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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

Which of the matrices that follow are in row echelon form? Which are in reduced row echelon form? (a) \(\left[\begin{array}{llll}1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 2\end{array}\right]\) (b) \(\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)\) (c) \(\left(\begin{array}{lll}1 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right)\) (d) \(\left(\begin{array}{ll}0 & 1 \\ 0 & 0 \\ 0 & 0\end{array}\right)\) (e) \(\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 3\end{array}\right)\) (f) \(\left(\begin{array}{lll}1 & 4 & 6 \\ 0 & 0 & 1 \\ 0 & 1 & 3\end{array}\right)\) (g) \(\left(\begin{array}{lllll}1 & 0 & 0 & 1 & 2 \\ 0 & 1 & 0 & 2 & 4 \\ 0 & 0 & 1 & 3 & 6\end{array}\right)\) (h) \(\left(\begin{array}{llll}0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0\end{array}\right)\) (f) \(\left(\begin{array}{ccc|c}1 & -1 & 3 & 8 \\ 0 & 1 & 2 & 7 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0\end{array}\right)\)

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 \\[ A=\left(\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right) \\] Show that if \(d=a_{11} a_{22}-a_{21} a_{12} \neq 0,\) then \\[ A^{-1}=\frac{1}{d}\left(\begin{array}{rr} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{array}\right) \\]

Let \(A\) be an \(n \times n\) matrix and let \(\mathbf{x}\) and \(\mathbf{y}\) be vectors in \(\mathbb{R}^{n} .\) Show that if \(A \mathbf{x}=A \mathbf{y}\) and \(\mathbf{x} \neq \mathbf{y},\) then the matrix \(A\) must be singular.

Let \(D\) be an \(n \times n\) diagonal matrix whose diagonal entries are either 0 or 1 (a) Show that \(D\) is idempotent. (b) Show that if \(X\) is a nonsingular matrix and \(A=X D X^{-1},\) then \(A\) is idempotent.
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