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 25E from Chapter 7.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 25E
Chapter:
Problem:
Show that if B is m × n, then BTB is positive semidefinite; and if B is n × n and invertible, then BTB is positive definite.
Step-by-Step Solution
Given Information
We have to show that if $B$ is $m \times n ,$ then $B ^ { T } B$ is positive semidefinite; and if $B$ is $n \times n$ and invertible, then $B ^ { T } B$ is positive definite.
Step-1:
If we can prove that $B ^ { T } B$ is a symmetric matrix, and the quadratic form $\mathbf { x } ^ { T } \left( B ^ { T } B \right) \mathbf { x } \geq 0$ , for all $\mathbf { x }$ . Then it proves that $B ^ { T } P$ is positive semi definite,
Step-2:
Consider: \[ \begin{aligned} \left( B ^ { T } B \right) ^ { T } & = B ^ { T } \left( B ^ { T } \right) ^ { T } \\ & = B ^ { T } B \end{aligned} \] This proves that the matrix $B ^ { T } B$ is a symmetric matrix.
Step-3:
Consider the quadratic form: $$\begin{aligned} \mathbf { x } ^ { T } \left( B ^ { T } B \right) \mathbf { x } & = \left( \mathbf { x } ^ { T } B ^ { T } \right) B \mathbf { x } \\ & = ( B \mathbf { x } ) ^ { T } B \mathbf { x } \\ & = \| B \mathbf { x } \| ^ { 2 } \end{aligned}$$ Hence, \[ \mathbf { x } ^ { T } \left( B ^ { T } B \right) \mathbf { x } \geq 0 \] Therefore,
The matrix $B ^ { T } B$ is positive semi definite.
Step-4:
Let: \[ \begin{aligned} \mathbf { x } ^ { T } \left( B ^ { T } B \right) \mathbf { x } & = 0 \\ \left( \mathbf { x } ^ { T } B ^ { T } \right) B \mathbf { x } & = 0 \\ ( B \mathbf { x } ) ^ { T } B \mathbf { x } & = 0 \\ \| B \mathbf { x } \| ^ { 2 } & = 0 \end{aligned} \] It is given that the matrix $B$ is invertible, hence the equation, $B \mathbf { x } = 0$ has trivial solution only. Therefore, \[\begin{array}{l} {{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} = 0,{\rm{ then }}{\bf{x}} = {\bf{0}}\\ {\bf{x}} \ne {\bf{0}},{{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} > 0 \end{array}\] Therefore,
The matrix $B ^ { T } B$ is positive definite.
We have to show that if $B$ is $m \times n ,$ then $B ^ { T } B$ is positive semidefinite; and if $B$ is $n \times n$ and invertible, then $B ^ { T } B$ is positive definite.
Step-1:
If we can prove that $B ^ { T } B$ is a symmetric matrix, and the quadratic form $\mathbf { x } ^ { T } \left( B ^ { T } B \right) \mathbf { x } \geq 0$ , for all $\mathbf { x }$ . Then it proves that $B ^ { T } P$ is positive semi definite,
Step-2:
Consider: \[ \begin{aligned} \left( B ^ { T } B \right) ^ { T } & = B ^ { T } \left( B ^ { T } \right) ^ { T } \\ & = B ^ { T } B \end{aligned} \] This proves that the matrix $B ^ { T } B$ is a symmetric matrix.
Step-3:
Consider the quadratic form: $$\begin{aligned} \mathbf { x } ^ { T } \left( B ^ { T } B \right) \mathbf { x } & = \left( \mathbf { x } ^ { T } B ^ { T } \right) B \mathbf { x } \\ & = ( B \mathbf { x } ) ^ { T } B \mathbf { x } \\ & = \| B \mathbf { x } \| ^ { 2 } \end{aligned}$$ Hence, \[ \mathbf { x } ^ { T } \left( B ^ { T } B \right) \mathbf { x } \geq 0 \] Therefore,
The matrix $B ^ { T } B$ is positive semi definite.
Step-4:
Let: \[ \begin{aligned} \mathbf { x } ^ { T } \left( B ^ { T } B \right) \mathbf { x } & = 0 \\ \left( \mathbf { x } ^ { T } B ^ { T } \right) B \mathbf { x } & = 0 \\ ( B \mathbf { x } ) ^ { T } B \mathbf { x } & = 0 \\ \| B \mathbf { x } \| ^ { 2 } & = 0 \end{aligned} \] It is given that the matrix $B$ is invertible, hence the equation, $B \mathbf { x } = 0$ has trivial solution only. Therefore, \[\begin{array}{l} {{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} = 0,{\rm{ then }}{\bf{x}} = {\bf{0}}\\ {\bf{x}} \ne {\bf{0}},{{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} > 0 \end{array}\] Therefore,
The matrix $B ^ { T } B$ is positive definite.