If $A$ is positive definite then so is $A^k$

Question

I know how to show the inverse of positive definite is positive definite but I don't know how to expand that.

Suppose $A$ is positive definite then $A$ is invertible, so define $y=Ax$ for $x\neq 0$. Then $y^TA^{-1}y=x^TA^TA^{-1}Ax=x^TAx>0$, so the inverse of $A$ is positive definite.

How can I show that for other powers of $A$?

thanasissdr · Accepted Answer · 2015-04-26T23:37:02.423

4

One way to prove this is through eigenvalues. Since $A$ is positive definite, it is a symmetric matrix. $$A \text{ is positive definite } \iff \text{ all eigenvalues are positive. }$$

It is known that for any matrix $M$: $$\lambda \text{ eigenvalue of matrix } M \implies \lambda^k \text{ eigenvalue of matrix } M^k,\quad k=1,2,\ldots$$

Also, you can use the fact that for any invertible matrix $M$: $$\lambda \text{ eigenvalue of matrix } M \implies \dfrac 1{\lambda^k} \text{ eigenvalue of matrix } M^{-k}, \quad k = 1,2,\ldots$$

edited Apr 26 '15 at 23:37

answered Apr 26 '15 at 12:52

thanasissdr

6,348

Got it. Thanks ! – shimura Apr 26 '15 at 13:00
@shimura You are welcome! – thanasissdr Apr 26 '15 at 13:03
@thanasissdr matrix A may not be symmetric,what would you say in case A is not symmetric? – Ibrahim Islam Jul 26 '20 at 11:49
@Ibs By definition a positive definite matrix is symmetric. – thanasissdr Jul 26 '20 at 21:44
@thanasissdr it may be convention of different books,I have studied definition which says that A is positive definite if x'Ax >0 for all x (x' being transpose of X) – Ibrahim Islam Jul 27 '20 at 02:23
@Ibs Maybe this could be a bit more insightful. – thanasissdr Jul 27 '20 at 08:54
@Ibs Or even this. – thanasissdr Jul 27 '20 at 08:59

If $A$ is positive definite then so is $A^k$

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