Proving Rank Relationship Between Matrices A and B

In summary: As for the original problem, suppose that the statement: rank(A) + rank(B) - s <= rank(AB) is false. Then there is a counterexampleA = (a_ij), i = 1, 2, ... m, j = 1, 2, ... s,B = (b_ij), i = 1, 2, ... s, j = 1, 2, ... n,such that:m < = n < = s,rank(A) = s,rank(B) = s,rank(AB) = s + 1.Since we know that s < = min(m,n), we can also assume that n > s.Let's define C =
  • #1
D.K.
12
0

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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  • #2
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.
 
Last edited:
  • #3
JonF said:
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.
 
  • #4
sorry I'm half asleep, after thinking about it a bit more i'd use the rank-nullity theorem
 
  • #5
JonF said:
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?
 
  • #6
you have it backwards:

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

Can you think of why?
 
  • #7
JonF said:
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.
 
  • #8
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)
 
  • #9
JonF said:
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.
 
  • #10
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.
 
  • #11
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.
 
  • #12
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.
 
Last edited:

What is the purpose of proving the rank relationship between matrices A and B?

The purpose of proving the rank relationship between matrices A and B is to determine the similarity or difference between the two matrices. This can help in understanding the properties and characteristics of the matrices, as well as identifying any connections or patterns between them.

What is the definition of rank in matrices?

The rank of a matrix is defined as the maximum number of linearly independent rows or columns in the matrix. In other words, it is the number of rows or columns that cannot be expressed as a linear combination of other rows or columns in the matrix.

How can the rank of a matrix be determined?

The rank of a matrix can be determined by performing row operations on the matrix and reducing it to its row echelon form. The number of non-zero rows in the row echelon form will be equal to the rank of the matrix.

What are the different types of rank relationships between matrices A and B?

The different types of rank relationships between matrices A and B are equality, inequality, and no relationship. If the rank of matrix A is equal to the rank of matrix B, they have an equality relationship. If the rank of matrix A is greater than the rank of matrix B, they have an inequality relationship. If the rank of matrix A is less than the rank of matrix B, there is no relationship between them.

What are some methods for proving the rank relationship between matrices A and B?

Some methods for proving the rank relationship between matrices A and B include using the definition of rank, performing row operations, and using the rank-nullity theorem. Other methods include using the determinant, eigenvalues, and singular value decomposition of the matrices.

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