$\dim RS(A) = \dim CS(A)$ from $\dim W^\perp + \dim W = \dim V$

Question

Please help me with this:

From the formula $\dim W^\perp + \dim W = \dim V$ for general subspaces $W \subset V$ of an inner product space, deduce that the row rank of $A$ is equal to its column rank:

$\dim RS(A) = \dim CS(A)$.

Thank you in advance! Any help would be great appreciated!

I tried to use rank-nullity and the fact that ker(A) = (RS(A))perp, but I am stuck. — Loulou EC, Nov 22 '17 at 12:27

score 0 · Accepted Answer · answered Nov 22 '17 at 12:54

0

Is this proof what you were looking for?

https://math.stackexchange.com/a/1337691/60023

I am not sure if you are allowed to use the rank-nullity theorem, but it is basically what you were trying to do.

answered Nov 22 '17 at 12:54

Zach Boyd

908

In my proof, I have to use dimW⊥+dimW=dimV. In this one, it isn't used. – Loulou EC Nov 22 '17 at 13:20
It is used in the line: dim(ker(A)) = n - row rank, which can also be written as dim(ker(A)) = n - dim(ker(A))^perp, so ker(A) is the subspace. – Zach Boyd Nov 22 '17 at 13:22

$\dim RS(A) = \dim CS(A)$ from $\dim W^\perp + \dim W = \dim V$

1 Answers1