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 one year ago
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
 one year ago
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

This Question is Closed

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Given: 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
 one year ago
Best ResponseYou've already chosen the best response.2Let'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
 one year ago
Best ResponseYou've already chosen the best response.2We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^1BQ^1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361430166380:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^1AQ^1 because P and Q are products of nonsingular matrices.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361430248917:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361430265828:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0it looks like your on the right track, but i can not remember this stuff

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2@joemath314159 and @hartnn

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2thank goodness! I was searching everywhere for someone to help me see if I'm on the right track.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0I'm afraid this might be over my head :c

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2AWWW ffdsfkadjfksldsjfkadsjfoiweuro

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2have you done proof writing before or no?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0mmmm not so much... and i don't have a very good grasp on matrices yet i'm afraid :P darn.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2crud. This thing is like due in 1112 days! I like to plan ahead and such. wahhh

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2@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
 one year ago
Best ResponseYou've already chosen the best response.2This seems to be on the right track to me.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ok so what do I need to write next for this proof or is that it?

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2have you learned that a matrix A is nonsingular if and only if it's invertible?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2oh that! yes square matrix and the AB=Ba=In. yeah I dealt with that before.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2if a matrix is singular, there's no inverse, so it's impossible for AB=BA=In to work

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2Alright. Suppose A is nonsingular, and \(A=PBQ\) for some nonsingular 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 nonsingular. This is how I would do it. Just switch A and B to prove the other statement you need.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ok the switching is a bit nerve wrecking. omg... that's a lot of letters.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2Just work through it slowly. You had most of it yourself. You're only missing the last couple lines.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ok let me try ummm... B is nonsingular and B =PAQ for some nonsingular matrices P,Q. This means that B, P, and Q are invertiable

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ok so that means that dw:1361502760161:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ok so B inverse must exist

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2You already know that B inverse exists. You need to show that A inverse exists.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2What can you multiply \[P^{1}BQ^{1}\]by to get the identity?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2to get if B is nonsingular A is nonsingular. dw:1361503034980:dw

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2I'm not understanding what you're trying to do there. Would you mind elaborating?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I'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
 one year ago
Best ResponseYou've already chosen the best response.2Theorem 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
 one year ago
Best ResponseYou've already chosen the best response.2so much letters going oN!

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2So can you find some product of matrices such that \(P^{1}BQ^{1}\) multiplied by that product of matrices gives you the identity?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2hmmm... B = PAQ A = P^1BQ^1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2(PAQ)(P^1BQB^1) = (P^1BQB^1)(PAQ)

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2Did that make sense?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2omg no. I'm lost... :/

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ok wait so A = P^1BQ^1 that must be Ax = Id

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2o.o wahhh this is torture.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2That'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
 one year ago
Best ResponseYou've already chosen the best response.2yeah more work and torture D:

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but how do I find x in terms of P? Won't that be P = whatever it is X.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361504529420:dw

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2That's what I was going to suggest next actually.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361504588920:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361504640692:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361504663129:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361504710718:dw

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2Hold 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361504883640:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361504957573:dw

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2Sorry, 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
 one year ago
Best ResponseYou've already chosen the best response.2yeah sorry...I feel a bit weak...need to eat dinner

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.2No worries. I'll probably be gone when you get back though. Good luck!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2@KingGeorge are you going to be on openstudy tomorrow? I had a busy day today and feel tired.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Oh well can I email it to you tomorrow?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361519596240:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361519710516:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1361519769721:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ughhhh so close!!! X( I'm trying to prove the second statement

ArkGoLucky
 one year ago
Best ResponseYou've already chosen the best response.0Sorry don't know this stuff

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Ok. I may have got it. Suppose B is nonsingular and B=PAQ for some nonsingular 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
 one year ago
Best ResponseYou've already chosen the best response.2@KingGeorge I may have gotten it

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2can you please check if I got this right?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1don't divide by matrices, division is not defined. Inverses and division are different.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ok so how do I prove the converse of this

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Given: 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

wio
 one year ago
Best ResponseYou've already chosen the best response.0The converse of what statement?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2these are the two situations

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2King George and I did the first one yesterday

wio
 one year ago
Best ResponseYou've already chosen the best response.0Okay, what is meant by A and B are equivalent?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2just the second one. I attempted through the drawings last night

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1use 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
 one year ago
Best ResponseYou've already chosen the best response.0Hi can you help me after you help this person please

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but the letters are different on the second

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I drew the attempt on it last night

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I'm not sure if that's it or something

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2no and I can't use fancy stuff that my class didn't learn. I already got chewed once by the professor

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2like he will give points but taunt you in the comments

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Theorem 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
 one year ago
Best ResponseYou've already chosen the best response.2We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^1BQ^1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^1AQ^1 because P and Q are products of nonsingular matrices.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1I think you earlier asked the question AB is invertible iff both A and B are invertible. Isn't this same as proving it?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.2? ? ? I think that's a different question that involves something different.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1this is same, B = PAQ what did you prove above and what do you want to prove next? write in short.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2THere'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
 one year ago
Best ResponseYou've already chosen the best response.2I want to prove through the usage of Theorem 2. 13 whatever I posted previously

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I already have one done and I attempted the second one (it's drawn)

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1\[ B = PAQ => P^{1}BQ^{1}=A \] Let \( P^{1}=P'\) and \(Q^{1}=Q'\) isn't this same as before?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2THat's if B is nonsingular then A is nonsingular since we have the product of nonsingular matrices P and Q

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1sorry, I didn't get you ... where do you have problem understanding it?

wio
 one year ago
Best ResponseYou've already chosen the best response.0If matrix equivalence is an equivalence relation, then it follows the transitive property.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.2really? so A=PBQ then B=P^1AQ^1

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1yes P and Q are invertible matrices by definition of Equivalent Matrices. So their inverses are also invertible matrices making the the relation symmetric.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2omg I've been freaking out and stressing for nothing then! ________

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but ok I should put that for if A is nonsingular then B is nonsingular

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1No ... use symmetry argument. A = PBQ B = P'AQ' < this is (similar) to first.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0you can do this with elementry matricies.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1Theorem 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
 one year ago
Best ResponseYou've already chosen the best response.1write it up like that.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2what 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
 one year ago
Best ResponseYou've already chosen the best response.1sure that works too.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so. 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
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