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Let $T:V\rightarrow V$ be a linear transformation and dimension of $V$ be infinite. Then show that either the nullity or the rank of $T$ is infinite.

The problem is that I cannot use rank nullity theorem. Thanks for any help.

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Does "I cannot use" mean you are not allowed or do you mean that you think it doesn't apply? – rschwieb Aug 14 '12 at 18:33
I meant that I thought that the theorem is true for finite dimensional spaces – Ester Aug 15 '12 at 17:55
When given an "A or B" statement, it would really be best to tell me "A" or "B" and not "C". I think you mean you think it doesn't apply. $V\cong \mathrm{Im}(f)\oplus\ker(f)$ in all cases. If the two on the right were finite dimensional, then the one on the left would be finite dimensional as well. – rschwieb Aug 15 '12 at 20:38
as you can see, it is also true in infinite dimension... However the answer of donantonio is nice. – Louis La Brocante Aug 15 '12 at 20:43
up vote 4 down vote accepted

Let $\,\{v_1,...\}\,$ be any (obviously infinite) basis of $\,V\,$ , then:

(1) if $\,\{Tv_1,...\}\,$ contains an infinite linearly independent subset then $\,\dim Im\,\,(T)=rank\,\,T=\infty\,$ , other wise:

(2) there only exists a finite linearly independent subset in $\,\{Tv_1,...\}\,$ ,so it must be that an infinite subset of $\,\{v_1,...\}\,$ is mapped to zero by $\,T\,$ and thus $\,\dim\ker T=null\,\,T=\infty\,$

Added: Please do read the comment below by Tunococ as it proves (2) accurately.

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I think (2) is not quite true. If $T(v_i) = v_1$ for all $i$ in the index set of the basis, then $\{Tv_1, \ldots\} = \{v_1\}$ has only one element, but $T(v_i) \ne 0$ for all $i$. However, counting linearly dependent subsets works. You know that if (1) is not true, there must be infinitely many disjoint subsets $S_j$ of $\{v_1, \ldots\}$ whose images are linearly dependent subsets. Each $S_j$ corresponds to at least one element in the kernel, and since $S_j$'s are disjoint, these elements are linearly independent. Therefore, the kernel has infinite dimension. – Tunococ Aug 14 '12 at 19:44
You're right and good point: that's exactly what should have been written and I'll edit it. Thanks.\ – DonAntonio Aug 15 '12 at 2:02

You try to use a too strong version of the rank nullity theorem. A more general version (true in infinite dimension) asserts that $V$ is isomorphic fo $Ker(T)\oplus Im(T)$. In this setting saying that both $ker(T)$ and $Im(T)$ are finite dimensional would imply that $V$ would also be finite dimensional.

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I think this is exactly the rank-nullity theorem, which the OP stated he can't use. – DonAntonio Aug 14 '12 at 17:37
@DonAntonio I think Rolando interpreted that Sopu was under the impression that it was not possible to use it rather than under some mandate not to use it. Based on the OP's response above, this seems to be the correct interpretation. – rschwieb Aug 15 '12 at 20:41
Yes, i though he did not dare use it, for instance because he only knew that "the sum of the dimensions is the dimension of V" (in the finite dimensional case). Looking at the comments of the question, it seems i was right. – Louis La Brocante Aug 15 '12 at 20:43
Apparently, indeed. A matter of interpretation of the OP's words... – DonAntonio Aug 15 '12 at 23:16

Suppose the rank and nullity are both finite. Then there's some positive integer $n$ which exceeds both of them. Let $W$ be a subspace of $V$ of dimension $2n+1$ (I take it you can see that such a thing exists). Let $T'$ be the restriction of $T$ to $W$, $T':W\to V$. The rank of $T'$ is at most the rank of $T$, right? And the nullity of $T'$ is at most the nullity of $T$, right? But this contradicts the finite form of the rank-nullity theorem. Are you allowed/able to use that?

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I have got it.Thanks to all of you for your kind help. – Ester Aug 15 '12 at 17:54

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