Suppose $T_i: V_i\to V_{i+1}$ is a sequence of vector spaces such that $KerT_{i+1} = Im T_i$ with $V_0=V_{n+1}=0$. Show that $\sum_{i=1}^n (-1)^i dimV_i=0$.

For this question, what I did is

$dimV_n=Rank(T_n)+Ker(T_n)$, $dimV_n-dimV_{n-1}=Rank(T_n)-(dimV_{n-2}-Ker(T_{n-2})$, $dimV_n-dimV_{n-1}+dimV_{n-2}=Rank(T_n)+Ker(T_{n-2})$......

Since $ImT_i=KerT_{i+1}$ with $V_{n+1}=0$, $Rank(T_n)=0$, but I don't know how to continue.

The basic thing we want to show is that $-dimV_1=0$, $dimV_2-dimV_1=0$ and extend to n, while from my above process we can show from n to 1 but cannot determine $dimV_2-dimV_1=0$.

Any hints will be helpful, thx!


Use induction. Here's a sketch of how the inductive step works.

For example, if you know it works for $0 \to V \to W \to 0$, an isomorphism, then we can consider

$$0 \to V \to W \to X \to 0$$ and split this as $$0 \to V \to (\operatorname{im}(V \to W)) \to 0$$ since the map $V \to W$ is injective and also on the right we have an isomorphism $$0 \to W/\operatorname{im}(V \to W) \to X \to 0$$ since the image of $V \to W$ is the kernel of $W \to X$. Now writing $W = \operatorname{im}(V \to W) \oplus \ker(V \to W)$ by the fundamental theorem of linear algebra we can write our total sequence as

$$0 \to V \to \operatorname{im}(V \to W)\oplus\ker(V \to W) \to X \to 0$$

But now we see how to split this as a sum of two exact sequences. The first one is simply:

$$0 \to V \to \operatorname{im}(V \to W) \to 0 \to 0$$

and the second one:

$$0 \to 0 \to \ker(V \to W) \to X \to 0$$

So showing that the property is stable under taking direct sums, and generalizing this argument, your proof will be complete.


Assume for the sake of easier notation that $V_0=V_{n+1}=0$.

Now, denote $d_k=\dim\,V_k$, and $\delta_k=\dim\,\ker\,T_k$.

Now, if $1 \leq k \leq n$, $d_k=\delta_k+\delta_{k+1}$, since the image of $T_k$ is the same as the kernel of $T_{k+1}$.

Then computing the sum works.

  • $\begingroup$ Hi since $T_i: V_i\to V_{i+1}$ and $V_0=V_{n+1}=0$, does that imply $ dim ImT_n = dim KerT_{n+1)=0 $? $\endgroup$ – WaterBro Jul 14 '19 at 21:06
  • $\begingroup$ Isn’t your exact sequence supposed to start and end at $0$? $\endgroup$ – Mindlack Jul 14 '19 at 21:10
  • $\begingroup$ As my induction, the sequence should start at 0, while if we can observe there is a +-+... or -+-... pattern over the $dimV_n$, and prove that the RHS is always 0, then it doesn't matter starts at 0 or 1, even $(-1)^i$ doesn't matter. $\endgroup$ – WaterBro Jul 14 '19 at 21:12
  • $\begingroup$ Answering your first comment: $T_{n+1}$ doe not exist since $V_{n+2}$ doesn’t. Since $V_{n+1}=0$, $T_n$ is the null operator. $\endgroup$ – Mindlack Jul 14 '19 at 21:16
  • $\begingroup$ Since finally, from the induction, we can get $dimV_n-dimV_{n-1}+dimV_{n-2}...+dimV_2-dimV_1+dimV_0=dim im(T_n)+dim ker(T_0)$, if we can prove $dimV_0=dim im(T_n)=dim ker(T_0)=0$ then we can ignore i starts at 0 or 1, and $-dimV_1 or +dimV_1$ doesn't matter since RHS is all 0. Is there any ways can prove this with the condition given in question? Thank u very much! $\endgroup$ – WaterBro Jul 14 '19 at 21:22

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