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anonymous
 4 years ago
Can someonehelp me prove the left cancellation property with matrices?
anonymous
 4 years ago
Can someonehelp me prove the left cancellation property with matrices?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL I know that by now

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1oh give me a week i will learn matrices before you! :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sound like hero! r u guys related

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can show u how they proved right cancellation Give me a sec

LollyLau
 4 years ago
Best ResponseYou've already chosen the best response.0not something worth lolling much.

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1yea if u show me maybe i can explain

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1that seems so simple!

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1which step do u dont get?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ya but i dont get what they did

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is right cancellation how do i change it for left cancellation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What 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
 4 years ago
Best ResponseYou've already chosen the best response.0If 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
 4 years ago
Best ResponseYou've already chosen the best response.0LOL Its that simple???????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks imperialist :P

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1imperialist is cool hehe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, thank you Akshay. Just trying to combine two of my great loves, teaching and math. :D

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1i 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
 4 years ago
Best ResponseYou've already chosen the best response.1rld613 is typing a reply…

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL Called it first to be ur student

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry lost connection for a second

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1now imperialist is typing a reply…

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1imperialist is "STILL" typing a reply…

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Akshay u r impatient LOL

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, 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
 4 years ago
Best ResponseYou've already chosen the best response.1i like things to be "FAST" :P

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1u got a great teacher rld.. go for it

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1as far as calc or geography or weather is concerned i may help u :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yay!!!!! Today is my lucky today, Its not as if imperialist ahs been helping me b4 LOL

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok i guess its official. ya dont worry akshay i will still be posting questions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am a slow student so like I hope u dont regret it :D

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1i 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
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, 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
 4 years ago
Best ResponseYou've already chosen the best response.1can i bug you as well?

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1say like why is 1+1=2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lol. Russell has a 347 page proof of that, I think he's covered all the bases. Go look it up there. :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL I dont like to bug :D I usually say thank you andhead over to another person LOL

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0neways I have another question LOL

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, sorry, 379 http://en.wikipedia.org/wiki/Principia_Mathematica#Quotations

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL akshay go read it now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I gotta prove another thing

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1LOL, i agree! 1+1 is indeed 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"The above proposition is occasionally useful."

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that A^2=A when A is singular and when A=matrix

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1neways its late here, 00;40, i need to go to sleep, what is the time there @rld :P

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1oh wow! where r u from guys?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess in the same time zone. We are from toronto

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1Lol yea we r from toronto :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Duke university in Durham, North Carolina

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1lemme google it, my geography is very bad

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess just a couple of hours away

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I mean more than a couple if u r driving LOL

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1LOL,, i am off now.. @razor are we gonna eat u if u tell us good morning?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lol, later Akshay. Anyway, could you repeat your question rld, I didn't really understand what you want to show.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327211029340:dw This can occur if A=identity matrix or singular y is this so?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Gmorning razor lol Awful username uve got there LOL

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0quarter to one here so i guess u can call it morning

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I posted it like scroll up a bit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Now 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
 4 years ago
Best ResponseYou've already chosen the best response.0what is teh zero matrix?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A matrix (of whatever size you want) where all of its entries are zero

razor99
 4 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Singular_value think this link would be a help rld

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, 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!
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