# More questions on psd and pd of matrices

#### crabchef

##### New member
First question: If A is pd, then A^-1 is pd.
1. If A is pd, then there exists a nonsingular matrix P st A=P'P
2. Then A^-1 = (P'P)^-1 = (P^-1) * (P^-1)' = (P^-1)' * (P^-1)
3. If (P^-1)' * (P^-1), then there exists a A^-1 that is positive definite

Second question: If rank(Anxp) = p<n, then A'A is pd.
1. x'(A'A)x = (Ax)'(Ax)
2. (Ax)'(Ax) is a sum of squares which is > 0 as long as Ax does not = 0

Thanks =)

EDIT: I think I figured it out if anyone wants me to post the answers. I deleted the attempts since they look plain stupid in retrospect

Last edited:

#### Fernando Revilla

##### Well-known member
MHB Math Helper
First question: If A is pd, then A^-1 is pd.
Outline of answer: 1. If A is pd, then there exists a nonsingular matrix P st A=P'P
Right.

2. Then A^-1 = (P'P)^-1 = (P^-1) * (P^-1)' = (P^-1)' * (P^-1)
Right. Now, denoting $Q=(P^{-1})^t$ we have $A^{-1}=Q^tQ=Q^tIQ.$

If (P^-1)' * (P^-1), then there exists a A^-1 that is positive definite
This has no sense. There exists a non singular matrix $Q$ such that $A^{-1}=Q^tIQ$ which implies $A^{-1}$ is positive definite.

---------- Post added at 09:56 AM ---------- Previous post was at 09:38 AM ----------

Second question: If rank(Anxp) = p<n, then A'A is pd.
Outline of answer:1. x'(A'A)x = (Ax)'(Ax)
Right.

2. (Ax)'(Ax) is a sum of squares which is > 0 as long as Ax does not = 0
More details: denote $x=(x_1,\ldots,x_p)^t$ and $y=(y_1,\ldots,y_n)^t=Ax$, then $x^t(A^tA)x=(Ax)^t(Ax)=y^ty=y_1^2+\ldots+y_n^2\geq 0$ for all $y$. But $y_1^2+\ldots+y_n^2=0$ if and only if $y=0$ and $Ax=0$ if and only if $x=0$ (why?) which implies $A^tA$ is definite positive.