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I have a questions related to the positive definite[PD] matrix and positive semi definite[PSD] matrix

I see and get the property about PD and PSD

1) PD + PD = PD
2) PSD+ PSD = PSD

how about the positive definite[PD] matrix plus positive semi definite matrix ?

(I mean sum of positive definite matrix and positive semi definite matrix : PD + PSD)

Is it right to be positive definite matrix?


For example, If matrix B is $R \times R$ and it is sum of identity matrix $I$ and symmetry matrix A

that is, $B=I+A$

1) $I=\det(I)=1>0 $ positive definite
2) $X^{T}AX=X^{T}L^{T}LX=U^{T}U=||U||\geqslant 0 $ positive semidefinite

I think that it would be positive definite, I am not so sure...

So I would like to get some help from you

Thank you very much in advance !

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  • $\begingroup$ Hint: $x^T (A+B) x = x^T A x + x^T B x$... $\endgroup$
    – user251257
    Commented Mar 9, 2016 at 14:36
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    $\begingroup$ @user251257 So it depends on $x^{T}(I+A)x= x^{T}Ix+x^{T}Ax >0 $ so it is positive definite! Thank you for your help $\endgroup$ Commented Mar 10, 2016 at 1:19

2 Answers 2

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Let us consider $A$ is PD and $B$ is PSD. Then $x^T(A + B)x = x^TAx + x^TBx > 0$ for $0 \neq x \in R^n.$ Hence $A+B$ is PD matrix.

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(1) PD+PD=PD is right. Let $A,B\in R^{p\times p}$ are PD. For any nonzero $x\in R^p$, $$ x^\top(A+B)x=x^\top Ax+x^\top B x>0\text. $$ (2) PSD+PSD=PSD is not right. One example is that $A=\operatorname{diag}(1,0)$ and $B=\operatorname{diag}(0,1)$. Both of them are PSD, but $A+B=I$ is identity matrix.

(3) PD+PSD=PD should be right. It can be shown as (1).

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