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Chapter 1: Problem 29

Let \(A\) and \(B\) be symmetric \(n \times n\) matrices. Prove that \(A B=B A\) if and only if \(A B\) is also symmetric.

Short Answer

Expert verified
To prove that the product of two symmetric matrices A and B is symmetric if and only if they commute, we show two things: 1. If A and B commute, i.e., AB = BA, then their product AB is symmetric. To show this, we find the transpose of AB, which equals to (AB)^T = (B^T)(A^T), and since A and B are symmetric, we have (AB)^T = BA. Given AB = BA, we can conclude that (AB)^T = AB, so AB is symmetric. 2. If AB is symmetric, i.e., (AB)^T = AB, then A and B commute. We have (AB)^T = B^TA^T by definition of transpose. By comparing AB and (BA)^T, we see that AB = (BA)^T, and taking the transpose of both sides, we find AB = BA. Thus, AB is symmetric if and only if A and B commute.

Step by step solution

01

Proof for AB is symmetric if A and B commute (A.B = B.A)

We are given that A.B = B.A. If A and B are symmetric, we know that A = A^T and B = B^T. To show AB is symmetric, we have to prove that (AB)^T = AB. Let's find the transpose of AB: (AB)^T = (B^TA^T), by definition of transpose, and also inverting the order of multiplication. Now, since A and B are symmetric, we have that A^T = A and B^T = B. Hence, (AB)^T = BA Given A.B = B.A, we can substitute BA for AB: (AB)^T = AB This shows that AB is symmetric if A and B commute.
02

Proof for commutativity of A and B if AB is symmetric

We are given that AB is symmetric, i.e., (AB)^T = AB. We must show that A and B commute if AB is symmetric. We know that: (AB)^T = B^TA^T, by definition of transpose. Given that AB is symmetric, AB = B^TA^T. Now, we will calculate (BA)^T: (BA)^T = A^TB^T, by definition of transpose. Let's compare AB and (BA)^T: - We know that AB = B^TA^T - We also know that (BA)^T = A^TB^T Notice that: AB = (BA)^T Taking the transpose of both sides of the equation, we get: (AB)^T = ((BA)^T)^T. Since AB is symmetric, we know: AB = ((BA)^T)^T. We also know that: ((BA)^T)^T = BA. Hence, we have: AB = BA. This shows that A and B commute if AB is symmetric. In conclusion, AB is symmetric if and only if A and B commute.

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Most popular questions from this chapter

Let \(A\) and \(B\) be \(n \times n\) matrices. Show that if \\[ A B=A \quad \text { and } \quad B \neq I \\] then \(A\) must be singular.

Let \(A\) be an \(n \times n\) matrix and \(\mathbf{x} \in \mathbb{R}^{n}\) (a) A scalar \(c\) can also be considered as a \(1 \times 1\) \(\operatorname{matrix} C=(c),\) and a vector \(\mathbf{b} \in \mathbb{R}^{n}\) can be considered as an \(n \times 1\) matrix \(B\). Although the matrix multiplication \(C B\) is not defined, show that the matrix product \(B C\) is equal to \(c \mathbf{b},\) the scalar multiplication of \(c\) times \(\mathbf{b}\) (b) Partition \(A\) into columns and \(\mathbf{x}\) into rows and perform the block multiplication of \(A\) times \(\mathbf{x}\). (c) Show that \\[ A \mathbf{x}=x_{1} \mathbf{a}_{1}+x_{2} \mathbf{a}_{2}+\cdots+x_{n} \mathbf{a}_{n} \\]

Let \(U\) be an \(n \times n\) upper triangular matrix with nonzero diagonal entries. (a) Explain why \(U\) must be nonsingular. (b) Explain why \(U^{-1}\) must be upper triangular.

Let \(C\) be a nonsymmetric \(n \times n\) matrix. For each of the following, determine whether the given matrix must be symmetric or could be nonsymmetric: (a) \(A=C+C^{T}\) (b) \(B=C-C^{T}\) (c) \(D=C^{T} C\) (d) \(E=C^{T} C-C C^{T}\) (e) \(F=(I+C)\left(I+C^{T}\right)\) (f) \(G=(I+C)\left(I-C^{T}\right)\)

Let \(A\) be a nonsingular matrix. Show that \(A^{-1}\) is also nonsingular and \(\left(A^{-1}\right)^{-1}=A\).
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