which of the followings are positive definite:

Question

(CSIR-UGC NET Mathematical Sciences-2011)

Suppose $A,B$ are $n \times n$ positive definite. Then which of the followings are positive definite:

$A+B$

$ABA^{*}$

$A^2+I$

$AB$

Maybe tell us your definition of positive definitenes and where you're stuck. It's probably not the best idea for us to solve the problem for you when it is most likely an immediate application of the definition. — Olivier Bégassat, May 21 '13 at 12:53
$A$ is positive definite $\iff X^tAX>0~\forall~X$ (column vector), that's what I know. Using it I can see that $1$ is true. But what about the others? — Sriti Mallick, May 21 '13 at 12:56
What can you say about the nullspace of a positive definite matrix? Use this to solve 2 and 3. As for 4, $AB$ will not necessarily be symmetric, and you can try and look at the $2\times 2$ case to find counterexamples. — Olivier Bégassat, May 21 '13 at 13:03
Are you sure that in your definition of positive definiteness, you only require that $x^TAx>0$ for all $x\ne0$, but not that $A$ is symmetric? The answer of (3) depends on whether symmetry of $A$ is required. — user1551, May 21 '13 at 13:04
Do studying positive definiteness using the concept of inner product help it to solve easily? — Sriti Mallick, May 21 '13 at 13:13
I've seen it's application here http://math.stackexchange.com/questions/158559/which-are-positive-definite-matrix?rq=1 which I didn't understand. Could anyone suggest any short lecture note/material? — Sriti Mallick, May 21 '13 at 13:15
@SritiMallick In general product need not be positive definite because because for Hermitian $A$ and $B$, $(AB)^{∗}=AB$ if and only if A and B commute. $ABA$ can be positive definite. — Srijan, May 21 '13 at 13:18
@robjohn: Could you tell me whether the concept of inner product help me anyhow? — Sriti Mallick, May 21 '13 at 13:18
@SritiMallick: since $M$ is positive definite when $\langle x,Mx\rangle\gt0$ for all $x\ne0$, I would say that inner products are key. — robjohn, May 21 '13 at 14:20

score 5 · Answer 1 · edited Apr 13 '17 at 12:21

5

Partial answer: Your complete answer can be given by combining these two links

link 1

link 2

and possible my partial answer

If $A, B \in \mathbb{R}^{n\times n}$ are positive definite then $x^t A x > 0$, $x^t B x > 0$, $\forall x \neq 0 $. This implies that $x^t (A + B)x = x^t A x+ x^t B x > 0$. Hence sum of two positive definite matrices positive definite.

edited Apr 13 '17 at 12:21

Community

1

answered May 21 '13 at 12:57

Srijan

12,518
10
73
115

score 1 · Answer 2 · edited May 23 '13 at 10:02

1

case #4 has a simple counterexample (if $A$ is not necessarily symmetric or Hermit):

$A = B = \begin{pmatrix} 1 & 1\\ -1&1 \end{pmatrix}$,

$x = (0, 1)$.

Since $z^{T}Az= z_1^2+z_2^2 ,A$ is positive defined; however

x^TABx = xTAAx = 0

edited May 23 '13 at 10:02

learner

6,726

answered May 21 '13 at 14:09

Dmitry Bychenko

316

which of the followings are positive definite:

2 Answers2