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

  • $\begingroup$ 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 '17 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 '17 at 16:56

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.

  • $\begingroup$ I get it, really simple. Thanks $\endgroup$
    – ivana14
    May 19 '17 at 17:00

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}


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.