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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?

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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.

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  • $\begingroup$ 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
    Commented Dec 5, 2013 at 8:25
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    $\begingroup$ @Gigili Maybe this could be of interest: en.wikipedia.org/wiki/Sylvester%27s_criterion $\endgroup$
    – hejseb
    Commented Dec 5, 2013 at 8:30
  • $\begingroup$ That answers my first question, thank you. $\endgroup$
    – Gigili
    Commented 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
    Commented 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
    Commented Dec 5, 2013 at 10:07

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