Linear Algebra and Its Applications · Problem 6.2.7

Positive Definite Matrices — Gilbert Strang

Mathematics · tests-for-positive-definiteness

Problem

A = Q\Lambda Q^T

is symmetric positive definite, then

R = Q\sqrt{\Lambda}Q^T

is its symmetric positive definite square root. Why does

R

have positive eigenvalues? Compute

R

and verify

R^2 = A

for

A = \begin{bmatrix} 10 & 6 \\ 6 & 10 \end{bmatrix} \quad \text{and} \quad A = \begin{bmatrix} 10 & -6 \\ -6 & 10 \end{bmatrix}.