Solution Found!
We say a symmetric matrix A is positive definite if Ax x > 0 for all x _= 0, negative
Chapter 6, Problem 12(choose chapter or problem)
We say a symmetric matrix A is positive definite if \(A \mathbf{x} \cdot \mathbf{x}>0\) for all \(\mathbf{x} \neq \mathbf{0}\), negative definite if \(A \mathbf{x} \cdot \mathbf{x}<0\) for all \(\mathbf{x} \neq \mathbf{0}\), and positive (resp., negative)semidefinite if \(A \mathbf{x} \cdot \mathbf{x} \geq 0\) (resp., \(\leq\) 0) for all x.
a. Show that if A and B are positive (negative) definite, then so is A + B.
b. Show that A is positive (resp., negative) definite if and only if all its eigenvalues are positive (resp., negative).
c. Show that A is positive (resp., negative) semidefinite if and only if all its eigenvalues are nonnegative (resp., nonpositive).
d. Show that if C is any \(m \times n\) matrix of rank n, then \(A=C^{\top} C\) has positive eigenvalues.
e. Prove or give a counterexample: If A and B are positive definite, then so is AB + BA
Questions & Answers
QUESTION:
We say a symmetric matrix A is positive definite if \(A \mathbf{x} \cdot \mathbf{x}>0\) for all \(\mathbf{x} \neq \mathbf{0}\), negative definite if \(A \mathbf{x} \cdot \mathbf{x}<0\) for all \(\mathbf{x} \neq \mathbf{0}\), and positive (resp., negative)semidefinite if \(A \mathbf{x} \cdot \mathbf{x} \geq 0\) (resp., \(\leq\) 0) for all x.
a. Show that if A and B are positive (negative) definite, then so is A + B.
b. Show that A is positive (resp., negative) definite if and only if all its eigenvalues are positive (resp., negative).
c. Show that A is positive (resp., negative) semidefinite if and only if all its eigenvalues are nonnegative (resp., nonpositive).
d. Show that if C is any \(m \times n\) matrix of rank n, then \(A=C^{\top} C\) has positive eigenvalues.
e. Prove or give a counterexample: If A and B are positive definite, then so is AB + BA
ANSWER:Step 1 of 5
(a) Suppose A and B are positive (resp. negative) definite.
To prove is positive definite.
Since A and B are positive (resp. negative) definite, and for all .
Now, let us consider a vector . For such choice of vector,
Since x is chosen arbitrarily, for all ,
Hence, is positive (resp. negative) definite.