Question

How can I prove that for all symmetric matrices, its inverse is also symmetric?

Answer 1

so we know
\[A=A^T\]

Answer 2

one sec

Answer 3

ok lets me think

Answer 4

yea we know \[A^{T}=A\] if symmetric. no problem, take your time. thanks.

Answer 5

god joe is awesome at this stuff

Answer 6

lol, i thought i had it, one sec <.<

Answer 7

i need to go get my linear algebra book

Answer 8

my linear algebra book doesn't have this proof. :(

Answer 9

no zarkon is here we won't be able to do it now because he will
i need to think faster

Answer 10

\[(A^T)^{-1}=(A^{-1})^T\]

Answer 11

add one step to what i have above and you have a proof

Answer 12

\[A^T(A^{-1})^T=(A^{-1}A)^T=I^T=I\]

Answer 13

that proves my statement

Answer 14

But this would only work if A is orthonormal isn't it?

Answer 15

\[(A^{-1})^TA^T=(AA^{-1})^T=I^T=I\]

Answer 16

that are only using the fact that A is symmetrical. It doesnt have to be orthonormal.

Answer 17

A has to be a n by n invertible matrix for the above to hold

Answer 18

it says i spelled invertible wrong

Answer 19

what am i missing

Answer 20

\[A^{-1}=(A^T)^{-1}=(A^{-1})^T\]

Answer 21

invertible i think tthats right

Answer 22

give me a sec...i'm trying to understand it...haha...

Answer 23

but i still don't understand how does this show that the inverse of a symmetric matrix will also be symmetric?

Answer 24

I just showed that the inverse matrix is symmetric...it is equal to it transpose

Answer 25

i would do it like this:
Lets call the inverse of A B instead of A inverse (that -1 floating around everywhere is pissing me off). Then we have:
\[AB = BA = I\]
\[AB= I \iff (AB)^T = I^T \iff B^TA^T = I \iff B^TA = I\]
so we have:
\[B^TA =I , BA = I\]
\[B^TA = BA \iff B^T = B\]
Thus B is symmetric.

Answer 26

that works too...just longer

Answer 27

yeah, yours is good. I just hate that -1 floating around. And I also wanted to do this for myself lolol

Answer 28

@zarkon, for the one you written: \[A^{-1}=(A^{T})^{-1}=(A^{-1})^{T}\]
How can do I know from start that \[A^{-1}\] is \[A^{T}\] from the start?

Answer 29

As in how is \[A^{-1}=(A^{T})^{-1}\]. I am not sure about this part.

Answer 30

@joemath, for the part \[B^TA =I , BA = I\], why is \[B^{T}=B\]?

Answer 31

B is symmetrical

Answer 32

\[A^{-1}=(A)^{-1}\]

\[A=A^T\]

thus

\[A^{-1}=(A)^{-1}=(A^T)^{-1}\]

I'm just replacing A by A^T since they are equal

Answer 33

oh ya you are so right! didn't thought of this! thanks for enlightening me!

Answer 34

for what Joe wrote

\[B^TA =I , BA = I\]

this tells us that

\[B^TA=BA\]

multiply by \[A^{-1}\]

on the right

\[B^TAA^{-1}=BAA^{-1}\]

thus

\[B^T=B\]

Answer 35

thanks for covering for me zarkon :)

Answer 36

ohh now i got what Joe wrote! that's equally brilliant too. thanks guys!! really appreciate your help.