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Question: T is linear map from X to X. $T^k$ is the k th power of T: $T^k = TTTT...$; $N_k$ is the null space of $T^k$. How to prove that the quotient space $N_{k+1}/N_k$ is isomorphic to a subspace of $N_k/N_{k-1}$?

  • Is it enough that I show the dimension of $N_{k+1}/N_k$ is smaller than dimension of $N_k/N_{k-1}$?
  • If not, then how to prove?
  • What are the elements in $N_{k+1}/N_k$ and $N_k/N_{k-1}$? It is just really hard to understand...

Thank you!

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    Yes, it's enough that you show the dimension is smaller – 2017-02-27
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    But wouldn't it necessary to show linearity and bijection to show isomorphism? Also, how can we be sure about the dimension of Nk+1/Nk and Nk/Nk−1? Thank you so much! @Omnomnomnom – 2017-02-27
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    any two vector spaces of the same dimension are isomorphic. – 2017-02-27
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    Hi @Omnomnomnom, your answer is really helpful for me. But how to prove the dimension of Nk+1/Nk is smaller than Nk/Nk−1? Sorry about wasting too much of your time. Can you give me a hint on that? Thank you so so much! – 2017-02-28
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    Do you know about Jordan normal form? – 2017-02-28
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    @Omnomnomnom Not really...this is a linear algebra course as pre-requisite for economics master so I guess it won't be that hard? I can definitely look it up, but do you know any other approach? Thank you!!! – 2017-02-28
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    Can't think of any yet. I'll let you know if I do. – 2017-02-28
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54468/discussion-between-elmer-li-and-omnomnomnom). – 2017-02-28
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    In fact, using Jordan form is a cheat, since this result provides a nice proof of the Jordan from result. In any case, see my answer below. – 2017-02-28

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First, note that $N_k \subset N_{k+1}$ for all $k$. Next, we have the following:

For each $k$, $T(N_{k+1}) \subset N_{k}$. Moreover, the induced map $T_{\sim}:N_{k+1}/N_k \to N_{k}/N_{k-1}$ defined by $$ T_{\sim}(v + N_k) = T(v) + N_{k-1} $$ is (well-defined and) injective.

To see this, you need a lemma like the following:

Lemma: Suppose that $T:V \to W$ with $V' \subset V$ and $W' \subset W$, with $T(V') \subset W'$. Then the map $T_{\sim}:V/V' \to W/W'$ defined by $$ T(v + V') = T(v) + W' $$ is well defined

you'll probably find something like this in your textbook. With that out of the way, we want to show that the map we've defined is injective (one to one). To see this, note that for $k \geq 1$, we have $\ker(T) \subset N_k$. So, we have $$ v + N_k \in \ker(T_\sim) \implies\\ T(v) + N_{k-1} = 0 + N_{k-1} \implies\\ T(v) \in N_{k-1} \implies\\ v \in N_k \implies\\ v + N_k = 0 + N_k $$ So that $\dim \ker(T_\sim) = 0$. Since $\dim \ker T_\sim = 0$, $T_\sim$ is injective.

Since $T_\sim$ is injective, the map $T_{\sim}:N_{k+1}/N_k \to T(N_{k + 1}/N_{k}) \subset N_k/N_{k-1}$ is an isomorphism between $N_{k+1}/N_k$ and a subspace of $N_{k}/N_{k-1}$, as desired.

To answer your surrounding questions: it turns out the direct proof above is easier than any sort of "dimension counting" approach, unless we assume the existence of Jordan form.

For any vector space $V$ with subspace $U$, the elements of $V/U$ are all of the form $v + U$, with $v \in V$. These elements are best visualized as the "subspace parallel to $U$, passing through $v$" (technically, $v+U$ is not a "linear subspace"; it's an "affine subspace"). Note that we add two elements of the quotient space by $$ (v_1 + U) + (v_2 + U) = (v_1 + v_2) + U $$ that is, we combine the offsets from the origin. It is notable that $\dim(V/U) = \dim(V) - \dim(U)$.

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    Let me know if you have any questions about this – 2017-03-03