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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.


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up vote 2 down vote accepted

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.

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Dear sir B may depend on $A$. Thanks for your answer. – srijan Aug 11 '12 at 7:43

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.)

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I have assumed $A$ to be non zero matrix. I have to edit. – srijan Aug 11 '12 at 7:44
@srijan: being $A$ non zero do not change to much, given the $AB$, $(-A)B$ argument of joriki. – enzotib Aug 11 '12 at 8:06
@enzotib: That's an argument against a universal $B$. If the $B$ is allowed to depend on $A$, as the edited post says, then if $B$ works for $A$, $-B$ works for $-A$. – André Nicolas Aug 11 '12 at 8:11

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.

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Thank you very much sir. It was helpful to me. – srijan Aug 12 '12 at 10:56

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