Chapter 5: Problem 5
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
Expert verified
AB = 0 if and only if the columns of B are in the null space of A.
Step by step solution
01
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.
02
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 \( B \).
03
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., \( \operatorname{col} B \subseteq \operatorname{null} A \).
04
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, \( AB = [A\mathbf{b}_1, A\mathbf{b}_2, \ldots, A\mathbf{b}_m] = 0 \).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( A(B+C) = AB + AC \).
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 \( A \).
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 \( B \).
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.