0
$\begingroup$

I have the matrix $A$ which is of size $m \times n$, with $m > n$. Let $Q = A^T A$.

How do I determine, if $Q$ is Semi-Positive-definite?

  • 0
    Such $Q = A^T A$ is always positive semi-definite. Did you perhaps mean how to tell if it is not positive definite?2011-01-18

1 Answers 1

3

Recall that $B$ is semi-positive-definite iff $x^T B x\ge 0$ for all vectors $x$, where the superscript $T$ denotes the transpose. In this case, $x^TQx=x^T A^T A x.$ Now, recall that $(Bv)^T=v^T B^T$, so $x^TQ x=(Ax)^T(Ax).$ Finally, note that the square of the norm $\|v\|$ of a vector $v$ is just $\|v\|^2=v^T v$, so $x^T Qx=\|Ax\|^2 \ge0,$ and this gives you that $Q$ is as wanted.

Note that in general this won't be positive definite, because there may be values of $x\ne0$ such that $Ax=0$. (This would be guaranteed if $m. If $m>n$, it may or may not happen, it depends on $A$.)

  • 0
    If $A$ is a matrix of zeros,$Q$is singular whether m < n or not! In general, even if m > n, $A$ must have full row rank for $Q$ to be nonsingular. When m < n, $Q$ is always singular.2011-10-26