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The rank nullity theorem states that for vector spaces $V$ and $W$ with $V$ finite dimensional, and $T: V \to W$ a linear map,

$$\dim V = \dim \ker T + \dim \operatorname{im} T.$$

Does this hold for infinite dimensional $V$? According to this, the statement is false. But according to this, page 4, the statement is still true. I'm thoroughly confused.

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    $\begingroup$ Where do you deduce from that in your second link, page 4, it is said the theorem remains as it was in the infinite dimensional case?? $\endgroup$
    – DonAntonio
    Apr 13 '14 at 17:22
  • $\begingroup$ @DonAntonio It says that if $V$ is infinite dimensional, one of $N(T)$ and $T(V)$ is infinite dimensional, in which case the equation still holds. $\endgroup$
    – user142870
    Apr 13 '14 at 17:24
  • $\begingroup$ Where does it say "the equation still holds", @user142870 ? $\endgroup$
    – DonAntonio
    Apr 13 '14 at 17:28
  • $\begingroup$ @DonAntonio If $\dim V = \dim ker T + \dim im T$, in the infininte dimensional case, we have $\infty = \dim N(T) + \dim T(V)$. One of $N(T)$ and $T(V)$ is infinite dimensional, so that $\infty = \infty + n$ where $n \in \mathbb{N} \cup \{\infty\}$. Can you explain what's wrong with my reasoning? $\endgroup$
    – user142870
    Apr 13 '14 at 17:30
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    $\begingroup$ Ahh..very frustrating question. The fact that two reputable users give conflicting answers does not help either. $\endgroup$
    – user142870
    Apr 13 '14 at 17:39
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The rank formula also holds in infinite dimensions, whether you use cardinal arithmetic for the dimensions, or just say $\infty + n = \infty$, and $\infty + \infty = \infty$ (but one should use cardinal arithmetic). The proof is basically the same as in the finite-dimensional case, you choose a basis $\mathcal{B}_1$ of $\ker T$, a basis $\mathcal{B}_2$ of $\operatorname{im} T$, let $\mathcal{B}_3$ consist of preimages of the elements of $\mathcal{B}_2$ (choose one preimage per element), then $\mathcal{B}_1 \cup \mathcal{B}_3$ is a basis of $V$. In the infinite-dimensional case, some form of the axiom of choice is required, while the finite-dimensional case can be proved without that.

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  • $\begingroup$ Can you then explain what the first link is saying, when it asks to give a counterexample for when $V$ is infinite dimensional. $\endgroup$
    – user142870
    Apr 13 '14 at 17:27
  • $\begingroup$ Hmmm...some problems may arise with the interpretation of "it is true", @Daniel. Read my answer to see what I mean. $\endgroup$
    – DonAntonio
    Apr 13 '14 at 17:27
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    $\begingroup$ @DonAntonio Actually, I don't understand what you mean. The equivalence of injectivity and surjectivity of endomorphisms is of course a specialty of the finite-dimensional situation. But the rank formula just says $$\dim V = \dim \ker T + \dim \operatorname{im} T,$$ it doesn't say anything about surjectivity. $\endgroup$ Apr 13 '14 at 17:30
  • $\begingroup$ @DanielFischer, it says that $\;\ker T=\{0\}\iff \text{Im}T=V\;$ ... $\endgroup$
    – DonAntonio
    Apr 13 '14 at 17:33
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    $\begingroup$ @DonAntonio Since the rank-nullity theorem (Gah, who thought up such a disgusting un-word), let's correctly call it the rank formula, is concerned with linear maps between (possibly) different spaces, it cannot say such a thing. A certain corollary of it, which is valid only in the finite-dimensional case, links injectivity and surjectivity. $\endgroup$ Apr 13 '14 at 17:39
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In the finite case, for an operator $\;T:V\to V\;$ , the dimension theorem says that

$$T\;\;\text{is surjective}\;\implies \dim\ker T=0\iff \ker T=\{0\}$$

But if we define

$$V:=\left\{\{x_n\}_{n\in\Bbb N}\subset\Bbb R\right\}\;,\;\;T:V\to V\;,\;\;T\{x_1,x_2,...\}:=\{x_2,x_3,...\}$$

then clearly $\;T\;$ is onto, yet

$$\ker T:=\left\{\{x_n\}\in V\;;\;x_i=0\;\;\forall\,i\ge 2\right\}\neq\{0\}$$

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  • $\begingroup$ What you have have shown is that there exists onto linear maps between infinite dimensional spaces that are not one-to-one. This was not my question. $\endgroup$
    – user142870
    Apr 13 '14 at 17:29
  • $\begingroup$ No @user142870, I think I've shown that one straighforward deduction from the dim. theorem in the finite dim. case is not true in the infinite dimensional one...and also what you say, of course. $\endgroup$
    – DonAntonio
    Apr 13 '14 at 17:32
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    $\begingroup$ You are using your own version of the dimension theorem to explain why it is false. Again, please look at my statement of the rank-nullity theorem that I wrote down. $\endgroup$
    – user142870
    Apr 13 '14 at 17:36
  • $\begingroup$ Well, and again we come back to the interpretation thing, as I wrote below Daniel's answer. This may well have been what they meant in that link of yours (I can't tell for sure, of course). Yet, yes: with your interpretation then it seems to be the theorem remains true. $\endgroup$
    – DonAntonio
    Apr 13 '14 at 17:36
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    $\begingroup$ @DonAntonio In the light of Daniel Fischer's answer, the problem here is simply that $\alpha + \beta = \alpha \Rightarrow \beta = 0$ holds only for finite cardinals, not for infinite ones. In your particular case, it comes down to $\textrm{dim ker } T + \textrm{dim im } T = 1 + \aleph_0 = \aleph_0 = \textrm{dim } V$. $\endgroup$
    – fgp
    Apr 13 '14 at 17:43
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Indeed there might be some problem, as $\dim V$ is undefined if $V$ is not finite dimensional. Unless one wants to identify $\dim V$ with the cardinality of any of its bases, that is. But then one should also define algebraic operations with cardinalities to make sense of the term $\dim \ker T + \dim \text{im} T$, and this is complicated. (I don't even know if it is possible in some meaningful way).

Instead, in the general case (both finite and infinite dimensional) you can use the following result: the map $\tilde T$, defined by $$ \tilde{T}(v+\ker T):=T(v), $$ is a vector space isomorphism of $V/\ker T$ onto $\text{im}(T)$. In the finite dimensional case, the rank-nullity theorem is an immediate corollary.

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    $\begingroup$ It is possible: there is Cardinal Arithmetic. The sum is simply defined as cardinality of the union of two sets of the cardinalities which are to be added. $\endgroup$
    – J.R.
    Apr 13 '14 at 17:29
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    $\begingroup$ @YourAdHere That definition is true in spirit, but a bit troublesome for finite cardinals. If you construct the cardinals as specific von-Neumann ordinals, you have $\mathbb{n} \subset \mathbb{n+1}$ (where $\mathbb{n}$ is the $n$-th finite cardinal), and thus $\mathbb{n} + (\mathbb{n+1}) = \mathbb{n+1}$. But that isn't what you want... $\endgroup$
    – fgp
    Apr 13 '14 at 17:54
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    $\begingroup$ @fgp: what strange kind of cardinal arithmetic is leading you to that equality? $\endgroup$ Sep 7 '18 at 13:07
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    $\begingroup$ @darijgrinberg Taking "The sum is simply defined as cardinality of the union of two sets of the cardinalities which are to be added." from the first comment literally leads to the strange cardinal arithmetic in fgp's comment. They hinted at the need to consider the disjoint union to get the right arithmetic for finite cardinals. $\endgroup$ Sep 7 '18 at 19:50
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    $\begingroup$ It's one of the fundamental isomorphism theorem in algebra. Maybe you have seen it for groups. $\endgroup$ Nov 10 '20 at 11:59

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