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If a matrix is symmetric and positive definite, determine if it is invertible and if its inverse matrix is symmetric and positive definite.

I know that "if a matrix is symmetric and positive definite, then its inverse matrix is also positive definite", based on a theorem. But I am not sure how to prove that the matrix even is invertible or that its inverse matrix is also symmetric.

It would really help if someone explained this a bit. Thanks

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  • $\begingroup$ How do you talk about its inverse matrix according to your theorem then you asked if it's invertible or no? $\endgroup$
    – user296113
    May 19, 2017 at 16:54
  • $\begingroup$ If a matrix is not invertible, then it send some nonzero vector to zero, preventing it from being positive definite. Further, $(A^{-1})^T = (A^T)^{-1} = A^{-1}$, if $A$ is symmetric and invertible. $\endgroup$
    – florence
    May 19, 2017 at 16:56

3 Answers 3

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We have $(A^{-1})^T = (A^T)^{-1}$ for any invertible matrix. It follows from this that if $A$ is invertible and symmetric $$(A^{-1})^T = (A^T)^{-1} = A^{-1}$$ so $A^{-1}$ is also symmetric. Further, if all eigenvalues of $A$ are positive, then $A^{-1}$ exists and all eigenvalues of $A^{-1}$ are positive since they are the reciprocals of the eigenvalues of $A$. Thus $A^{-1}$ is positive definite when $A$ is positive definite.

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  • $\begingroup$ I get it, really simple. Thanks $\endgroup$
    – ivana14
    May 19, 2017 at 17:00
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If A is positive definite matrix, then its eigenvalues are $\lambda_1, \dotsc, \lambda_n >0$ so,

\begin{equation} |A| = \prod_{i=1}^n \lambda_i > 0 \end{equation} and A is invertible. Moreover, eigenvalues of $A^{-1}$ are $\frac{1}{\lambda_i}>0$, hence $A^{-1}$ is positive definite. To see $A^{-1}$ is symmetric consider \begin{equation} A^{-1} = (A^T)^{-1}=(A^{-1})^T \end{equation}

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If $Q$ is psd, $$y^T Q y ≥ 0 \quad ∀ y \qquad \& \qquad I = Q^{-1}Q$$because all eigenvalues are positive.

Let $x = Q^{-1}y$ for any y,

$$(Q^{-1}y)^T Q Q^{-1}y \geq 0 ⇒ y^T (Q^{-1})^T y \geq 0$$

and we know psd matrices are symmetric.

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  • $\begingroup$ I know this isn't crazy different or simpler than the previous. I wanted to leave my mark on the world. $\endgroup$
    – Will_sharp
    Mar 3 at 20:19
  • $\begingroup$ This sort of manipulation that avoids explicit decomposition into eigenvalues generalizes to infinite dimensions, where one has much weaker spectral decompositions. I think this is different enough that it does more than "leave [your] mark on the world." $\endgroup$ Mar 3 at 23:28

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