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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
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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
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.
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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Other answers:

|dw:1361430166380:dw|
Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^-1AQ^-1 because P and Q are products of nonsingular matrices.
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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
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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
@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.
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
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.
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.
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
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?
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
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.
so much letters going oN!
So can you find some product of matrices such that \(P^{-1}BQ^{-1}\) multiplied by that product of matrices gives you the identity?
hmmm... B = PAQ A = P^-1BQ^-1
(PAQ)(P^-1BQB^-1) = (P^-1BQB^-1)(PAQ)
:/
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 \(X\)?
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.
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?
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.
|dw:1361504529420:dw|
That's what I was going to suggest next actually.
|dw:1361504588920:dw|
|dw:1361504640692:dw|
|dw:1361504663129:dw|
|dw:1361504710718:dw|
Hold up. Look at it as an algebraic equation like this where you solve for x.\[\frac{1}{p}b\frac{1}{q}x=1\]
|dw:1361504883640:dw|
|dw:1361504957573:dw|
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?
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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|dw:1361519710516:dw|
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ughhhh so close!!! X( I'm trying to prove the second statement
Sorry don't know this stuff
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
@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
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
The converse of what statement?
If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular
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
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
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
no and I can't use fancy stuff that my class didn't learn. I already got chewed once by the professor
like he will give points but taunt you in the comments
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.
We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^-1BQ^-1
Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^-1AQ^-1 because P and Q are products of nonsingular matrices.
I think you earlier asked the question AB is invertible iff both A and B are invertible. Isn't this same as proving it?
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
? ? ? 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.
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
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?
THat's if B is nonsingular then A is nonsingular since we have the product of nonsingular matrices P and Q
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.
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.
really? so A=PBQ then B=P^-1AQ^-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.
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?
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.
write it up like that.
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.
sure that works too.
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
yes
:D thank you

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