# Thread: Dimension of null space of two matrix multiplication

1. ## Dimension of null space of two matrix multiplication

let S and T be linear transformation on n-dimensional vector space
why is dim ker(ST) less than dim ker(S) + dim ker(T)

2. ## Re: Dimension of null space of two matrix multiplication

Originally Posted by parklover
let S and T be linear transformation on n-dimensional vector space
why is dim ker(ST) less than dim ker(S) + dim ker(T)
define $f: \ker(ST) \longrightarrow \ker S$ by $f(x)=T(x),$ for all $x \in \ker(ST)$ and apply the rank-nulity theorem to $f$.

Spoiler:
if you tried and still couldn't finish the proof, see the proof of the "claim" in my blog.

3. ## Re: Dimension of null space of two matrix multiplication

Actually, would it be always less than? I always thought it could be less than or equals, am I right on this?
Say like if matrix A times B, then rank(AB) <= min(rank(A), rank(B)), is this right? And then apply this same idea to the nullspace too, would it all come out right?

4. ## Re: Dimension of null space of two matrix multiplication

Originally Posted by xEnOn
Actually, would it be always less than?
The equality is possible:

$\dim \ker (SI)=\dim \ker (S)=\dim \ker (S)+0=\dim \ker (S)+\dim \ker (I)$

5. ## Re: Dimension of null space of two matrix multiplication

ohh...
So am I right to say about $rank(AB) \leq min(rank(A), rank(B))$? That's the rank of $AB$ would always be less than or equals to the minimum rank between rank(A) and rank (B)?

6. ## Re: Dimension of null space of two matrix multiplication

Originally Posted by xEnOn
ohh...
So am I right to say about $rank(AB) \leq min(rank(A), rank(B))$? That's the rank of $AB$ would always be less than or equals to the minimum rank between rank(A) and rank (B)?
Yes, you are right.