finite dimensional subspace of $E$ is closed due to the fact that all norms on a finite dimensional space are equivalent. Hence $L$ is complete, therefore closed. – Ryszard Szwarc May 20 '23 at 15:55
  • @RyszardSzwarc Thank you so much for your verification! – Analyst May 20 '23 at 16:18
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    Using the rank nullity theorem when dimensions are infinite tends to require care... – Mariano Suárez-Álvarez May 20 '23 at 17:16
  • @MarianoSuárez-Álvarez I see. In my below answer, I don't mention cardinality anymore. Do you think it's fine now? – Analyst May 20 '23 at 18:28
  • I understand that the codimension of a subspace $L$ of $E$ is the dimension of the quotient space $E/L.$ This space is isomorphic to $G,$ where $G$ is a complement space. Since $\dim G +\dim L=\dim E,$ we get that ${\rm codim} ,L+\dim L=\dim E.$ – Ryszard Szwarc May 20 '23 at 20:06
  • @RyszardSzwarc I'm sorry, but could you explain the purpose of your previous comment? – Analyst May 20 '23 at 20:13
  • There was a comment of yours asking me to take a look at some related post. I do not see this comment anymore. So I will delete my previous comment. – Ryszard Szwarc May 20 '23 at 20:55
  • @RyszardSzwarc Ah Thank you so much for your elaboration! – Analyst May 20 '23 at 20:56
  • @RyszardSzwarc I have a recent question about Fredholm alternative. If you don't mind, please have a look at it. Thank you so much! – Analyst May 22 '23 at 08:37

1 Answers1

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Consider the quotient map $J:E \to E/G$. Let $B_1$ be a basis of $G$. Let $B_2$ be a basis of $E/G$. By rank-nullity theorem, $B_1 \cup J^{-1} (B_2)$ is a basis of $E$. Notice that $B_1 \cap J^{-1} (B_2) = \emptyset$. Then $L := \operatorname{span} (J^{-1} (B_2))$ is the required complement of $G$.

Analyst
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