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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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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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  • $\begingroup$ This is a perfect proof, just what I was searching for. $\endgroup$ – Kent Munthe Caspersen Mar 11 '16 at 10:29
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    $\begingroup$ The best proof, exactly where I was puzzled when figuring the proof of OLS estimator properties, thanks!. $\endgroup$ – commentallez-vous Jan 3 at 18:48
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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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$$ (AA^{-1})^T=I^T=I=(A^{-1})^TA^T=(A^{-1})^TA=I\quad\Rightarrow\quad(A^{-1})^T=IA^{-1}\quad\Rightarrow\quad(A^{-1})^T=A^{-1} $$ By definition $$A=A^T$$ Used linear algebra equations $$(AB)^T=B^TA^T;AA^{-1}=I;I^T=I$$

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All the proofs here use algebraic manipulations. But I think it may be more illuminating to think of a symmetric matrix as representing an operator consisting of a rotation, an anisotropic scaling and a rotation back. This is provided by the Spectral theorem, which says that any symmetric matrix is diagonalizable by an orthogonal matrix. With this insight, it is easy to see that the inverse of the operator is a similar three-step sequence. Finally, we need to establish that any such three-step sequence produces a symmetric matrix: given any orthogonal matrix $V$ and diagonal matrix $D$, $(V D V^T)^T = V D V^T$. Hence the inverse of a symmetric matrix is also symmetric.

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Inverse of $A$ can be expressed as a polynomial $p(A)$ of $A$ (from Cayley-Hamilton theorem).

So it is sufficient to prove that if $A$ is symmetric then power $A^k$ is symmetric, sum of symmetric matrices is symmetric and multiply by scalar is symmetric.

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  • $\begingroup$ I.e. for symmetric $A,B$ $(A^k)^T=A^k,(tA)^T=tA, (A+B)^T=A+B, $ $\endgroup$ – Widawensen yesterday

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