Inverse Matrix of Positive-Definite Symmetric Matrix is Positive-Definite
Problem 397
Suppose $A$ is a positive definite symmetric $n\times n$ matrix.
(a) Prove that $A$ is invertible.
(b) Prove that $A^{-1}$ is symmetric.
(c) Prove that $A^{-1}$ is positive-definite.
(MIT, Linear Algebra Exam Problem)
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Proof.
(a) Prove that $A$ is invertible.
Recall that a symmetric matrix is positive-definite if and only if its eigenvalues are all positive.
Thus, since $A$ is positive-definite, the matrix does not have $0$ as an eigenvalue.
Hence $A$ is invertible.
(b) Prove that $A^{-1}$ is symmetric.
By part (a), we know that $A$ is invertible. We have
\[A^{-1}A=I,\]
where $I$ is the $n\times n$ identity matrix.
Taking the transpose, we have
\begin{align*}
&I=I^{\trans}=(A^{-1}A)^{\trans}\\
&=A^{\trans}(A^{-1})^{\trans}\\
&=A(A^{-1})^{\trans} && \text{since $A$ is symmetric}.
\end{align*}
It follows that $A^{-1}=(A^{-1})^{\trans}$, and hence $A^{-1}$ is a symmetric matrix.
(c) Prove that $A^{-1}$ is positive-definite.
Again we use the fact that a symmetric matrix is positive-definite if and only if its eigenvalues are all positive.
(See the post “Positive definite real symmetric matrix and its eigenvalues” for a proof.)
All eigenvalues of $A^{-1}$ are of the form $1/\lambda$, where $\lambda$ is an eigenvalue of $A$.
Since $A$ is positive-definite, each eigenvalue $\lambda$ is positive, hence $1/\lambda$ is positive.
So all eigenvalues of $A^{-1}$ are positive, and it yields that $A^{-1}$ is positive-definite.
Comment.
Note that the above proof of (b) shows the following.
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