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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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6 Answers

up vote 28 down vote accepted

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

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

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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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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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