CSIR JUNE 2011 PART C QUESTION 73 SOLUTION (which are positive definite? 1) $A+B$, 2) $ABA^{*}$, 3) $A^2+I$, 4) $AB$.)

Suppose $A$ and $B$ are $n \times n$ positive definite matrices and $I$ be the $n \times n$ identity matrix. Then which of the following matrices are positive definite?
1) $A+B$,
2) $ABA^{*}$,
3) $A^2+I$,
4) $AB$.
Solution:
Let $A$ be a real symmetric matrix, then $A$ is said to positive definite if it satisfies any of the following equivalent conditions.
i) all its eigenvalues are positive.
ii) $x^t A x > 0$ for all vectors $x \ne 0$.
iii)$<x,Ax> > 0$ for all vectors $x \ne 0$.
iii) A is positive definite if and only if it can be written as $A = R^tR$, where $R$ is possibly a rectangular matrix, with independent columns.
iv) All the principal minors of $A$ are positive. (Please share if you know any other equivalent conditions in the comment below)
Let $A$ and $B$ be two $n \times n$ positive definite matrices. We have $x^t A x>0$ and $x^t B x>0$ for $x \ne 0$. We will solve each given option by each of the above given definition of positive definiteness in order to understand them clearly.
option 1. (True)We have for $x \ne 0$, $$x^t (A+B) x = x^t A x + x^t B x > 0.$$
Therefore option 1 is true.
option 2. (True) We have for $x \ne 0$, $Ax \ne 0$ since $A$ is invertible.
$$<x,ABA^*x> = <A^*x,BA^*x> = <B^*A^*x,A^*x>  \\ = <BAx,Ax> > 0 .$$
Therefore option 2 is true. Since $A$ and $B$ are real symmetric, we have $A^* = A$ and $B^* = B$.
option 3. (True) Let the eigen values of $A$ be $\lambda_1,\lambda_2,\dots,\lambda_n$. Since $A$ is positive definite we have all these eigen values are positive. We observe that the eigen values of $A^2+1$ are $\lambda_1^2+1,\lambda_2^2+1,\dots,\lambda_n^2+1$ which are all positive. Hence $A^2+1$ is positive definite. 
option 4. (False)
The product of two positive definite matrices need not be even symmetric.  In particular, we have $AB$ is symmetric if and only if $A$ and $B$ commutes with each other. Because $$(AB)^* = B^*A^* = A^*B^* = AB.$$ Note that, if $A$ and $B$ commutes with each other, then $A^*$ and $B^*$ commutes with each other. 

To illustrate this, consider the positive definite matrices $$A = \begin{bmatrix}11 & 10 \\ 10 & 10\end{bmatrix}$$ and $$B = \begin{bmatrix}11 & 5 \\ 5 & 10\end{bmatrix}$$ 
text{ Then their product } $$AB = \begin{bmatrix}171&155 \\ 160&150\end{bmatrix}$$ which is not symmetric.

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