# Inverse of a symmetric positive definite matrix

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

• How do you talk about its inverse matrix according to your theorem then you asked if it's invertible or no? May 19, 2017 at 16:54
• 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. May 19, 2017 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.

• I get it, really simple. Thanks May 19, 2017 at 17:00

If A is positive definite matrix, then its eigenvalues are $\lambda_1, \dotsc, \lambda_n >0$ so,

$$|A| = \prod_{i=1}^n \lambda_i > 0$$ 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 $$A^{-1} = (A^T)^{-1}=(A^{-1})^T$$

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.

• I know this isn't crazy different or simpler than the previous. I wanted to leave my mark on the world. Mar 3 at 20:19
• 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." Mar 3 at 23:28