Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
UsukiDoll
Group Title
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
 one year ago
UsukiDoll Group Title
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
 one year ago

This Question is Closed

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

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

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

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

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
@UnkleRhaukus
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
?????????????
 one year ago

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
@zepdrix
 one year ago

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

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

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

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
@Risadinha
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Poor lil pony :o
 one year ago

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

UsukiDoll Group TitleBest 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.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
This seems to be on the right track to me.
 one year ago

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
invertible?
 one year ago

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
Alright. 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.
 one year ago

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
ok 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
 one year ago

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

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

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

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

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

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

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

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
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 \(X\)?
 one year ago

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

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

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
yeah more work and torture D:
 one year ago

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

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

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

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

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

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
Hold up. Look at it as an algebraic equation like this where you solve for x.\[\frac{1}{p}b\frac{1}{q}x=1\]
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
dw:1361504883640:dw
 one year ago

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
yeah sorry...I feel a bit weak...need to eat dinner
 one year ago

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

UsukiDoll Group TitleBest 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.
 one year ago

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
@ArkGoLucky
 one year ago

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

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
@AriPotta
 one year ago

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
Ok. 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
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
@KingGeorge I may have gotten it
 one year ago

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

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

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

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

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

toxicsugar22 Group TitleBest ResponseYou've already chosen the best response.0
Hi can you help me after you help this person please
 one year ago

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

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

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

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

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

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

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

experimentX Group TitleBest 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?
 one year ago

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

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

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

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

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
symmetry argument?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
sure that works too.
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

UsukiDoll Group TitleBest ResponseYou've already chosen the best response.2
:D thank you
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.