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This question already has an answer here:

Suppose that , $A$ and $B$ are $n\times n$ positive definite matrices and > $I$ be $n\times n$ identity matrix. Then which of the followings are positive definite ?

(i) $A+B$

(ii) $ABA$

(iii) $A^2+I$

(iv) $AB$

I know the definition of positive definite as : $\color{red}{A_{n\times n}}$ $\color{red}{\text{is positive definite if it's quadratic form}} $ $\color{red}{x^TAx>0}$

Since $A$ and $B$ are positive definite so, $x^TAx>0$ and $x^TBx>0$.

Then, $x^T(A+B)x=x^TAx+x^TBx>0.$ So $A+B$ is positive definite.

I am confused about the product.. I saw a lot of questions in this site about the product of positive definiteness. But the answer in those questions it is assume that the matrices are symmetric. For example see the answer of this question. I want to know whether the product of any two arbitrary positive definite matrices is positive definite or NOT with a valid proof or counter example....

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marked as duplicate by Jean-Claude Arbaut, N. F. Taussig, Batman, Community Apr 25 '15 at 11:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ To be clear: by your definition, positive definite matrices are not necessarily symmetric. Is that correct? $\endgroup$ – Omnomnomnom Apr 24 '15 at 20:27
  • $\begingroup$ Yes.. I know the definition of positive definite as $A$ is positive definite if it's quadratic form $x^TAx>0$ $\endgroup$ – Empty Apr 25 '15 at 2:16
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If you use your definition of 'positive definitive' for matrices,

$$A \text{ is P.D. } \quad\stackrel{def}{\iff}\quad \forall x \ne 0, x^T A x > 0$$

then neither $(ii)$, $(iii)$ nor $(iv)$ works in general.

Let $K = \begin{bmatrix}1.1 & -1\\3 & 1.1\end{bmatrix}$ which is P.D. by your definition.

  • Case $(ii)$, consider $(A,B) = (K,I)$, we have $$ABA = KIK = K^2 = \begin{bmatrix}-1.79 & -2.2\\6.6 & -1.79\end{bmatrix} \implies ABA \text{ not P.D.}$$
  • Case $(iii)$, consider $A = K$, then $$A^2 + I = K^2 + I = \begin{bmatrix}-0.79 & -2.2\\6.6 & -0.79\end{bmatrix} \implies A^2+1 \text{ not P.D.}$$
  • Case $(iv)$, consider $(A,B) = (K,K)$, then $$AB = K^2 =\begin{bmatrix}-1.79 & -2.2\\ 6.6 & -1.79\end{bmatrix} \implies AB \text{ not P.D.}$$

If you include the requirement 'symmetric' to you defintion of 'positive definitive', the situation is much better. Both $(ii)$ and $(iii)$ will be true. However, $(iv)$ is still false, as demonstrated in Ebearr's answer.

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  • $\begingroup$ Ok.Thanks...But what is the actual definition of positive definite matrix ? Is it include ' symmetric ' ? In comment of Batman's answer he says that my definition of positive definiteness is false/... Is it true ? I am confused about the definition.. $\endgroup$ – Empty Apr 25 '15 at 3:29
  • $\begingroup$ @S.Panja-1729 The common "convention" is a real matrix $A$ is positive definitive iff $A$ is symmetric and $x^T A x > 0$ for $x \ne 0$. If you see some paper/book declare a matrix $A$ is positive definitive w/o extra qualification, then $A$ is assumed to be symmetric (or Hermitian in the complex case). $\endgroup$ – achille hui Apr 25 '15 at 3:33
  • $\begingroup$ @achille hui sir the matrix $K $ you considered is not positive definite . so will you please edit your answer? Actually I was looking for a counter example for option 3, and then I saw your answer, but then I saw $K $ is not even positive definite. $\endgroup$ – suchanda adhikari May 9 at 18:51
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    $\begingroup$ @suchandaadhikari fixed, the original $K$ is "positive semi-definite" in OP sense, if you perturb it a little bit to make it "positive definite", it will work. $\endgroup$ – achille hui May 9 at 19:06
  • $\begingroup$ Thank you so much sir, with regards $\endgroup$ – suchanda adhikari May 9 at 19:21
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I think you can rule out $AB$. Consider $$A= \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \quad \text{ and } \quad B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix}.$$ Then $$AB = \begin{bmatrix} 12 & 5 \\ 7 & 3 \end{bmatrix},$$ which isn't even symmetric. It is clear that $A$ and $B$ are both positive definite since they are symmetric and have positive $(1,1)$ entries and determinants.

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    $\begingroup$ My question is for any two arbitrary matrices..Why you take particularly Hermitian matrix..?? That's my problem!!! In every answer matrices are considered as either symmetric or positive definite...Forget symmetric, skew-symmetric, Hermitian, Skew-hermitian all such matrices. Just think for arbitrary matrices $\endgroup$ – Empty Apr 25 '15 at 2:09
  • $\begingroup$ Indeed, according to the definition used in the question, $AB$ in this example is positive definite. $\endgroup$ – celtschk Dec 3 '17 at 5:10
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In general, for symmetric $A$ and $B$, $(AB)^T = B^T A^T = B A \neq AB$. You can generate some random positive definite matrices and see that they don't commute:

In Matlab:

A=randn(5); A=A+A.'; A=A+(abs(min(eig(A)))+1)*eye(size(A))

(and do the same code for getting a matrix B)

Then, calculate AB and BA and chances are, they won't be the same. Thus, product of p.d. matrices is not necessarily p.d. (or even symmetric).

However, if $A,B,A$ are p.d., then $ABA$ is as well since it is clearly symmetric and $x^T A B A x = (A x)^T B (Ax) >0$ for $x \neq 0$ since $B$ is p.d. and $A x \neq 0$ since $A$ is full rank (since it is p.d.).

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  • $\begingroup$ Read my question carefully....My question is for any two arbitrary matrices..Why you take particularly Hermitian matrix..?? That's my problem!!! In every answer matrices are considered as either symmetric or positive definite...Forget symmetric, skew-symmetric, IHermitian, Skew-hermitian all such matrices. Just think for arbitrary matrices. I know what happen for symmetric matrices..That is not necessary in your answer.. $\endgroup$ – Empty Apr 25 '15 at 2:10
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Let $A = \begin{pmatrix} 10 & 6 \\ 0 & 1\end{pmatrix}$ and $B = \mathrm{diag}(10,1)$. Let $x = (x_1,x_2)$, then $x^T Ax = 10 x_1^2 + 6 x_1 x_2 + x_2^2 = x_1^2 + (3x_1 + x_2)^2 \geq 0$ and we have positive definiteness.

$$ B A = \begin{pmatrix} 100 & 60 \\ 0 & 1 \end{pmatrix} $$

Let $y = (1,-2)$ we have

$$ y^T BA y = 100 - 120 + 4 < 0 $$

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    $\begingroup$ $A$ isn't even symmetric. $\endgroup$ – Batman Apr 24 '15 at 19:52
  • $\begingroup$ @Batman Agreed, however see this and this. Bad habit, if you want my opinion :) $\endgroup$ – Jean-Claude Arbaut Apr 24 '15 at 20:20
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I answered this in the comments the last time you asked.

It particular, take $$ A = B = \pmatrix{2&3\\0&2} $$ $A,B$ are positive definite, but $AB$ is not.

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  • $\begingroup$ OK...$AB$ is NOT positive definite for arbitrary matrices..But how $ABA$ and $A^2+I$ are positive definite for arbitrary matrices ? In your this answer $ABA$ and $A^2+I$ are positive definite for $A$ and $B$ are symmetric matrices.. $\endgroup$ – Empty Apr 25 '15 at 2:03
  • $\begingroup$ @S.Panja-1729 check $ABA$ again, with these matrices $A,B$. $A^2+I$ will necessarily be positive definite. $\endgroup$ – Omnomnomnom Apr 25 '15 at 3:16

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