Using rank-nullity theorem to show alternating sum of dimensions = 0

[bmanbs2]

Homework Statement

Consider integer sequence $$n_{1},...,n_{r}$$ and matrices $$A_{1},...,A_{n-1}$$. Assume $$im\left(A_{i}\right) = ker\left(A_{i+1}\right)$$

Using the rank-nullity theorem, show that $$\sum^{n}_{i=1}\left(-1\right)^{i}d_{i} = 0$$

Homework Equations

The rank-nullity theorem states that if $$v$$ and $$w$$ are vector spaces and $$A$$ is the linear map A: v -> w, then
$$dim\left(im\left(A\right)\right) + dim\left(ker\left(A\right)\right) = dim\left(v\right)$$

The Attempt at a Solution

I know that the relation $$im\left(A_{i}\right) = ker\left(A_{i+1}\right)$$ means that $$\sum^{n}_{i even}d_{i}$$ = $$\sum^{n}_{i odd}d_{i}$$, but I don't know how to get there.

 

[micromass]

First, apply rank-nullity on A₁,...,A_n. You'll get n equations. Adding up those equations should get you the answer...

 

[bmanbs2]

Thanks for the reply, but I want to make sure I'm setting up the equations properly.

So far I have that
$$im\left(A_{i-1}\right) + im\left(A_{i}\right) = d_{i}$$
$$im\left(A_{i}\right) + im\left(A_{i+1}\right) = d_{i+1}$$

Subtracting the system of equations gives
$$im\left(A_{i-1}\right) - im\left(A_{i+1}\right) = d_{i} - d_{i+1}$$

That may then be continued for $$2 \leq i \leq n-1$$, but I'm not sure how to make it all equal zero.

 

[micromass]

OK, I was wrong about adding up the equations. But what you have looks good:

$$im(A_{i-1})+im(A_i)=d_i$$.

Now, you have to calculate

$$-d_1+d_2-d_3+d_4-...$$

Just substitute every dⁱ in the above addition by $$im(A_{i-1})+im(A_i)$$, and make a little calculation.