# Rank of matrix

## Homework Statement

Prove that for any m x s matrix A and any s x n matrix B it holds that:

rank(A) + rank(B) - s
is less or equal to:
rank(AB)

## The Attempt at a Solution

Obviously, the following are true:

- rank(A) is less or equal to s,
- rank(B) is less or equal to s,
- rank(AB) is less or equal to both rank(A) and rank(B).

So it is possible to prove:

rank(A) + rank(B) - 2s is less or equal to rank(AB).

Really don't know what can be done next. Thanks for any help on this.

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These should be easy to prove:
Rank(A) <= min(m,s,)
Rank(B) <= min(s,n)
Rank(AB) <= min (m,n)

and put you in the right direction

edit: If you have to prove these cause you haven’t proven it in class, just prove the general form of:
Rank(A) <= min (m,n) where A is a m x n matrix.

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These should be easy to prove:
Rank(A) <= min(m,s,)
Rank(B) <= min(s,n)
Rank(AB) <= min (m,n)

and put you in the right direction
I'm afraid that doesn't suffice.

For suppose that m <= n < = s, e.g. m = 4, n = 4, s = 5. Then:

rank(A) <= 4,
rank(B) <= 4,
rank(AB) <= 4.

That, however, doesn't make rule out as impossible following:

rank(A) = 4,
rank(B) = 4,
rank(AB) = 2,

which would make the statement:

rank(A) + rank(B) - s <= rank(AB)

entirely false.

sorry i'm half asleep, after thinking about it a bit more i'd use the rank-nullity theorem

sorry i'm half asleep, after thinking about it a bit more i'd use the rank-nullity theorem
Well, I see a way to use the rank-nullity theorem here successfully provided it is known that:

for any A, B, AB it is the case that: nullity A + nullity B <= nullity AB.

But is the above true?

you have it backwards:

nullity(AB) <= nullity(A) + nullity(B)

Can you think of why?

you have it backwards:

nullity(AB) <= nullity(A) + nullity(B)

Can you think of why?

Sorry, I've meant to write it the way you did. I'm aware that it'll do the trick but don't know why is it true.

sorry it took me so long to respond, i'm at work and didn’t have a chance to get back to you.

Do you know Rank(AB) <= min(Rank(A),Rank(B))? If so:
Since Null(A) >= 0, Rank(AB) >=0, if we use the rank-nullity theorem a few times we get:

Null(A) + Null(B) =
Null(A) + n – Rank(B) =
Null(A) + Rank (AB) + Null(AB) – Rank(B) >=
Rank (AB) + Null(AB) – Rank(B) >=
Rank(B) – Rank(B) + Null(AB) =
Null(AB)

sorry it took me so long to respond, i'm at work and didn’t have a chance to get back to you.

Do you know Rank(AB) <= min(Rank(A),Rank(B))? If so:
Since Null(A) >= 0, Rank(AB) >=0, if we use the rank-nullity theorem a few times we get:

Null(A) + Null(B) =
Null(A) + n – Rank(B) =
Null(A) + Rank (AB) + Null(AB) – Rank(B) >=
Rank (AB) + Null(AB) – Rank(B) >=
Rank(B) – Rank(B) + Null(AB) =
Null(AB)
Would you be so kind as to explain how did you get

Rank (AB) + Null(AB) – Rank(B) >= Rank(B) – Rank(B) + Null(AB)?

The rest - including Rank(AB) <= min(Rank(A),Rank(B)) - is perfectly clear to me.

With an inequality error, sigh I’m doing poorly with this problem and algebraic manipulation tonight.

I know: nullity(AB) <= nullity(A) + nullity(B) is true. I can't for the life of me remember why.

For sure the inequality holds:

max(null(A), null(B)) <= null(AB).

Oh, and null(A) + null(B) >= null(AB) turns out to be simply equivalent to the statement we were trying to prove from the beginning.

I found it easy to prove:

null(A) + null(B) >= null(AB),

using maps instead of matrices. Indeed, take /phi_A, /phi_B and /phi_AB defined by matrices A, B and AB respectively.

If we restrict the domain of /phi_AB to just those vectors X of which images \phi_B(X) lies in the Ker(A), it's not hard to apply rank-null theorem in it's full generality once again and get:

null(AB) = rank C + null(B), where C \in null(A).

That concludes the proof.

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