# Homework Help: Block triangular matix has rank >= ranks of diagonal blocks?

1. May 7, 2013

### Constantinos

Hey!
I found this interesting theorem in a textbook, but I was unable to find a proof for it neither in the web nor on my own

1. The problem statement, all variables and given/known data
The rank of a block triangular matrix is at least and can be greater than the triangular blocks. proof?
specificaly, look here: pp. 25
http://books.google.gr/books?id=PlY...ce=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

2. Relevant equations
$M_n$ is the field $R^{n \times n}$ in the book.
3. The attempt at a solution
Well I proved the equality claim if the matrix is block diagonal. The proof should go like this:

Take a maximal set of linear independent rows from each diagonal block and combine them. The resulting set is a set of linear independent rows for the whole matrix. Moreover it is maximal since if it could grow larger, then some of the sets we chose for the diagonal blocks would not be maximal. (I can get more precise if need be)

For the block triangular matrix though, I'm pretty clueless... I tried a similar strategy as the above, but didn't work, tried proving by contradiction but again failed (and I don't think it can happen, the resulting inequalities didn't lead me anywhere) and also tried using the subadditivity property of ranks but this also got me nowhere.

Can anyone help me with that? No homework question, independent research.

Thanks!

2. May 7, 2013

### Office_Shredder

Staff Emeritus
For intuition into the "could be greater than", imagine an upper triangular matrix whose zeros are all diagonal. It could have rank n-1, but the diagonal 1x1 blocks all have rank 0, so the rank can definitely be arbitrarily larger.

Your proof for block diagonal does extend to "at least as large" in the block triangular case. If I pick out rows corresponding to linearly independent vectors from each block, then all those rows combined are linearly independent. Of course, in general it will not be a maximally linearly independent set (the reason why is because if in one block I pull out an extra vector, in the block by itself it may be linearly dependent on the rest of the block's vectors, but once I tack on the rest of the upper triangular entries it may become linearly independent)

3. May 8, 2013

### Constantinos

hmm yes I think you are right, thanks for your help! I couldn't get anywhere because I was thinking in terms of whole rows, instead of their parts that are into a diagonal block. Of course if a set of block-rows(or how else to call that?) from the diagonal blocks are L.I then the corresponding rows of the matrix have to be L.I. We take all these sets generated by the diagonal blocks and combine them, they are L.I and their number can't exceed the maximal independent set of the matrix. So I think it is proved!

Thanks again!

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