A basic question related with the positive definite matrix I have a one doubt related with positive definite matrices. 
Suppose that we have an arbitrary non zero matrix $A$ . Can we find such matrix $B$ which may depend on $A $such that product $AB$ is always a positive definite matrix irrespective of the nature of matrix $A$? 
I need help with this.
Thanks 
 A: It's not entirely clear from the question whether $B$ can depend on $A$.
If it can, the answer is "almost"; you can choose $B=A^\top$; then
$$x^\top ABx=x^\top AA^\top x=(A^\top x)^\top(A^\top x)\ge0\;,$$
so $AB$ is positive semi-definite, though not definite if $A$ is singular.
If it can't, the answer is "no", since if $AB$ is positive definite then $(-A)B$ is negative definite.
A: Surely not, as you specify,  irrespective of the nature of $A$. Take $A$ to be the zero matrix.
If, as edited, it is specified that $A$ is not the zero matrix, let $A$ be non-invertible.  (But we can find a $B$ such that $AB$ is positive semi-definite.)
A: As a rephrasing of joriki's answer, if you relax the condition "positive definite" to "positive semidefinite", note that any square matrix $\mathbf A$ possesses a (left) polar decomposition, $\mathbf A=\mathbf S\mathbf Q$, where $\mathbf S$ is positive semidefinite and $\mathbf Q$ is unitary. Letting $\mathbf A=\mathbf U\mathbf \Sigma\mathbf V^\ast$ be the singular value decomposition of $\mathbf A$, we have the relations $\mathbf S=\mathbf U\mathbf \Sigma\mathbf U^\ast$ and $\mathbf Q=\mathbf U\mathbf V^\ast$. Thus, $\mathbf B=\mathbf Q^\ast$ is the matrix you are looking for.
