anonymous
  • anonymous
Can someonehelp me prove the left cancellation property with matrices?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Akshay_Budhkar
  • Akshay_Budhkar
Not me.. sorry
anonymous
  • anonymous
LOL I know that by now
Akshay_Budhkar
  • Akshay_Budhkar
oh give me a week i will learn matrices before you! :P

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anonymous
  • anonymous
Sound like hero! r u guys related
anonymous
  • anonymous
I can show u how they proved right cancellation Give me a sec
LollyLau
  • LollyLau
not something worth lolling much.
Akshay_Budhkar
  • Akshay_Budhkar
yea if u show me maybe i can explain
anonymous
  • anonymous
1 Attachment
Akshay_Budhkar
  • Akshay_Budhkar
that seems so simple!
Akshay_Budhkar
  • Akshay_Budhkar
which step do u dont get?
anonymous
  • anonymous
Ya but i dont get what they did
anonymous
  • anonymous
This is right cancellation how do i change it for left cancellation?
anonymous
  • 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
  • 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
  • anonymous
LOL Its that simple???????
Akshay_Budhkar
  • Akshay_Budhkar
LOL
anonymous
  • anonymous
Thanks imperialist :P
anonymous
  • anonymous
No problem :)
Akshay_Budhkar
  • Akshay_Budhkar
imperialist is cool hehe
anonymous
  • anonymous
hehe
anonymous
  • anonymous
Haha, thank you Akshay. Just trying to combine two of my great loves, teaching and math. :D
Akshay_Budhkar
  • 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
  • Akshay_Budhkar
rld613 is typing a reply…
anonymous
  • anonymous
Or not
Akshay_Budhkar
  • Akshay_Budhkar
Lol
anonymous
  • anonymous
LOL Called it first to be ur student
anonymous
  • anonymous
Sorry lost connection for a second
Akshay_Budhkar
  • Akshay_Budhkar
1 second o_O ?
Akshay_Budhkar
  • Akshay_Budhkar
now imperialist is typing a reply…
Akshay_Budhkar
  • Akshay_Budhkar
imperialist is "STILL" typing a reply…
anonymous
  • anonymous
Akshay u r impatient LOL
anonymous
  • 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
  • Akshay_Budhkar
i like things to be "FAST" :P
Akshay_Budhkar
  • Akshay_Budhkar
oh cool!
Akshay_Budhkar
  • Akshay_Budhkar
u got a great teacher rld.. go for it
Akshay_Budhkar
  • Akshay_Budhkar
as far as calc or geography or weather is concerned i may help u :P
anonymous
  • anonymous
Yay!!!!! Today is my lucky today, Its not as if imperialist ahs been helping me b4 LOL
anonymous
  • anonymous
Lol
anonymous
  • anonymous
Ok i guess its official. ya dont worry akshay i will still be posting questions
anonymous
  • anonymous
I am a slow student so like I hope u dont regret it :D
Akshay_Budhkar
  • 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
  • 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
  • Akshay_Budhkar
can i bug you as well?
Akshay_Budhkar
  • Akshay_Budhkar
say like why is 1+1=2?
anonymous
  • anonymous
AKSHAY!!!!!!! BEHAVE
anonymous
  • 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
  • Akshay_Budhkar
ok ok sorry :P
anonymous
  • anonymous
LOL I dont like to bug :D I usually say thank you andhead over to another person LOL
anonymous
  • anonymous
neways I have another question LOL
anonymous
  • anonymous
Haha, sorry, 379 http://en.wikipedia.org/wiki/Principia_Mathematica#Quotations
anonymous
  • anonymous
LOL akshay go read it now
anonymous
  • anonymous
I gotta prove another thing
Akshay_Budhkar
  • Akshay_Budhkar
LOL, i agree! 1+1 is indeed 2
anonymous
  • anonymous
"The above proposition is occasionally useful."
Akshay_Budhkar
  • Akshay_Budhkar
lol
anonymous
  • anonymous
that A^2=A when A is singular and when A=matrix
anonymous
  • anonymous
*identity matrix
Akshay_Budhkar
  • Akshay_Budhkar
neways its late here, 00;40, i need to go to sleep, what is the time there @rld :P
anonymous
  • anonymous
LOL the same as u
anonymous
  • anonymous
Same time here
Akshay_Budhkar
  • Akshay_Budhkar
oh wow! where r u from guys?
anonymous
  • anonymous
I guess in the same time zone. We are from toronto
Akshay_Budhkar
  • Akshay_Budhkar
Lol yea we r from toronto :D
anonymous
  • anonymous
Duke university in Durham, North Carolina
Akshay_Budhkar
  • Akshay_Budhkar
lemme google it, my geography is very bad
anonymous
  • anonymous
Lol, the US
Akshay_Budhkar
  • Akshay_Budhkar
oh ok :P
anonymous
  • anonymous
I guess just a couple of hours away
Akshay_Budhkar
  • Akshay_Budhkar
really?
anonymous
  • anonymous
I mean more than a couple if u r driving LOL
razor99
  • razor99
good morning rld
Akshay_Budhkar
  • Akshay_Budhkar
LOL,, i am off now.. @razor are we gonna eat u if u tell us good morning?
anonymous
  • anonymous
Lol, later Akshay. Anyway, could you repeat your question rld, I didn't really understand what you want to show.
razor99
  • razor99
good morning rld
anonymous
  • anonymous
|dw:1327211029340:dw| This can occur if A=identity matrix or singular y is this so?
anonymous
  • anonymous
Gmorning razor lol Awful username uve got there LOL
razor99
  • razor99
is it mornig to you
anonymous
  • anonymous
quarter to one here so i guess u can call it morning
razor99
  • razor99
so whats the question
anonymous
  • anonymous
I posted it like scroll up a bit
anonymous
  • 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
  • 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
  • anonymous
what is teh zero matrix?
anonymous
  • anonymous
A matrix (of whatever size you want) where all of its entries are zero
anonymous
  • anonymous
LOL oh ok will do:D
anonymous
  • anonymous
ok that is it Thanks
anonymous
  • anonymous
No problem :)
LollyLau
  • LollyLau
LOL count: 22
razor99
  • razor99
http://en.wikipedia.org/wiki/Singular_value think this link would be a help rld
anonymous
  • 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
  • anonymous
Thanks @razor
razor99
  • razor99
i think singular y

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