# Is the inverse of a symmetric matrix also symmetric?

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.

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

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You can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more complete proof. Given A is nonsingular and symmetric, show that $A^{-1} = (A^{-1})^T$:

$$I = I^T$$

since $AA^{-1} = I$,

$$AA^{-1} = (AA^{-1})^T$$

since $(AB)^T = B^TA^T$,

$$AA^{-1} = (A^{-1})^TA^T$$

since $AA^{-1} = A^{-1}A = I$, we rearrange the left side

$$A^{-1}A = (A^{-1})^TA^T$$

since $A = A^T$, we substitute the right side

$$A^{-1}A = (A^{-1})^TA$$ $$A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$ $$A^{-1}I = (A^{-1})^TI$$ $$A^{-1} = (A^{-1})^T$$

and we are done.

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Yes.

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

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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.

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

\begin{align} A & = A^T \tag{Assumption: $A$ is symmetric}\\ \\ A^{-1} & = (A^T)^{-1} \tag{$A$ invertible $\implies A^T = A$ invertible}\\ \\ A^{-1} & = (A^{-1})^T \tag{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}

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We have two properties to use: $A^{T}=A$ and $A^{-1}$exist, here we go!

$$A^{-1}A=I$$since $A^{-1}$exist $$(A^{-1}A)^{T}=A^{T}(A^{-1})^{T}=A(A^{-1})^{T}=I^{T}=I$$since $A^{T}=A$ $$A^{-1}A(A^{-1})^{T}=A^{-1}I$$left multiple by $A^{-1}$

Thus $$I(A^{-1})^{T}=(A^{-1})^{T}=A^{-1}$$

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