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

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

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

UnkleRhaukus
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2@joemath314159 and @hartnn

UsukiDoll
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0I'm afraid this might be over my head :c

UsukiDoll
 3 years ago
Best ResponseYou've already chosen the best response.2AWWW ffdsfkadjfksldsjfkadsjfoiweuro

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

zepdrix
 3 years 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
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2This seems to be on the right track to me.

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

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

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

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

KingGeorge
 3 years 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
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2so much letters going oN!

KingGeorge
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2hmmm... B = PAQ A = P^1BQ^1

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

KingGeorge
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2Did that make sense?

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

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

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

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

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

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

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

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

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

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

KingGeorge
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1361504883640:dw

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

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

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

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

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry don't know this stuff

UsukiDoll
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2@KingGeorge I may have gotten it

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

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

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

UsukiDoll
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The converse of what statement?

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

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

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

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

experimentX
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Hi can you help me after you help this person please

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

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

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

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

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

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

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

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

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

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

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

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

experimentX
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1write it up like that.

UsukiDoll
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1sure that works too.

UsukiDoll
 3 years 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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