Prove that inverse of matrix A is symmetric

Question

Suppose $A^T = A$ is a real, $n$ by $n$ matrix.

We want to show that $A^{-1} = (A^{-1})^T$, that is, the inverse is symmetric.

$A^T = A$

$ (A^T)A^{-1} = A A^{-1} = I $. Thus $A^{-1}$ is the right inverse of $A^T$.

$(A^{-1} A)^T = A^T (A^{-1})^T$. Thus $(A^{-1})^T$ is the right inverse of $A^T$.

I'm not quite sure this proof is right since we aren't given that there is a unique right inverse of $(A^T)$.

Is this proof okay, or is there another way to do this?

Answer 1

It is enough to prove this: for any invertible square matrix $A$, we have $$(\mkern1mu{}^{\mathrm t\!}A)^{-1}={}^{\mathrm t\mkern-1.5mu}(A^{-1}) $$

Indeed $$^{\mathrm t\mkern-1.5mu}(A^{-1}){\,}^{\mathrm t\!}A={}^{\mathrm t\mkern-2mu}(A\,A^{-1})={}^{\mathrm t\mkern-1.5mu}I=I,$$ and similarly $$^{\mathrm t\!}A{\,}^{\mathrm t\mkern-1.5mu}(A^{-1})={}^{\mathrm t\mkern-1.5mu}(A^{-1}A)={}^{\mathrm t\mkern-1.5mu}I=I.$$ In particular, if $A$ is symmetric, $\;\mkern1mu{}^{\mathrm t\!}A=A$, so $$A^{-1}={}^{\mathrm t\mkern-1.5mu}(A^{-1}), $$ which means $A^{-1}$ is symmetric.

Answer 2

Let $B = A^{-1}$ so that $A_{ij}B_{jk}=\delta_{ik}$ (where by convention, indices appearing twice are to be summed over).

$$A_{ij}B_{jk}=\delta_{ik} \wedge A_{ji}=A_{ij} \implies \\ A_{ji}B_{jk}=\delta_{jk} \implies B_{jk}A_{ji}=\delta_{ik} \implies \\ B_{jk}A_{ji}B_{in} = \delta_{ik}B_{in} = B_{kn} \implies \\ B_{jk}\left(A_{ji}B_{in}\right) = B_{kn} \implies \\ \implies\delta_{jn} = B_{kn} \implies \\ B_{nk} = B_{kn} $$ so $B$ is symmetric.