All main minors of a positive definite matrix are positive definite as well and therefore $A$ is strictly invertible.

All I know about positive-definiteness is that for the symmetric matrix $A$ the following inequality holds:

$$x^TAx>0, \forall x(\neq 0) \in \mathbb R^3$$

Could you please give me some hints as to how can I prove the above theorem, specially its second part, what does positive-definiteness have to do with being invertible?


1 Answer 1


I think it's fairly straight-forward actually, naturally depending on how much you want to show. Do you take the following as given?

If $\mathbf{A}$ is non-invertible, then there exists a non-zero vector $\mathbf{x}$ such that $\mathbf{Ax}=\boldsymbol{0}$.

If you do, then it shouldn't be too problematic.

If $\mathbf{A}$ is not invertible, then that means there is a vector such that $\mathbf{Ax}=\boldsymbol{0}$. Thus, $\mathbf{x}^\prime \mathbf{Ax}=\mathbf{x}^\prime \boldsymbol{0}=0$. But, if $\mathbf{A}$ is positive definite, then $\mathbf{x}^\prime \mathbf{Ax}>0$. So $\mathbf{A}$ cannot be both non-invertible and positive definite. Hence, it must be invertible if it is positive definite.

Edit: This only addresses the second part.

  • $\begingroup$ Great, thank you for your detailed answer. I have no problem with the second part now, but the problem with the first part still remains. $\endgroup$
    – Gigili
    Dec 5, 2013 at 8:25
  • 1
    $\begingroup$ @Gigili Maybe this could be of interest: en.wikipedia.org/wiki/Sylvester%27s_criterion $\endgroup$
    – hejseb
    Dec 5, 2013 at 8:30
  • $\begingroup$ That answers my first question, thank you. $\endgroup$
    – Gigili
    Dec 5, 2013 at 9:42
  • $\begingroup$ One last question, when we write $A=\begin{bmatrix}a_{11}&A_{21}^T\\A_{21}&A_{22}\end{bmatrix}$, then $A_{21}$ and $A_{22}$ are minors, right? Could you tell me how they're constructed? $\endgroup$
    – Gigili
    Dec 5, 2013 at 9:47
  • $\begingroup$ @Gigili All of them are minors, but $a_{11}$ and $A_{22}$ are principal minors and $a_{11}$ and $A$ are leading principal minors. A minor $M_{i, j}$ is the matrix with row $i$ and column $j$ deleted. If $i=j$ then it's a principal minor. The leading principal minors are $k\times k$ squares beginning at the top left corner of the original matrix. $a_{11}$ is the $(1, 1)$ element, $A_{21}$ a row vector with the elements from $(2, 1)$ to $(n, 1)$ and $A_{22}$ is the $(n-1)\times (n-1)$ square matrix consisting of the bottom right $(n-1)\times (n-1)$ square of $A$. $\endgroup$
    – hejseb
    Dec 5, 2013 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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