Is the inverse of a symmetric matrix also symmetric?

Question

Let $A$ be a symmetric invertible matrix, $A^T=A$, $A^{-1}A = A A^{-1} = I$ Can it be shown that $A^{-1}$ is also symmetric?

I seem to remember a proof similar to this from my linear algebra class, but it has been a long time, and I can't find it in my text book.

Answer 1

In fact, $(A^T)^{-1}=(A^{-1})^T$. Indeed, $A^T(A^{-1})^T=(A^{-1}A)^T=I$.

Answer 2

Yes.

$$ AB=BA=I\quad\Rightarrow\quad B^TA^T=A^TB^T=I\quad\Rightarrow\quad B^TA=AB^T=I $$

Answer 3

Another way to see that is to recall the formula $$A^{-1} = \frac{1}{\det(A)} \mathrm{Adj}(A)^T$$ and to note that the adjoint matrix of a symmetric matrix is by construction symmetric.

Answer 4

Yes. The inverse $A^{-1}$ of invertible symmetric matrix is also symmetric:

\begin{align} A & = A^T &&\text{(Assumption: $A$ is symmetric)}\\ \\ A^{-1} & = (A^T)^{-1} &&\text{($A$ invertible $\implies A^T = A$ invertible)}\\ \\ A^{-1} & = (A^{-1})^T &&\text{(Identity: $(A^T)^{-1} = (A^{-1})^T$)} \\ \\ {\large \therefore}\quad \rlap{\text{If $A$ is symmetric and invertible, then $A^{-1}$ is symmetric.}} \end{align}

Answer 5

Every symmetric matrix $A$ can be decomposed as $VDV^T$, where $V$ is orthogonal($V^T=V^{-1}$)

$$\begin{split} A^{-1}&=(VDV^T)^{-1} \\ &=(V^{T})^{-1}D^{-1}V^{-1}\\ &=VD_{1}V^{T} (D_{1}=D^{-1}) \end{split}$$ Inverse of a diagonal matrix can be found by taking reciprocals of all the entries of diagonal matrix. We can see that $A^{-1}$ has the same form as $A$ except that $D$ is replaced by $D_{1}$. So,inverse of a symmetrix matrix is symmetric.