Prove that if the quadratic form $x^{T}Ax$ is positive definite, so is $x^{T}A^{-1}x$

Question

Is it because the values in the inverse matrix will have the same positive/negative values?

Inverse will have eigenvalues $1/\lambda$ where $\lambda$ is an eigenvalues of $A$. — AnyAD, Jul 04 '14 at 07:12

score 1 · Answer 1 · answered Jul 04 '14 at 07:12

1

Well, A is p.d. implies that A=BB' for some square matrix B (follows immediately from the spectral theorem). Hence, det(A) = det(B).det(B), which being non-zero, det(B) is non-zero too, hence B has an inverse, say C. Then, inv(A)=C'C, hence is p.d.

answered Jul 04 '14 at 07:12

Somabha Mukherjee

2,570

score 1 · Accepted Answer · answered Jul 04 '14 at 07:15

$A$ is positive definite if its eigenvalues are positive. If $\lambda$ is an eigenvalue of $A$ you have

$$Ax=\lambda x\tag{1}$$

for any vector $x$. Multiplying (1) from the left with $A^{-1}$ you obtain

$$A^{-1}Ax=\lambda A^{-1}x\Longrightarrow A^{-1}x=\frac{1}{\lambda}x$$

So the eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$, and if all eigenvalues of $A$ are positive, then so are the eigenvalues of $A^{-1}$. Consequently, $A^{-1}$ is positive definite.

Prove that if the quadratic form $x^{T}Ax$ is positive definite, so is $x^{T}A^{-1}x$

2 Answers2