Sure, let's break this down:

1. **Col AB ⊆ Col A**:
Let's take any vector v in Col AB. By definition of matrix multiplication, v can be written as ABu for some vector u in R^n. Now, ABu is a linear combination of the columns of A, with the entries of Bu as weights. Therefore, v is in the column space of A. Hence, Col AB is a subset of Col A.

2. **Nul B ⊆ Nul AB**:
Let's take any vector u in Nul B. By definition, Bu = 0. Therefore, ABu = A0 = 0, which means u is in Nul AB. Hence, Nul B is a subset of Nul AB.

3. **rank (AB) ≤ rank A**:
Since Col AB ⊆ Col A, the dimension of Col AB (which is rank AB) is less than or equal to the dimension of Col A (which is rank A).

4. **rank (AB) ≤ rank B**:
This follows from the Rank-Nullity Theorem. The Rank-Nullity Theorem states that for any matrix, the sum of its rank and the dimension of its null space is equal to the number of its columns. Therefore, rank B = n - dim(Nul B) ≥ n - dim(Nul AB) = rank AB.

Therefore, we have shown that rank (AB) ≤ rank A and rank (AB) ≤ rank B.

Question

Sure, let's break this down:

1. **Col AB ⊆ Col A**: 
   Let's take any vector v in Col AB. By definition of matrix multiplication, v can be written as ABu for some vector u in R^n. Now, ABu is a linear combination of the columns of A, with the entries of Bu as weights. Therefore, v is in the column space of A. Hence, Col AB is a subset of Col A.

2. **Nul B ⊆ Nul AB**: 
   Let's take any vector u in Nul B. By definition, Bu = 0. Therefore, ABu = A0 = 0, which means u is in Nul AB. Hence, Nul B is a subset of Nul AB.

3. **rank (AB) ≤ rank A**: 
   Since Col AB ⊆ Col A, the dimension of Col AB (which is rank AB) is less than or equal to the dimension of Col A (which is rank A).

4. **rank (AB) ≤ rank B**: 
   This follows from the Rank-Nullity Theorem. The Rank-Nullity Theorem states that for any matrix, the sum of its rank and the dimension of its null space is equal to the number of its columns. Therefore, rank B = n - dim(Nul B) ≥ n - dim(Nul AB) = rank AB.

Therefore, we have shown that rank (AB) ≤ rank A and rank (AB) ≤ rank B.

Knowee AI · Accepted Answer

Sure, let's break this down:

1. **Col AB ⊆ Col A**: 
   Let's take any vector v in Col AB. By definition of matrix multiplication, v can be written as ABu for some vector u in R^n. Now, ABu is a linear combination of the columns of A, with the entries of Bu as weights. Therefore, v is in the column space of A. Hence, Col AB is a subset of Col A.

2. **Nul B ⊆ Nul AB**: 
   Let's take any vector u in Nul B. By definition, Bu = 0. Therefore, ABu = A0 = 0, which means u is in Nul AB. Hence, Nul B is a subset of Nul AB.

3. **rank (AB) ≤ rank A**: 
   Since Col AB ⊆ Col A, the dimension of Col AB (which is rank AB) is less than or equal to the dimension of Col A (which is rank A).

4. **rank (AB) ≤ rank B**: 
   This follows from the Rank-Nullity Theorem. The Rank-Nullity Theorem states that for any matrix, the sum of its rank and the dimension of its null space is equal to the number of its columns. Therefore, rank B = n - dim(Nul B) ≥ n - dim(Nul AB) = rank AB.

Therefore, we have shown that rank (AB) ≤ rank A and rank (AB) ≤ rank B.

Suppose A is an m × n matrix and B is an n × p matrix. Show thatCol AB ⊆ Col A and Nul B ⊆ Nul AB.Hence, rank (AB) ≤ rank A and rank (AB) ≤ rank B.

Question

Suppose A is an m × n matrix and B is an n × p matrix. Show that

Solution

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