38

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.

Jendrik Stelzner
  • 17,081
user142870
  • 389
  • 1
    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?? – DonAntonio Apr 13 '14 at 17:22
  • @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. – user142870 Apr 13 '14 at 17:24
  • Where does it say "the equation still holds", @user142870 ? – DonAntonio Apr 13 '14 at 17:28
  • 1
    @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? – user142870 Apr 13 '14 at 17:30
  • you're aware of the fact there are several kinds of infinity, right? So it could also be $;\infty=\infty+\infty;$ ... – DonAntonio Apr 13 '14 at 17:35
  • 3
    Ahh..very frustrating question. The fact that two reputable users give conflicting answers does not help either. – user142870 Apr 13 '14 at 17:39
  • Well @user, I tried to address the conflict between the two links you put. In last analysis, and relying on what you wrote, I must say I'd agree with Daniel: as that, the theorem remains true in infinite dimension. Also think of the fact that we use that theorem many times to deduce stuff about the dimension of either the image or the kernel of a matrix/operator, which could render compeltely useless in the infinite dimensional case. – DonAntonio Apr 13 '14 at 17:41

4 Answers4

37

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.

Daniel Fischer
  • 211,575
  • Can you then explain what the first link is saying, when it asks to give a counterexample for when $V$ is infinite dimensional. – user142870 Apr 13 '14 at 17:27
  • Hmmm...some problems may arise with the interpretation of "it is true", @Daniel. Read my answer to see what I mean. – DonAntonio Apr 13 '14 at 17:27
  • 7
    @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. – Daniel Fischer Apr 13 '14 at 17:30
  • @DanielFischer, it says that $;\ker T={0}\iff \text{Im}T=V;$ ... – DonAntonio Apr 13 '14 at 17:33
  • ...and that's what I meant when I wrote about the interpretation thing. – DonAntonio Apr 13 '14 at 17:34
  • 4
    @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. – Daniel Fischer Apr 13 '14 at 17:39
  • The rank-nullity name is terrible indeed, and it reminds me directly of matrices, which are a rather finite dimensional notion (at least in the first courses of linear algebra). As I wrote below my answer, in a most purely (?) and naive understanding of things, I agree with your answer. – DonAntonio Apr 13 '14 at 17:44
  • So, rank-nullity theorem still holds in infinite-dimensional cases. However, one-to-one suggests onto while onto doesn’t suggests one-to-one due to cardinal arithmetic when considering infinite-dimensional cases. Here one-to-one is equivalent with nullity(T) = 0 or Ker(T) = {0} – gpanda Mar 14 '21 at 15:55
  • @DonAntonio I think one confusion here is the following: If $\ker (T)={0}$, then $\dim V=\dim (\mbox{im} T)$. Does this imply that $V=\mbox{im} T$? Not necessarily for Hilbert spaces, say. – chhro Aug 15 '21 at 06:25
  • I think that you have proved general situation that says: dim(V)=dim[ker(T)]+dim[Rank(T)], Since B1 and B3 have empty intersection and dim(V)=card(B1 U B2)=card(B1)+card(B3). Actually if dim(V) be infinite then the rank-nullity theorem holds and we work with cardinal numbers. – Amirhossein Haddadian Jan 09 '23 at 10:50
2

Daniel Fischer's answer is correct, but there is some confusion in the other answers and comments as to the meaning of "rank-nullity theorem" and its connection to the equivalence of injectivity/surjectivity on finite-dimensional spaces. To clear this up, I'm adding this answer as an appendix to Daniel Fischer's.

Let $V$ and $W$ be vector spaces (of arbitrary dimension, over the same field), and let $T\in L(V, W).$

Direct Sum Theorem. If $V = X \oplus Y$, then $\dim V = \dim X + \dim Y.$

Rank-Nullity Theorem. $\dim V = \dim \ker T + \dim \operatorname{im} T.$

Surjective-Injective-Bijective (SIB) Theorem.$^1$ If $\dim V = \dim W < \infty$, then surjectivity, injectivity, and bijectivity of $T$ are equivalent.

Since the SIB theorem is a corollary of the rank-nullity theorem, one may wonder: Since the rank-nullity theorem holds in arbitrary dimension, why does the SIB theorem hold only in finite dimension? The answer, as @fgp points out in a comment to @DonAntonio's answer, is the following lemma.

Lemma. Let $\kappa$ and $\lambda$ be cardinals and suppose $\kappa + \lambda = \kappa.$ If $\kappa < \infty,$ then $\lambda = 0$.

Proof of SIB theorem from rank-nullity theorem. It suffices to prove equivalence of injectivity and surjectivity (both directions require finite dimension). Let $\kappa = \dim V = \dim W \; (< \infty).$

  • $(\text{Injectivity} \implies \text{surjectivity})$ If $T$ is injective, then $\ker T = 0,$ hence $\dim \operatorname{im} W = \kappa$ by the rank-nullity theorem. Write $W = \operatorname{im} T \oplus U$. By the direct sum theorem, $\kappa = \kappa + \dim U.$ By the lemma, $\dim U = 0$, so $\operatorname{im} T = W$, thus $T$ is surjective.
  • $(\text{Surjectivity} \implies \text{injectivity})$ If $T$ is surjective, then $\operatorname{im} T = W$, so the rank-nullity theorem gives $\kappa = \dim \ker T + \kappa$. By the lemma, $\dim \ker T = 0$, so $T$ is injective. $\square$

$^1$ This is not a standard name. I am also not arguing that my designation of $\dim V = \dim \ker T + \dim \operatorname{im} T$ as the rank-nullity theorem is the "correct" terminology. The important thing is to see how the theorems are related.

WillG
  • 7,382
-1

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.

Giuseppe Negro
  • 34,508
  • 5
    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. – J.R. Apr 13 '14 at 17:29
  • 1
    @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... – fgp Apr 13 '14 at 17:54
  • 1
    @fgp: what strange kind of cardinal arithmetic is leading you to that equality? – darij grinberg Sep 07 '18 at 13:07
  • 3
    @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. – Daniel Fischer Sep 07 '18 at 19:50
  • Ah, if that's what he meant :) – darij grinberg Sep 07 '18 at 19:57
  • What is $\tilde{T}$? – Filippo Nov 10 '20 at 11:03
  • @Filippo: That formula is the definition of $\tilde T$. I edited the answer. – Giuseppe Negro Nov 10 '20 at 11:26
  • 1
    Thank you, now I understand :) Does this theorem have a name (just in case I decide to search for a proof)? – Filippo Nov 10 '20 at 11:41
  • 2
    It's one of the fundamental isomorphism theorem in algebra. Maybe you have seen it for groups. – Giuseppe Negro Nov 10 '20 at 11:59
-2

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

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

DonAntonio
  • 214,715
  • 2
    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. – user142870 Apr 13 '14 at 17:29
  • 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. – DonAntonio Apr 13 '14 at 17:32
  • 3
    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. – user142870 Apr 13 '14 at 17:36
  • 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. – DonAntonio Apr 13 '14 at 17:36
  • 6
    @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 $\textrm{dim ker } T + \textrm{dim im } T = 1 + \aleph_0 = \aleph_0 = \textrm{dim } V$. – fgp Apr 13 '14 at 17:43
  • Yes @fgp, but not necessarily $;\aleph_0;$ . It can be bigger, too. – DonAntonio Apr 13 '14 at 17:45
  • 1
    @DonAntonio Sure. With "your particular case" I simply meant the particular $V$ and $T$ in the second part of your answer, where indeed $\textrm{dim }V = \aleph_0$. – fgp Apr 13 '14 at 17:47