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Chapter 6: Problem 9

Show that if \(A\) is a symmetric positive definite matrix, then \(A\) is nonsingular and \(A^{-1}\) is also positive definite.

Short Answer

Expert verified
We can show that if A is a symmetric positive definite matrix, then it is nonsingular and its inverse A^{-1} is also positive definite by following these steps: 1. Prove that A is nonsingular by showing its determinant is nonzero. This is because if A were singular, there would exist a nonzero vector x such that Ax = 0 and x^T A x = 0, which contradicts the positive definiteness of A. 2. Show that A^{-1} exists since A is nonsingular and has a nonzero determinant. 3. Prove that A^{-1} is symmetric by showing (A^{-1})^T = A^{-1}. We use the properties (AB)^T=B^TA^T and AA^{-1}=I, along with A's symmetry. 4. Prove that A^{-1} is positive definite. Let y be an arbitrary nonzero vector, and let x = A^{-1}y. Then y^T A^{-1} y = x^T A x, and since A is positive definite, x^T A x > 0, so y^T A^{-1} y > 0. Thus, both A^{-1}'s symmetric and positive definiteness are proved.

Step by step solution

01

Prove that A is nonsingular

In order to prove that matrix A is nonsingular, we need to show that it has a nonzero determinant. Let us consider the determinant of x^T A x for any nonzero vector x. Since A is positive definite, we know that x^TAx > 0. Now, if A were singular, then its determinant det(A) would be zero, which implies that there exists a nonzero vector x such that Ax = 0. In this case, x^T A x = x^T (0) = 0, which is a contradiction with the property of positive definiteness. Therefore, A must be nonsingular, which means det(A) 鈮� 0.
02

Show that A^{-1} exists

Since we have proven that A is nonsingular, we know that its determinant is nonzero. Therefore, the inverse of A, denoted by A^{-1}, exists.
03

Prove that A^{-1} is symmetric

We know that A is symmetric, which means A = A^T. In order to show that A^{-1} is symmetric, we will show that (A^{-1})^T = A^{-1}. Using the property (AB)^T=B^TA^T and given that A is symmetric and AA^{-1}=I, we have: (A^{-1})^T = (A^{-1}A)^T = A^T(A^{-1})^T Since A^{-1}A=I, this becomes: (A^{-1})^T = A(A^{-1})^T Since AA^{-1}=I, and A is symmetric: (A^{-1})^T = A^{-1} Thus, A^{-1} is symmetric.
04

Prove that A^{-1} is positive definite

Finally, we need to show that A^{-1} is positive definite. Let y be an arbitrary nonzero vector. We need to prove that y^T A^{-1} y > 0. Using the definitions of inverse and positive definite, we have: A A^{-1} = I A^{-1} = A^{-T} Now, let x = A^{-1}y, which is a nonzero vector since both A^{-1} and y are non-zero. Then, we can rewrite y^TA^{-1}y as x^TAx, which is: y^T A^{-1} y = x^T A x Since A is positive definite, we know that x^T A x > 0. Therefore, y^T A^{-1} y > 0, proving that A^{-1} is positive definite.

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Most popular questions from this chapter

Let \(A\) and \(C\) be matrices in \(\mathbb{C}^{m \times n}\) and let \(B \in \mathbb{C}^{n \times r}\) Prove each of the following rules: (a) \(\left(A^{H}\right)^{H}=A\) (b) \((\alpha A+\beta C)^{H}=\bar{\alpha} A^{H}+\bar{\beta} C^{H}\) (c) \(\quad(A B)^{H}=B^{H} A^{H}\)

Let $$\begin{aligned} A &=\left(\begin{array}{rrrrr} -2 & 8 & 20 \\ 14 & 19 & 10 \\ 2 & -2 & 1 \end{array}\right) \\ &=\left(\begin{array}{rrr} \frac{3}{5} & -\frac{4}{5} & 0 \\ \frac{4}{5} & \frac{3}{5} & 0 \\ 0 & 0 & 1 \end{array}\right)\left(\begin{array}{rrr} 30 & 0 & 0 \\ 0 & 15 & 0 \\ 0 & 0 & 3 \end{array}\right)\left(\begin{array}{rrr} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \end{array}\right) \end{aligned}$$

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Let \(A\) be a singular \(n \times n\) matrix. Show that \(A^{T} A\) is positive semidefinite, but not positive definite.

Let \(A\) be a symmetric \(n \times n\) matrix with eigenvalues \(\lambda_{1}, \ldots, \lambda_{n} .\) Show that there exists an orthonormal set of vectors \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{n}\right\\}\) such that \\[ \mathbf{x}^{T} A \mathbf{x}=\sum_{i=1}^{n} \lambda_{i}\left(\mathbf{x}^{T} \mathbf{x}_{i}\right)^{2} \\] for each \(\mathbf{x} \in \mathbb{R}^{n}\)
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