Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 28E from Chapter 7.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 28E
Chapter:
Problem:
Suppose A is a symmetric n × n matrix and B is any n × m matrix. Show that BTAB, BTB, and BBT are symmetric matrices.
Step-by-Step Solution
Given Information
We are given that A is a symmetric $n \times n$ matrix and B is any $n \times m$ matrix. We have to show that $B^TAB$, $B^TB$, and $BB^T$ are symmetric matrices.
Step-1:
if matrix A is symmetric then, \[ A ^ { T } = A \] Let us assume, \[ C = B ^ { T } A B \] Apply transpose on both sides: \[\begin{array}{l} {C^T} \equiv {\left( {{B^T}AB} \right)^T}\\ = {B^T}{A^T}{\left( {{B^T}} \right)^T}\\ = {B^T}{A^T}B \end{array}\] Thus, \[ C ^ { T } = C \text { or } \left( B ^ { T } A B \right) ^ { T } = B ^ { T } A B \] Therefore,
The matrix $B^TAB$ is symmetric
Step-2:
Let us assume, \[ D = B ^ { T } B \] Apply transpose on both sides: \[\begin{array}{l} {D^T} = {\left( {{B^T}B} \right)^T}\\ = {B^T}{\left( {{B^T}} \right)^T}\\ = {B^T}B \end{array}\] Thus, \[ \left( B ^ { T } B \right) ^ { T } = B ^ { T } B \] Therefore,
The matrix $B^TB$ is symmetric
Step-3:
Let us assume, \[ P = B B ^ { T } \] Apply transpose on both sides: \[\begin{array}{l} {P^T} = {\left( {B{B^T}} \right)^T}\\ = {\left( {{B^T}} \right)^T}{B^T}\\ = B{B^T}\\ = P \end{array}\] Thus, \[ \left( B B ^ { T } \right) ^ { T } = B B ^ { T } \] Therefore,
The matrix $BB^T$ is symmetric
We are given that A is a symmetric $n \times n$ matrix and B is any $n \times m$ matrix. We have to show that $B^TAB$, $B^TB$, and $BB^T$ are symmetric matrices.
Step-1:
if matrix A is symmetric then, \[ A ^ { T } = A \] Let us assume, \[ C = B ^ { T } A B \] Apply transpose on both sides: \[\begin{array}{l} {C^T} \equiv {\left( {{B^T}AB} \right)^T}\\ = {B^T}{A^T}{\left( {{B^T}} \right)^T}\\ = {B^T}{A^T}B \end{array}\] Thus, \[ C ^ { T } = C \text { or } \left( B ^ { T } A B \right) ^ { T } = B ^ { T } A B \] Therefore,
The matrix $B^TAB$ is symmetric
Step-2:
Let us assume, \[ D = B ^ { T } B \] Apply transpose on both sides: \[\begin{array}{l} {D^T} = {\left( {{B^T}B} \right)^T}\\ = {B^T}{\left( {{B^T}} \right)^T}\\ = {B^T}B \end{array}\] Thus, \[ \left( B ^ { T } B \right) ^ { T } = B ^ { T } B \] Therefore,
The matrix $B^TB$ is symmetric
Step-3:
Let us assume, \[ P = B B ^ { T } \] Apply transpose on both sides: \[\begin{array}{l} {P^T} = {\left( {B{B^T}} \right)^T}\\ = {\left( {{B^T}} \right)^T}{B^T}\\ = B{B^T}\\ = P \end{array}\] Thus, \[ \left( B B ^ { T } \right) ^ { T } = B B ^ { T } \] Therefore,
The matrix $BB^T$ is symmetric