Product of any two arbitrary positive definite matrices is positive definite or NOT?

Question

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....

Answer 1

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.

Answer 2

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.).

Answer 3

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.

Answer 4

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

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

Answer 5

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.