Question

If \(A\) is \(m \times n\) and \(B\) is \(n \times m,\) show that \(A B=0\) if and only if \(\operatorname{col} B \subseteq\) null \(A .\)

Short answer

AB = 0 if and only if the columns of B are in the null space of A.

Step-by-step solution

Understand the Problem Statement

We need to show that \( AB = 0 \) if and only if \( \operatorname{col} B \subseteq \operatorname{null} A \). This involves showing both directions: If \( AB = 0 \), then \( \operatorname{col} B \subseteq \operatorname{null} A \), and vice versa.

Express the Multiplication

Consider the product \( AB \). Since \( A \) is \( m \times n \) and \( B \) is \( n \times m \), their product \( AB \) is an \( m \times m \) matrix composed of columns \( A \mathbf{b}_1, A \mathbf{b}_2, \ldots, A \mathbf{b}_m \), where \( \mathbf{b}_i \) are the columns of \( B \).

Prove 'If' Direction

Assume \( AB = 0 \). This implies \( A \mathbf{b}_i = \mathbf{0} \) for every column \( \mathbf{b}_i \) of \( B \). Therefore, each column \( \mathbf{b}_i \in \operatorname{col} B \) is in the null space of \( A \), i.e., \( \operatorname{col} B \subseteq \operatorname{null} A \).

Prove 'Only If' Direction

Assume \( \operatorname{col} B \subseteq \operatorname{null} A \). This means every column \( \mathbf{b}_i \) of \( B \) satisfies \( A \mathbf{b}_i = \mathbf{0} \). Therefore, \( AB = [A\mathbf{b}_1, A\mathbf{b}_2, \ldots, A\mathbf{b}_m] = 0 \).

Conclusion

Both directions have been proven: If \( AB = 0 \), then \( \operatorname{col} B \subseteq \operatorname{null} A \), and if \( \operatorname{col} B \subseteq \operatorname{null} A \), then \( AB = 0 \). Therefore, the statement 'AB = 0 if and only if \( \operatorname{col} B \subseteq \operatorname{null} A \)' is true.

Key concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra. It allows us to combine two matrices to produce a new matrix. When multiplying a matrix \( A \) of size \( m \times n \) with a matrix \( B \) of size \( n \times m \), the resulting matrix \( AB \) will be \( m \times m \). The multiplication is performed by taking each column of matrix \( B \) and doing a dot product with the rows of matrix \( A \). The result is a series of computations that fill in the entries of the resulting matrix.

• Non-Commutative: Remember that, generally, \( AB eq BA \).
• Associative: Matrix multiplication is associative, which means \((AB)C = A(BC)\).
• Distributive: It is distributive over addition, so \( A(B+C) = AB + AC \).

Matrix multiplication can be tricky since it is not commutative like number multiplication. In this problem, we are interested in the particular scenario when \( AB = 0 \), and what this says about the relationship between the matrices \( A \) and \( B \).

Null Space

The null space of a matrix \( A \) is a set of vectors that, when multiplied by \( A \), result in the zero vector. It is sometimes called the kernel of \( A \). To find the null space, you solve the equation \( A \mathbf{x} = 0 \) for \( \mathbf{x} \).

• Properties: It is a linear subspace of the vector space defined by the columns of \( A \).
• Dimension: The dimension of the null space is called the nullity of \( A \).

In our problem, if \( AB = 0 \), each column of \( B \) becomes a solution for the equation \( Ax = 0 \). Thus, those columns of \( B \) that satisfy this are considered a part of the null space of \( A \). Therefore, \( \operatorname{col} B \subseteq \operatorname{null} A \).

Column Space

The column space of a matrix \( B \), denoted as \( \operatorname{col}(B) \), is the set of all possible linear combinations of its columns. This space represents all the possible outputs if you multiply \( B \) by compatible vectors.

• Subspace: The column space is a subspace of \( \mathbb{R}^m \) for matrix \( B \) which is from \( \mathbb{R}^{n \times m} \).
• Span: It is spanned by the columns of \( B \).

In the exercise, the interaction between the column space of \( B \) and the null space of \( A \) is key. **If** it holds that \( \operatorname{col}(B) \subseteq \operatorname{null}(A) \), it implies that every column vector in \( B \) lies within the null space of \( A \). Hence, when \( AB = 0 \), the entire column space of \( B \) must belong to the null space of \( A \). This effectively creates a zero matrix upon multiplication.

Linear Algebra Proofs

Linear algebra proofs are a logical sequence of deductions used to demonstrate the truth of a given statement. In the context of this exercise, we employ a typical proof type known as 'if and only if'. This involves proving two implications: the forward direction ("if") and the reverse direction ("only if").

• Forward Proof: We proved that if \( AB = 0 \), then \( \operatorname{col} B \subseteq \operatorname{null} A \). This involves showing that each column of \( B \) turns into zero vectors when multiplied by \( A \).
• Reverse Proof: Conversely, assuming \( \operatorname{col} B \subseteq \operatorname{null} A \), we show it leads to \( AB = 0 \). Both directions together confirm the equivalence.

In linear algebra proofs, especially with matrices, understanding the geometric meaning can be valuable. The proof seen here reveals how different matrix properties interrelate logically and geometrically, cementing the understanding of matrix spaces. In essence, the solving of the forward and reverse proves not just logical equivalence but also enriches comprehension of linear mappings and subspace relationships.