UsukiDoll
  • UsukiDoll
Can anyone help me with this proof? Am I on the right track? Let A and B be equivalent square matrices. Prove that A is nonsingular if and only if B is nonsingular
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UsukiDoll
  • UsukiDoll
Given: Matrix A and Matrix B are equivalent square matrices. Since we are given a biconditional statement, we need to prove two situations: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular
UsukiDoll
  • UsukiDoll
Let's see.... Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.
UsukiDoll
  • UsukiDoll
We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^-1BQ^-1

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UsukiDoll
  • UsukiDoll
|dw:1361430166380:dw|
UsukiDoll
  • UsukiDoll
Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^-1AQ^-1 because P and Q are products of nonsingular matrices.
UsukiDoll
  • UsukiDoll
|dw:1361430248917:dw|
UsukiDoll
  • UsukiDoll
|dw:1361430265828:dw|
UsukiDoll
  • UsukiDoll
@UnkleRhaukus
UsukiDoll
  • UsukiDoll
?????????????
UnkleRhaukus
  • UnkleRhaukus
it looks like your on the right track, but i can not remember this stuff
UsukiDoll
  • UsukiDoll
@joemath314159 and @hartnn
UsukiDoll
  • UsukiDoll
@zepdrix
UsukiDoll
  • UsukiDoll
thank goodness! I was searching everywhere for someone to help me see if I'm on the right track.
zepdrix
  • zepdrix
I'm afraid this might be over my head :c
UsukiDoll
  • UsukiDoll
AWWW ffdsfkadjfksldsjfkadsjfoiweuro
UsukiDoll
  • UsukiDoll
: (
UsukiDoll
  • UsukiDoll
have you done proof writing before or no?
zepdrix
  • zepdrix
mmmm not so much... and i don't have a very good grasp on matrices yet i'm afraid :P darn.
UsukiDoll
  • UsukiDoll
@Risadinha
zepdrix
  • zepdrix
Poor lil pony :o
UsukiDoll
  • UsukiDoll
crud. This thing is like due in 11-12 days! I like to plan ahead and such. wahhh
UsukiDoll
  • UsukiDoll
@KingGeorge do you know how to write proofs in Linear Algebra? I've done some and I was wondering if you could tell me if I'm on the right track.
KingGeorge
  • KingGeorge
This seems to be on the right track to me.
UsukiDoll
  • UsukiDoll
ok so what do I need to write next for this proof or is that it?
KingGeorge
  • KingGeorge
have you learned that a matrix A is nonsingular if and only if it's invertible?
UsukiDoll
  • UsukiDoll
invertible?
UsukiDoll
  • UsukiDoll
oh that! yes square matrix and the AB=Ba=In. yeah I dealt with that before.
UsukiDoll
  • UsukiDoll
if a matrix is singular, there's no inverse, so it's impossible for AB=BA=In to work
KingGeorge
  • KingGeorge
Alright. Suppose A is nonsingular, and \(A=PBQ\) for some non-singular matrices P,Q. This means that A, P, Q are invertible. Then \[B=P^{-1}AQ^{-1}\]But since \(A^{-1}\) exists, look at the product\[P^{-1}AQ^{-1}QA^{-1}P =BQA^{-1}P=\text{Id}\]So B has an inverse and must be non-singular. This is how I would do it. Just switch A and B to prove the other statement you need.
UsukiDoll
  • UsukiDoll
ok the switching is a bit nerve wrecking. omg... that's a lot of letters.
KingGeorge
  • KingGeorge
Just work through it slowly. You had most of it yourself. You're only missing the last couple lines.
UsukiDoll
  • UsukiDoll
ok let me try ummm... B is nonsingular and B =PAQ for some non-singular matrices P,Q. This means that B, P, and Q are invertiable
KingGeorge
  • KingGeorge
right.
UsukiDoll
  • UsukiDoll
ok so that means that |dw:1361502760161:dw|
UsukiDoll
  • UsukiDoll
ok so B inverse must exist
KingGeorge
  • KingGeorge
You already know that B inverse exists. You need to show that A inverse exists.
KingGeorge
  • KingGeorge
What can you multiply \[P^{-1}BQ^{-1}\]by to get the identity?
UsukiDoll
  • UsukiDoll
to get if B is nonsingular A is nonsingular. |dw:1361503034980:dw|
KingGeorge
  • KingGeorge
I'm not understanding what you're trying to do there. Would you mind elaborating?
UsukiDoll
  • UsukiDoll
I'm trying to prove the other statement. I know there's two statements If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular
UsukiDoll
  • UsukiDoll
Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.
UsukiDoll
  • UsukiDoll
so much letters going oN!
KingGeorge
  • KingGeorge
So can you find some product of matrices such that \(P^{-1}BQ^{-1}\) multiplied by that product of matrices gives you the identity?
UsukiDoll
  • UsukiDoll
hmmm... B = PAQ A = P^-1BQ^-1
UsukiDoll
  • UsukiDoll
(PAQ)(P^-1BQB^-1) = (P^-1BQB^-1)(PAQ)
UsukiDoll
  • UsukiDoll
:/
KingGeorge
  • KingGeorge
I think you're on the right track. You want \[P^{-1}BQ^{-1}X=\text{Id}\]though. So \(BQ^{-1}X=P\) and \(Q^{-1}X=B^{-1}P\). If you continue this one more step, what do you get to be for \(X\)?
KingGeorge
  • KingGeorge
Did that make sense?
UsukiDoll
  • UsukiDoll
omg no. I'm lost... :/
UsukiDoll
  • UsukiDoll
ok wait so A = P^-1BQ^-1 that must be Ax = Id
UsukiDoll
  • UsukiDoll
o.o wahhh this is torture.
KingGeorge
  • KingGeorge
That's the right idea. We're finding an \(X\) such that \(AX=\text{Id}\). So if \[AX=P^{-1}BQ^{-1}X=\text{Id},\]you need to find \(X\) in terms of \(P,Q,\) and \(B\) (and/or their inverses). Make more sense?
UsukiDoll
  • UsukiDoll
yeah more work and torture D:
UsukiDoll
  • UsukiDoll
but how do I find x in terms of P? Won't that be P = whatever it is X.
UsukiDoll
  • UsukiDoll
|dw:1361504529420:dw|
KingGeorge
  • KingGeorge
That's what I was going to suggest next actually.
UsukiDoll
  • UsukiDoll
|dw:1361504588920:dw|
UsukiDoll
  • UsukiDoll
|dw:1361504640692:dw|
UsukiDoll
  • UsukiDoll
|dw:1361504663129:dw|
UsukiDoll
  • UsukiDoll
|dw:1361504710718:dw|
KingGeorge
  • KingGeorge
Hold up. Look at it as an algebraic equation like this where you solve for x.\[\frac{1}{p}b\frac{1}{q}x=1\]
UsukiDoll
  • UsukiDoll
|dw:1361504883640:dw|
UsukiDoll
  • UsukiDoll
|dw:1361504957573:dw|
KingGeorge
  • KingGeorge
Sorry, internet glitched out on me there. But yes. That right. Converting that back into matrix form, you get \[X=PB^{-1}Q\]That make sense?
UsukiDoll
  • UsukiDoll
yeah sorry...I feel a bit weak...need to eat dinner
KingGeorge
  • KingGeorge
No worries. I'll probably be gone when you get back though. Good luck!
UsukiDoll
  • UsukiDoll
@KingGeorge are you going to be on openstudy tomorrow? I had a busy day today and feel tired.
UsukiDoll
  • UsukiDoll
Oh well can I email it to you tomorrow?
UsukiDoll
  • UsukiDoll
@ArkGoLucky
UsukiDoll
  • UsukiDoll
|dw:1361519596240:dw|
UsukiDoll
  • UsukiDoll
|dw:1361519710516:dw|
UsukiDoll
  • UsukiDoll
|dw:1361519769721:dw|
UsukiDoll
  • UsukiDoll
@AriPotta
UsukiDoll
  • UsukiDoll
ughhhh so close!!! X( I'm trying to prove the second statement
anonymous
  • anonymous
Sorry don't know this stuff
UsukiDoll
  • UsukiDoll
Ok. I may have got it. Suppose B is nonsingular and B=PAQ for some non-singular matrices P and Q. That means B, P, and Q are invertible. |dw:1361526378870:dw| |dw:1361526405075:dw| |dw:1361526433824:dw| |dw:1361526465398:dw| Since B^-1 exist we look at the product of nonsingular matrices. |dw:1361526502682:dw| Therefore A has an inverse and it must be nonsingular
UsukiDoll
  • UsukiDoll
@KingGeorge I may have gotten it
UsukiDoll
  • UsukiDoll
@wio
UsukiDoll
  • UsukiDoll
can you please check if I got this right?
anonymous
  • anonymous
Hold on.
experimentX
  • experimentX
don't divide by matrices, division is not defined. Inverses and division are different.
UsukiDoll
  • UsukiDoll
ok so how do I prove the converse of this
UsukiDoll
  • UsukiDoll
Given: Matrix A and Matrix B are equivalent square matrices. Since we are given a biconditional statement, we need to prove two situations: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular
anonymous
  • anonymous
The converse of what statement?
UsukiDoll
  • UsukiDoll
If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular
UsukiDoll
  • UsukiDoll
these are the two situations
UsukiDoll
  • UsukiDoll
King George and I did the first one yesterday
anonymous
  • anonymous
Okay, what is meant by A and B are equivalent?
UsukiDoll
  • UsukiDoll
just the second one. I attempted through the drawings last night
experimentX
  • experimentX
use the same argument as you used on the first one, P and Q are invertible by definition.http://en.wikipedia.org/wiki/Matrix_equivalence
toxicsugar22
  • toxicsugar22
Hi can you help me after you help this person please
UsukiDoll
  • UsukiDoll
but the letters are different on the second
UsukiDoll
  • UsukiDoll
I drew the attempt on it last night
UsukiDoll
  • UsukiDoll
I'm not sure if that's it or something
UsukiDoll
  • UsukiDoll
no and I can't use fancy stuff that my class didn't learn. I already got chewed once by the professor
UsukiDoll
  • UsukiDoll
like he will give points but taunt you in the comments
UsukiDoll
  • UsukiDoll
Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.
UsukiDoll
  • UsukiDoll
We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^-1BQ^-1
UsukiDoll
  • UsukiDoll
Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^-1AQ^-1 because P and Q are products of nonsingular matrices.
experimentX
  • experimentX
I think you earlier asked the question AB is invertible iff both A and B are invertible. Isn't this same as proving it?
UsukiDoll
  • UsukiDoll
the second part. the one with conversely..... I don't know if I got it right even though it says to flip the letters A and B
UsukiDoll
  • UsukiDoll
? ? ? I think that's a different question that involves something different.
experimentX
  • experimentX
this is same, B = PAQ what did you prove above and what do you want to prove next? write in short.
UsukiDoll
  • UsukiDoll
THere's two situations to this: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular
UsukiDoll
  • UsukiDoll
I want to prove through the usage of Theorem 2. 13 whatever I posted previously
UsukiDoll
  • UsukiDoll
I already have one done and I attempted the second one (it's drawn)
experimentX
  • experimentX
\[ B = PAQ => P^{-1}BQ^{-1}=A \] Let \( P^{-1}=P'\) and \(Q^{-1}=Q'\) isn't this same as before?
UsukiDoll
  • UsukiDoll
THat's if B is nonsingular then A is nonsingular since we have the product of nonsingular matrices P and Q
experimentX
  • experimentX
sorry, I didn't get you ... where do you have problem understanding it?
anonymous
  • anonymous
If matrix equivalence is an equivalence relation, then it follows the transitive property.
experimentX
  • experimentX
you don't have to go through the rigorous way like you did the first one, Both P^-1 and Q^-1 are invertible matrices. If you take P and Q on the side of B, you will get same expression as you did on the first one.
UsukiDoll
  • UsukiDoll
really? so A=PBQ then B=P^-1AQ^-1
experimentX
  • experimentX
yes P and Q are invertible matrices by definition of Equivalent Matrices. So their inverses are also invertible matrices making the the relation symmetric.
UsukiDoll
  • UsukiDoll
omg I've been freaking out and stressing for nothing then! -________-
UsukiDoll
  • UsukiDoll
but ok I should put that for if A is nonsingular then B is nonsingular
experimentX
  • experimentX
No ... use symmetry argument. A = PBQ B = P'AQ' <-- this is (similar) to first.
zzr0ck3r
  • zzr0ck3r
you can do this with elementry matricies.
UsukiDoll
  • UsukiDoll
symmetry argument?
experimentX
  • experimentX
Theorem 1: A = PBQ if B, P, and Q is non invertible then A is non invertible. B = P'AQ',P', Q' is non invertible, then B is non invertible from theorem 1. Guess not symmetry argument.
experimentX
  • experimentX
write it up like that.
UsukiDoll
  • UsukiDoll
what about... A = PBQ. IF B, P, and Q are non invertible, then A is non invertible. B = P^-1AQ^-1. If P^-1 and Q^-1 are non invertible then B is non invertible.
experimentX
  • experimentX
sure that works too.
UsukiDoll
  • UsukiDoll
so. If B=PAQ. If A, P, and Q are non invertible then B is non invertible A = P^-1BQ^-1. If P^-1 and Q^-1 are non invertible then A is non invertible
experimentX
  • experimentX
yes
UsukiDoll
  • UsukiDoll
:D thank you

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