Can someonehelp me prove the left cancellation property with matrices?

- anonymous

Can someonehelp me prove the left cancellation property with matrices?

- Stacey Warren - Expert brainly.com

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- chestercat

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- Akshay_Budhkar

Not me.. sorry

- anonymous

LOL I know that by now

- Akshay_Budhkar

oh give me a week i will learn matrices before you! :P

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## More answers

- anonymous

Sound like hero! r u guys related

- anonymous

I can show u how they proved right cancellation Give me a sec

- LollyLau

not something worth lolling much.

- Akshay_Budhkar

yea if u show me maybe i can explain

- anonymous

##### 1 Attachment

- Akshay_Budhkar

that seems so simple!

- Akshay_Budhkar

which step do u dont get?

- anonymous

Ya but i dont get what they did

- anonymous

This is right cancellation how do i change it for left cancellation?

- anonymous

What they are doing is saying that if C is invertible, then by definition a matrix C^-1 must exist. Thus, they multiply both sides on the right by C^-1, then use the associative property and kill it. The proof for left is basically the same.

- anonymous

If C is invertible, then \[CA=CB \implies C^{-1}(CA)=C^{-1}(CB)\] \[\implies (C^{-1}C)A=(C^{-1}C)B \implies IA=IB\] \[\implies A=B\] QED :)

- anonymous

LOL Its that simple???????

- Akshay_Budhkar

LOL

- anonymous

Thanks imperialist :P

- anonymous

No problem :)

- Akshay_Budhkar

imperialist is cool hehe

- anonymous

hehe

- anonymous

Haha, thank you Akshay. Just trying to combine two of my great loves, teaching and math. :D

- Akshay_Budhkar

i like the way you tackle tough problems :D won't be surprised to see u a legend soon. Yea this is the best place to do that!

- Akshay_Budhkar

rld613 is typing a reply…

- anonymous

Or not

- Akshay_Budhkar

Lol

- anonymous

LOL Called it first to be ur student

- anonymous

Sorry lost connection for a second

- Akshay_Budhkar

1 second o_O ?

- Akshay_Budhkar

now imperialist is typing a reply…

- Akshay_Budhkar

imperialist is "STILL" typing a reply…

- anonymous

Akshay u r impatient LOL

- anonymous

Haha, if you really want to be my student, I guess I'll take you on board.
I have just been doing math for a long time Akshay. I don't particularly see problems as tough or not, I just like to see them as new ways of looking at things. For example, if I had never done this problem before and saw a proof of right cancellation, my natural inclination would be to wonder if left also works. The solution just grows naturally out of the problem.
And yes, I think a lot when I type, so I type slow, haha :)

- Akshay_Budhkar

i like things to be "FAST" :P

- Akshay_Budhkar

oh cool!

- Akshay_Budhkar

u got a great teacher rld.. go for it

- Akshay_Budhkar

as far as calc or geography or weather is concerned i may help u :P

- anonymous

Yay!!!!! Today is my lucky today, Its not as if imperialist ahs been helping me b4 LOL

- anonymous

Lol

- anonymous

Ok i guess its official. ya dont worry akshay i will still be posting questions

- anonymous

I am a slow student so like I hope u dont regret it :D

- Akshay_Budhkar

i want to help u as well ;) :P anyways gotta help that poor guy there so u guys can discuss the terms and conditions :P bye bye

- anonymous

Haha, there are no slow students. People just learn in different ways; it's the responsibility of the teacher to find the best way for each individual student to learn. My way of teaching is often very rigorous and methodical, and that is very helpful for some people. For instance, I could have just said "the proof for the left is the same of the right by symmetry" but I instead chose to actually write out the whole proof. I'll also often show my entire thought process as I'm solving a problem, like saying why I think of things in certain ways, so that might be of some help too.
And rld, if I ever explain something and you don't understand, feel free to bug me until you do. :)

- Akshay_Budhkar

can i bug you as well?

- Akshay_Budhkar

say like why is 1+1=2?

- anonymous

AKSHAY!!!!!!! BEHAVE

- anonymous

Lol. Russell has a 347 page proof of that, I think he's covered all the bases. Go look it up there. :p

- Akshay_Budhkar

ok ok sorry :P

- anonymous

LOL I dont like to bug :D I usually say thank you andhead over to another person LOL

- anonymous

neways I have another question LOL

- anonymous

Haha, sorry, 379 http://en.wikipedia.org/wiki/Principia_Mathematica#Quotations

- anonymous

LOL akshay go read it now

- anonymous

I gotta prove another thing

- Akshay_Budhkar

LOL, i agree! 1+1 is indeed 2

- anonymous

"The above proposition is occasionally useful."

- Akshay_Budhkar

lol

- anonymous

that A^2=A when A is singular and when A=matrix

- anonymous

*identity matrix

- Akshay_Budhkar

neways its late here, 00;40, i need to go to sleep, what is the time there @rld :P

- anonymous

LOL the same as u

- anonymous

Same time here

- Akshay_Budhkar

oh wow! where r u from guys?

- anonymous

I guess in the same time zone. We are from toronto

- Akshay_Budhkar

Lol yea we r from toronto :D

- anonymous

Duke university in Durham, North Carolina

- Akshay_Budhkar

lemme google it, my geography is very bad

- anonymous

Lol, the US

- Akshay_Budhkar

oh ok :P

- anonymous

I guess just a couple of hours away

- Akshay_Budhkar

really?

- anonymous

I mean more than a couple if u r driving LOL

- razor99

good morning rld

- Akshay_Budhkar

LOL,, i am off now.. @razor are we gonna eat u if u tell us good morning?

- anonymous

Lol, later Akshay.
Anyway, could you repeat your question rld, I didn't really understand what you want to show.

- razor99

good morning rld

- anonymous

|dw:1327211029340:dw|
This can occur if A=identity matrix or singular
y is this so?

- anonymous

Gmorning razor lol
Awful username uve got there LOL

- razor99

is it mornig to you

- anonymous

quarter to one here so i guess u can call it morning

- razor99

so whats the question

- anonymous

I posted it like scroll up a bit

- anonymous

Oh, this one isn't so bad. Suppose that A is invertible. Then we have \[A^2=A \implies A^2A^{-1}=AA^{-1}\] \[\implies A(AA^{-1})=AA^{-1} \implies AI=I\] \[\implies A=I\] Thus, if A is invertible and satisfies A^2=A, then A MUST be the identity matrix.

- anonymous

Now just find an example, such as the zero matrix, that is singular and satisfies A^2=A, and you have shown that A^2=A implies that A is either singular or the identity matrix and you are done.

- anonymous

what is teh zero matrix?

- anonymous

A matrix (of whatever size you want) where all of its entries are zero

- anonymous

LOL oh ok will do:D

- anonymous

ok that is it Thanks

- anonymous

No problem :)

- LollyLau

LOL count: 22

- razor99

http://en.wikipedia.org/wiki/Singular_value
think this link would be a help rld

- anonymous

No, that is not the same thing. There are many many uses of the word singular in mathematics and not all of them are really related to each other. This question uses singular to mean noninvertible, much simpler than dealing with operators on Hilbert spaces!

- anonymous

Thanks @razor

- razor99

i think singular y

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