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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.

Martin Sleziak
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gregmacfarlane
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9 Answers9

184

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 detailed and complete proof.

Given A is nonsingular and symmetric, show that $ A^{-1} = (A^{-1})^T $.

Since $A$ is nonsingular, $A^{-1}$ exists. Since $ I = I^T $ and $ 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 to obtain

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

Since $A$ is symmetric, $ A = A^T $, and we can substitute this into the right side to obtain

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

From here, we see that

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

thus proving the claim.

D.Deriso
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117

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

Seirios
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Yes.

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

Julien
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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.

Zev Chonoles
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Manos
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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}

cqfd
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amWhy
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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.

Lei
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  • Very nice intuitive answer. The reason this is getting fewer upvotes is that it’s posted much later than the others. That doesn’t stop this from being a good answer though. – Benjamin Wang Oct 18 '20 at 20:22
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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 $$I(A^{-1})^{T}=(A^{-1})^{T}=A^{-1}$$

user120050
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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.

Widawensen
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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 $$(AB)^T=B^TA^T;AA^{-1}=I;I^T=I$$

Marat Zakirov
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