Prove Isomorphism between Quotient Spaces of Null Space

Question

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!

Answer 1

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.

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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 $\dim(V/U) = \dim(V) - \dim(U)$.