# Show that $\sum_{i=1}^n (-1)^i dimV_i=0$ [duplicate]

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!

• – klirk Jul 14 '19 at 21:31

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.

• Hi since $T_i: V_i\to V_{i+1}$ and $V_0=V_{n+1}=0$, does that imply $dim ImT_n = dim KerT_{n+1)=0$? – WaterBro Jul 14 '19 at 21:06
• Isn’t your exact sequence supposed to start and end at $0$? – Mindlack Jul 14 '19 at 21:10
• 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. – WaterBro Jul 14 '19 at 21:12
• 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. – Mindlack Jul 14 '19 at 21:16
• 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! – WaterBro Jul 14 '19 at 21:22