1
$\begingroup$

Supposing I have a square real matrix M. If $MM^{T}$ is positive definite, is $M$ invertible?

I came up with the proof $MM^{T}=M\times I \times M^{T}$, that is equal to say $MM^{T}$ is congruent to an identity matrix, where $M$ is the transformation matrix. However, I couldn't prove why $M$ should be non-degenerating. Can anyone gives me a hint on this?

Thanks a lot.

$\endgroup$
1

2 Answers 2

7
$\begingroup$

$MM^T$ is positive definite $\Rightarrow (\det M)^2=\det(MM^T) >0\Rightarrow \det M\ne 0 \Rightarrow M$ is invertible.

$\endgroup$
2
  • $\begingroup$ Thank you Giraffe for your help. $\endgroup$
    – Jack2019
    Jul 31, 2013 at 20:03
  • $\begingroup$ @SoManyProb_for_a_broken_heart.: You are welcome. $\endgroup$ Jul 31, 2013 at 21:33
3
$\begingroup$

A more general approach:

Let $(\lambda ,v)$ be an eigenpair of $M$. Since $M$ is positive definite it holds that $0<[v^TMv]_{1\times 1}$.
Since $\color{grey}{0<}v^TMv=v^T\lambda v=\lambda \Vert v\Vert^2$, it follows that $0<\lambda $.

This proves that all eigenvalues are positive and hence not null.

$\endgroup$
2
  • $\begingroup$ Thanks Git! Now I won't forget to use the property that all eigenvalues are positive. $\endgroup$
    – Jack2019
    Jul 30, 2013 at 21:20
  • $\begingroup$ @SoManyProb_for_a_broken_heart. No problem. $\endgroup$
    – Git Gud
    Jul 30, 2013 at 21:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.