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.