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Chapter 13: Problem 47

Prove that a diagonal matrix \(A\) is positive (positive definite) if and only if every diagonal entry is a nonnegative (positive) real number.

Short Answer

Expert verified
In order to prove that a diagonal matrix A is positive definite if and only if every diagonal entry is a positive real number, we used the definition of positive definite matrices: \(x^T A x > 0\) for any non-zero vector x. For the given diagonal matrix A, we calculated \(x^T A x = x_1^2 a_{11} + x_2^2 a_{22} + x_3^2 a_{33} + \cdots + x_n^2 a_{nn}\) and analyzed the condition for the inequality to hold true. Since the elements \(x_i^2\) are positive, we need each diagonal entry \(a_{ii}\) to be positive in order to satisfy the inequality. Consequently, a diagonal matrix A is positive definite if and only if every diagonal entry is a positive real number.

Step by step solution

01

Understand the definition of positive definite matrix

A matrix A is said to be positive definite if for any non-zero vector x, the following inequality holds true: \[x^T A x > 0\] For diagonal matrices, we will work with the diagonal entries and the above inequality.
02

Consider a diagonal matrix A with elements a_{ij}

Let A be a diagonal matrix with elements a_{ij} such that: \[A = \begin{pmatrix} a_{11} & 0 & 0 & \cdots & 0 \\ 0 & a_{22} & 0 & \cdots & 0 \\ 0 & 0 & a_{33} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{pmatrix}\] Now, consider a non-zero vector x with elements x_i such that: \[x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n \end{pmatrix}\]
03

Calculate \( x^T A x \)

Now, we will calculate \( x^T A x \) and simplify the expression. \(x^T A x = (x_1, x_2, x_3, \dots, x_n) \begin{pmatrix} a_{11} & 0 & 0 & \cdots & 0 \\ 0 & a_{22} & 0 & \cdots & 0 \\ 0 & 0 & a_{33} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n \end{pmatrix} \) \(x^T A x = (x_1 a_{11}, x_2 a_{22}, x_3 a_{33}, \dots, x_n a_{nn}) \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n \end{pmatrix} \) \(x^T A x = x_1^2 a_{11} + x_2^2 a_{22} + x_3^2 a_{33} + \cdots + x_n^2 a_{nn}\)
04

Analyze the condition for positive definiteness

For the matrix A to be positive definite, we need: \[x_1^2 a_{11} + x_2^2 a_{22} + x_3^2 a_{33} + \cdots + x_n^2 a_{nn} > 0\] Since x is a non-zero vector, at least one of the elements \(x_i^2\) is nonzero (in fact, positive, since they are squares). Thus, for the above inequality to hold true, each element \(a_{ii}\) on the main diagonal must be positive, as we can't sum positive and nonpositive values and obtain a positive result. Therefore, we can conclude that a diagonal matrix A is positive definite if and only if every diagonal entry is a positive real number.

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Suppose \(P\) is both positive and unitary. Prove that \(P=I\)

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