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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

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

We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^-1BQ^-1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- UsukiDoll

|dw:1361430166380:dw|

- 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

|dw:1361430248917:dw|

- UsukiDoll

|dw:1361430265828:dw|

- UsukiDoll

@UnkleRhaukus

- UsukiDoll

?????????????

- UnkleRhaukus

it looks like your on the right track, but i can not remember this stuff

- UsukiDoll

@joemath314159 and @hartnn

- UsukiDoll

@zepdrix

- UsukiDoll

thank goodness! I was searching everywhere for someone to help me see if I'm on the right track.

- zepdrix

I'm afraid this might be over my head :c

- UsukiDoll

AWWW ffdsfkadjfksldsjfkadsjfoiweuro

- UsukiDoll

: (

- UsukiDoll

have you done proof writing before or no?

- zepdrix

mmmm not so much... and i don't have a very good grasp on matrices yet i'm afraid :P darn.

- UsukiDoll

@Risadinha

- zepdrix

Poor lil pony :o

- UsukiDoll

crud. This thing is like due in 11-12 days! I like to plan ahead and such. wahhh

- 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

This seems to be on the right track to me.

- UsukiDoll

ok so what do I need to write next for this proof or is that it?

- KingGeorge

have you learned that a matrix A is nonsingular if and only if it's invertible?

- UsukiDoll

invertible?

- UsukiDoll

oh that! yes square matrix and the AB=Ba=In. yeah I dealt with that before.

- UsukiDoll

if a matrix is singular, there's no inverse, so it's impossible for AB=BA=In to work

- 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

ok the switching is a bit nerve wrecking. omg... that's a lot of letters.

- KingGeorge

Just work through it slowly. You had most of it yourself. You're only missing the last couple lines.

- 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

right.

- UsukiDoll

ok so that means that |dw:1361502760161:dw|

- UsukiDoll

ok so B inverse must exist

- KingGeorge

You already know that B inverse exists. You need to show that A inverse exists.

- KingGeorge

What can you multiply \[P^{-1}BQ^{-1}\]by to get the identity?

- UsukiDoll

to get if B is nonsingular A is nonsingular. |dw:1361503034980:dw|

- KingGeorge

I'm not understanding what you're trying to do there. Would you mind elaborating?

- 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

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

so much letters going oN!

- 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

hmmm... B = PAQ
A = P^-1BQ^-1

- UsukiDoll

(PAQ)(P^-1BQB^-1) = (P^-1BQB^-1)(PAQ)

- UsukiDoll

:/

- 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

Did that make sense?

- UsukiDoll

omg no. I'm lost... :/

- UsukiDoll

ok wait so A = P^-1BQ^-1
that must be Ax = Id

- UsukiDoll

o.o wahhh this is torture.

- 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

yeah more work and torture D:

- UsukiDoll

but how do I find x in terms of P? Won't that be P = whatever it is X.

- UsukiDoll

|dw:1361504529420:dw|

- KingGeorge

That's what I was going to suggest next actually.

- UsukiDoll

|dw:1361504588920:dw|

- UsukiDoll

|dw:1361504640692:dw|

- UsukiDoll

|dw:1361504663129:dw|

- UsukiDoll

|dw:1361504710718:dw|

- 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

|dw:1361504883640:dw|

- UsukiDoll

|dw:1361504957573:dw|

- 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

yeah sorry...I feel a bit weak...need to eat dinner

- KingGeorge

No worries. I'll probably be gone when you get back though. Good luck!

- UsukiDoll

@KingGeorge are you going to be on openstudy tomorrow? I had a busy day today and feel tired.

- UsukiDoll

Oh well can I email it to you tomorrow?

- UsukiDoll

@ArkGoLucky

- UsukiDoll

|dw:1361519596240:dw|

- UsukiDoll

|dw:1361519710516:dw|

- UsukiDoll

|dw:1361519769721:dw|

- UsukiDoll

@AriPotta

- UsukiDoll

ughhhh so close!!! X( I'm trying to prove the second statement

- anonymous

Sorry don't know this stuff

- 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

@KingGeorge I may have gotten it

- UsukiDoll

@wio

- UsukiDoll

can you please check if I got this right?

- anonymous

Hold on.

- experimentX

don't divide by matrices, division is not defined. Inverses and division are different.

- UsukiDoll

ok so how do I prove the converse of this

- UsukiDoll

- anonymous

The converse of what statement?

- UsukiDoll

If Matrix A is nonsingular then Matrix B is nonsingular
If Matrix B is nonsingular then Matrix A is nonsingular

- UsukiDoll

these are the two situations

- UsukiDoll

King George and I did the first one yesterday

- anonymous

Okay, what is meant by A and B are equivalent?

- UsukiDoll

just the second one. I attempted through the drawings last night

- 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

Hi can you help me after you help this person please

- UsukiDoll

but the letters are different on the second

- UsukiDoll

I drew the attempt on it last night

- UsukiDoll

I'm not sure if that's it or something

- UsukiDoll

no and I can't use fancy stuff that my class didn't learn. I already got chewed once by the professor

- UsukiDoll

like he will give points but taunt you in the comments

- UsukiDoll

- UsukiDoll

- UsukiDoll

- 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

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

? ? ? I think that's a different question that involves something different.

- experimentX

this is same, B = PAQ what did you prove above and what do you want to prove next? write in short.

- 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

I want to prove through the usage of Theorem 2. 13 whatever I posted previously

- UsukiDoll

I already have one done and I attempted the second one (it's drawn)

- experimentX

\[ B = PAQ => P^{-1}BQ^{-1}=A \]
Let \( P^{-1}=P'\) and \(Q^{-1}=Q'\) isn't this same as before?

- UsukiDoll

THat's if B is nonsingular then A is nonsingular since we have the product of nonsingular matrices P and Q

- experimentX

sorry, I didn't get you ... where do you have problem understanding it?

- anonymous

If matrix equivalence is an equivalence relation, then it follows the transitive property.

- 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

really? so A=PBQ then B=P^-1AQ^-1

- 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

omg I've been freaking out and stressing for nothing then! -________-

- UsukiDoll

but ok I should put that for if A is nonsingular then B is nonsingular

- experimentX

No ... use symmetry argument.
A = PBQ
B = P'AQ' <-- this is (similar) to first.

- zzr0ck3r

you can do this with elementry matricies.

- UsukiDoll

symmetry argument?

- 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

write it up like that.

- 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

sure that works too.

- 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

yes

- UsukiDoll

:D thank you

Looking for something else?

Not the answer you are looking for? Search for more explanations.