Let A be a symmetric invertible matrix, AT=A, A−1A=AA−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.

**Answer**

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=IT

since AA−1=I,

AA−1=(AA−1)T

since (AB)T=BTAT,

AA−1=(A−1)TAT

since AA−1=A−1A=I, we rearrange the left side

A−1A=(A−1)TAT

since A=AT, we substitute the right side

A−1A=(A−1)TA

A−1A(A−1)=(A−1)TA(A−1)

A−1I=(A−1)TI

A−1=(A−1)T

and we are done.

**Attribution***Source : Link , Question Author : gregmacfarlane , Answer Author : D.Deriso*