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?????????????

it looks like your on the right track, but i can not remember this stuff

thank goodness! I was searching everywhere for someone to help me see if I'm on the right track.

I'm afraid this might be over my head :c

AWWW ffdsfkadjfksldsjfkadsjfoiweuro

: (

have you done proof writing before or no?

mmmm not so much... and i don't have a very good grasp on matrices yet i'm afraid :P darn.

Poor lil pony :o

crud. This thing is like due in 11-12 days! I like to plan ahead and such. wahhh

This seems to be on the right track to me.

ok so what do I need to write next for this proof or is that it?

have you learned that a matrix A is nonsingular if and only if it's invertible?

invertible?

oh that! yes square matrix and the AB=Ba=In. yeah I dealt with that before.

if a matrix is singular, there's no inverse, so it's impossible for AB=BA=In to work

ok the switching is a bit nerve wrecking. omg... that's a lot of letters.

Just work through it slowly. You had most of it yourself. You're only missing the last couple lines.

right.

ok so that means that |dw:1361502760161:dw|

ok so B inverse must exist

You already know that B inverse exists. You need to show that A inverse exists.

What can you multiply \[P^{-1}BQ^{-1}\]by to get the identity?

to get if B is nonsingular A is nonsingular. |dw:1361503034980:dw|

I'm not understanding what you're trying to do there. Would you mind elaborating?

so much letters going oN!

hmmm... B = PAQ
A = P^-1BQ^-1

(PAQ)(P^-1BQB^-1) = (P^-1BQB^-1)(PAQ)

:/

Did that make sense?

omg no. I'm lost... :/

ok wait so A = P^-1BQ^-1
that must be Ax = Id

o.o wahhh this is torture.

yeah more work and torture D:

but how do I find x in terms of P? Won't that be P = whatever it is X.

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That's what I was going to suggest next actually.

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yeah sorry...I feel a bit weak...need to eat dinner

No worries. I'll probably be gone when you get back though. Good luck!

@KingGeorge are you going to be on openstudy tomorrow? I had a busy day today and feel tired.

Oh well can I email it to you tomorrow?

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ughhhh so close!!! X( I'm trying to prove the second statement

Sorry don't know this stuff

@KingGeorge I may have gotten it

can you please check if I got this right?

Hold on.

don't divide by matrices, division is not defined. Inverses and division are different.

ok so how do I prove the converse of this

The converse of what statement?

these are the two situations

King George and I did the first one yesterday

Okay, what is meant by A and B are equivalent?

just the second one. I attempted through the drawings last night

Hi can you help me after you help this person please

but the letters are different on the second

I drew the attempt on it last night

I'm not sure if that's it or something

like he will give points but taunt you in the comments

? ? ? I think that's a different question that involves something different.

this is same, B = PAQ what did you prove above and what do you want to prove next? write in short.

I want to prove through the usage of Theorem 2. 13 whatever I posted previously

I already have one done and I attempted the second one (it's drawn)

\[ B = PAQ => P^{-1}BQ^{-1}=A \]
Let \( P^{-1}=P'\) and \(Q^{-1}=Q'\) isn't this same as before?

sorry, I didn't get you ... where do you have problem understanding it?

If matrix equivalence is an equivalence relation, then it follows the transitive property.

really? so A=PBQ then B=P^-1AQ^-1

omg I've been freaking out and stressing for nothing then! -________-

but ok I should put that for if A is nonsingular then B is nonsingular

No ... use symmetry argument.
A = PBQ
B = P'AQ' <-- this is (similar) to first.

you can do this with elementry matricies.

symmetry argument?

write it up like that.

sure that works too.

yes

:D thank you