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Let \(A\) be a symmetric positive definite \(n \times n\) matrix and let \(S\) be a nonsingular \(n \times n\) matrix. Show that \(S^{T} A S\) is positive definite

Short Answer

Expert verified
In summary, we showed that the quadratic form \(Q_B(x) = x^{T}(S^T A S)x = y^{T} A y\), where \(y = Sx\), is positive for any non-zero vector x. Since A is positive definite and S is nonsingular, we concluded that B = \(S^T A S\) is also positive definite.

Step by step solution

01

Define the quadratic form for B

Let x be a non-zero vector in 鈩漗n. Define the quadratic form for the matrix B as \(Q_B(x) = x^{T} B x = x^{T} (S^T A S) x\).
02

Substitute B

Now, substitute the expression for B. We have: \(Q_B(x) = x^{T} (S^T A S) x\).
03

Rearrange terms

Next, we will rearrange the terms in the quadratic form to isolate A: \(Q_B(x) = (x^{T} S^T) A (Sx)\).
04

Introduce new variable

Introduce a new variable, y, such that \(y = Sx\). Since S is a nonsingular matrix, this transformation is a one-to-one mapping and y is a non-zero vector if x is non-zero. Now, the quadratic form becomes: \(Q_B(x) = y^{T} A y\).
05

Positive definiteness of A

Given that A is positive definite, its quadratic form with respect to y is positive for any non-zero vector y: \(y^{T} A y > 0\).
06

Conclude positive definiteness of B

Since we have shown that the quadratic form of B is positive for any non-zero vector x, this implies that B is positive definite.

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