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Chapter 13: Problem 40

Show that the sum of two positive (positive definite) operators is positive (positive definite).

Short Answer

Expert verified
Given two positive (positive definite) operators A and B, their sum (A + B) is also a positive (positive definite) operator because for any arbitrary vector x, the inner product of (A(x) + B(x), x) is greater than or equal to zero (or strictly greater than zero for positive definite operators). This is due to the linearity of the inner product and the properties of positive operators: \( (A(x), x) + (B(x), x) \ge 0 \).

Step by step solution

01

Define the variables and operators

Let A and B be two positive (positive definite) operators acting on a linear space V with inner product.
02

Define what we want to show

We want to show that the sum of A and B, denoted as (A + B), is also a positive (positive definite) operator. In other words, we must show that the inner product of (A(x) + B(x), x) is greater than or equal to zero (or strictly greater than zero for positive definite operators) for any arbitrary vector x in V.
03

Calculate the inner product

To show this, we will calculate the inner product of (A(x) + B(x), x). We will use the linearity of the inner product and split the expression into the sum of two inner products as follows: \( (A(x) + B(x), x) = (A(x), x) + (B(x), x) \)
04

Apply properties of positive operators

Since A and B are positive (positive definite) operators, we know that \( (A(x), x) \ge 0 \) and \( (B(x), x) \ge 0 \) (or strictly greater than zero for positive definite operators). Therefore, their sum must also be greater than or equal to zero (or strictly greater than zero for positive definite operators): \( (A(x), x) + (B(x), x) \ge 0 \)
05

Conclude that the sum is a positive operator

Since we have shown that the inner product of (A(x) + B(x), x) is greater than or equal to zero (or strictly greater than zero for positive definite operators), we can conclude that their sum, (A + B), is indeed a positive (positive definite) operator.

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Most popular questions from this chapter

Suppose \(T\) is normal. Prove that (a) \(T\) is self-adjoint if and only if its eigenvalues are real. (b) \(T\) is unitary if and only if its eigenvalues have absolute value 1 (c) \(T\) is positive if and only if its eigenvalues are nonnegative real numbers.

Prove Theorem 13.1: Let \(T\) be a linear operator on an \(n\) -dimensional inner product space \(V\). Then (a) There exists a unique linear operator \(T^{*}\) on \(V\) such that $$\langle T(u), v\rangle=\left\langle u, T^{*}(v)\right\rangle \quad \text { for all } u, v \in V$$ (b) Let \(A\) be the matrix that represents \(T\) relative to an orthonormal basis \(S=\left\\{u_{i}\right\\} .\) Then the conjugate transpose \(A^{*}\) of \(A\) represents \(T^{*}\) in the basis \(S\)

Show that self-adjoint, skew-adjoint, and unitary (orthogonal) operators are normal.

Detcrmine which of the following matrices are positive (positive definitc): (i) \(\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right],(\text { ii })\left[\begin{array}{rr}0 & i \\ -i & 0\end{array}\right],\) (iii) \(\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right],\) (iv) \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right],\left(\begin{array}{ll}v & 1 \\ 1 & 2\end{array}\right],\) (vi) \(\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\)

For each of the following symmetric matrices \(A,\) find an orthogonal matrix \(P\) and a diagonal matrix \(D\) such that \(P^{\prime} A P\) is diagonal: (a) \(A=\left[\begin{array}{rr}1 & 2 \\ 2 & -2\end{array}\right]\) (b) \(A=\left[\begin{array}{rr}5 & 4 \\ 4 & -1\end{array}\right]\) (c) \(A=\left[\begin{array}{rr}7 & 3 \\ 3 & -1\end{array}\right]\)
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