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Let, $A$ & $B$ are $n\times n$ positive definite matrices & $I$ be the $n\times n$ identity matrix. Then which of the followings are positive definite?

(a) $A+B$

(b) $ABA$

(c) $A^{2}+I$

(d) $AB$

I know that, $A^{2}+I$ is positive definite, as if $\lambda$ is an eigen value of $A$ then $(1+\lambda^2)$ is an eigen value of $A^2+I$.

I think (d) is true.Suppose, $\lambda_{1}$ & $\lambda_{2}$ be two eigen values of $A_{2\times 2}$ matrix & $\beta_{1}$, $\beta_{2}$ be two eigen values of $B_{2\times 2}$ matrix.

Now, $det(AB)=det(A)det(B)=\lambda_{1}.\lambda_{2}.\beta_{1}.\beta_{2}.$

As, $\lambda_{1},\lambda_{2},\beta_{1},\beta_{2}$ are all positive so eigen values of $AB$ are all positive, so $AB$ is positive definite.

Similarly, $ABA$ is positive definite.But I am not sure & I have no idea about $A+B$.

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You have some of the right answers, but most of them appear to be for the wrong reasons.

A matrix is positive definite if(f) it is symmetric and has positive eigenvalues. Equivalently, we may state that a matrix $A$ is positive definite if(f) for every vector $x$, we have $$ x^TAx > 0 $$ The answer to your question is that (a), (b), (c) are all necessarily positive definite while (d) is not.

Some corrections to your reasoning: for any eigenvalue $\lambda$ of $A$, $\lambda^2 + 1$ is an eigenvalue of $A^2 + I$. Note also that $A^2 + I$ is symmetric.

Note that just because a matrix has a positive determinant, doesn't mean that the matrix has positive eigenvalues. Also, the eigenvalues of $A$ and $B$ are not generally the eigenvalues of $AB$.

A hint for (d): $AB$ is not necessarily symmetric.

Ben Grossmann
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    @copper.hat try $$ \pmatrix{1\&2},\pmatrix{2&1\1&2} $$ The product might not be Hermitian – Ben Grossmann Nov 07 '14 at 18:23
  • @copper.hat I don't think that set is closed under multiplication, but I'd have to put a bit more thought into that one. Note that positive eigenvalues don't guarantee positive definiteness. – Ben Grossmann Nov 07 '14 at 18:28
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    You are right. I found a counterexample at http://math.stackexchange.com/a/113859/27978. – copper.hat Nov 07 '14 at 18:30
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    @copper.hat I just thought of a nicer one: $$ A = B = \pmatrix{2&3\0&2} $$ – Ben Grossmann Nov 07 '14 at 18:35
  • All definitions I've seen of positive definite are with $x^TAx>0$ for $x\ne0$ (and $A$ symmetric); the fact that this is equivalent to having only positive eigenvalues relies on the deep theorem that normal matrices are diagonalizable with an orthogonal matrix. – egreg Nov 07 '14 at 18:38
  • @egreg in some fields, we forgo the requirement that $A$ be symmetric so that is enough for the symmetric part $\frac 12 (A + A^T)$ to have positive eigenvalues. – Ben Grossmann Nov 07 '14 at 18:40