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115.178.29.142 (
talk) 03:25, 20 August 2010 (UTC)
Subtract twice the second row from the first row
Subtract multiples of the third row from the other rows
This matrix is in reduced row echelon form. Bo Jacoby ( talk) 06:42, 20 August 2010 (UTC).
Staecker ( talk) 11:47, 20 August 2010 (UTC)
That doesn't work since the (2,2)th entry x 2 + (1,2)th entry isn't 0; it's 2. [ personal attack removed ] —Preceding unsigned comment added by 110.20.55.15 ( talk) 12:00, 20 August 2010 (UTC)
I've always wanted to find "the" Linear algebra book. A book that would serve me for my entire math life. But I never could find it. Hence I ask the gurus at this reference desk.
My ideal linear algebra book would: (a) Cover all the fundamentals of linear algebra such as what's covered in Herstein's Topics in Algebra in a nice and elegant way (canonical forms, bilinear forms, hermitian matrices, you know advanced topics + the basics) (b) Be significantly more advanced than this and cover topics in linear algebra that pop up deeply in other branches of math (say tensor algebra) (c) Be a pure math textbook.
So is there such a book? I like the style of Herstein's topics in algebra so a book with this style would be desirable. My problem with Herstein's topics in algebra is: it's great and solid but I want a more advanced book. I like my book to give some calculations but also lots of nice theory in a elegant way. + the book should cover as many advanced topics as possible. This is important. It should be comprehensive. If a series of texts is required, that's fine. I don't mind if my book requires math background in general algebra, in fact I want a book like that. Thanks guys ... Oh and I forget to add: this book should be primarily on finite dimensional vector spaces. It's alright if other vector spaces are covered, but I want the emphasis on finite. Thanks guys again. .. —Preceding unsigned comment added by 114.72.248.27 ( talk) 07:29, 20 August 2010 (UTC)
Please answer my question? I request you as politely as I can. Why do people avoid me? What have I done? Please answer my question. I've made my question sound as polite as possible Thanks guys ... 114.72.244.249 ( talk) 08:37, 20 August 2010 (UTC)
Find the values of for which the simultaneous equations:
have solutions other than x=0, y=0
--
220.253.222.146 (
talk) 09:05, 20 August 2010 (UTC)
[ vulgarisms removed ]
For the record, the answer is to use matrixes. It's as imple as 1,2,3 ... [ personal attack removed ] Your system of equations is
k+1 -3
2 -k
and your goal is to determine when this matrix above kills something to 0. (like maps (x,y) to 0) But it does so when it's determinent is 0 and this is like when -k (k+1) = -6 or -k^2 - k + 6 = 0. But the quadratic has a solution = 1 +,- sqrt(1 + 24)/-2. So k = 3 or k = -2. I repeat
<<<<<k = -3 or k = 2>>>>>
Write that on your homework assignment with the advanced matrix stuff and 'all. Your teacher will be very impressed and you'll get an A+. If she's a hot lady, maybe she'll go a bit further than that for your own pleasure ...—Preceding unsigned comment added by 110.20.55.15 ( talk) 12:01, 20 August 2010
I'm just speaking my voice. Maybe I was vulgar in the 80's but nowadays the words I used (I won't state them cause clearly people have problems with it) are common. If you get out more Meni Rosenfeld, you'll see that. And yes, the guy who asked the question is a ******* if he wants his homework done for him. But ******** are ******** and so we might as well give the ******* the answers. I CAN'T see what I've done wrong. I CAN see why other people *THINK* I've done wrong. For some odd reason society frowns down upon telling the truth ... But why shouldn't I voice my opinions? BTW, now that you've told me I'll stop, but all I'm saying is that you guys are all against me. First some dude called Tango removes my question for no particular reason. Then some annonymous guy tries to block me, and then you tell me to "sign my post". Well, I complied. But now people don't answer my questions anymore. I mean what do I need to do to get some service around here? Don't you guys get paid for answering my questions? I'll report you guys to Wikipedia for not doing your job right ... I mean I don't want to, but I just lost my temper. I'm ignored. And some guy called Gandalf insults me above. So everyone hates me. I think I should just leave and save you all the miseries. Goodbye. Block me, I don't really care. But I'm not posting here anymore until you guys start treating me with some respect. —Preceding unsigned comment added by 110.20.55.15 ( talk) 13:28, 20 August 2010 (UTC)
Stop talking bullcrap 92.230.70.110. If you want to see bullshit in it's finest form look above LOL. —Preceding unsigned comment added by 110.20.11.94 ( talk) 00:57, 21 August 2010 (UTC)
In a Yoplait Yogurt commercial, the development of a woman's bone strength is represented as a curve that starts at (0,0) and rises quickly before dropping at a slightly slower rate which decreases, and then approaching the X-axis. Whether this is accurate or not I don't know, particularly since the curve breaks like a bone when it reaches the point where a woman's bones would break easily. But I recall seeing a function like this, which for values of x less than 0, F(-x) = -F(x). What might this function be? Vchimpanzee · talk · contributions · 21:36, 20 August 2010 (UTC)
Hi all!
I'm trying to solve the following:
Let K be a field and c ∈ K. If m, n ∈ Z>0 are coprime, show that Xmn−c is irreducible if and only if both Xm − c and Xn − c are irreducible. (Hint: Use the Tower Law.)
Problem is, I don't really understand how you can use the tower law when you don't know something is irreducible: surely you require irreducibility of f(x) for K[x]/(f(x)) to actually be a field? In which case, you can't really say anything about the degree of the relevant field extensions before you know they're irreducible? And obviously you can't say they're irreducible until you've proved the result, at which point it's moot.
For example, if Kmn, Km and Kn represent K[x]/(xmn-c), and so on, respectively, and we assume Km, Kn are irreducible, then [Kn:K][Km:K]=[Kmn:Km][Km:K]=mn I believe, but we can't then say '[Kmn:K]=mn' because we don't know that's a field necessarily, do we? I'd appreciate a nudge in the right direction, as I've been stuck on this for a while and I think it's probably a very short argument, just one I'm not getting - I believe I've proved Xmn−c is irreducible implies both Xm − c and Xn − c are irreducible via the contrapositive, but I didn't use the tower law, or the fact that m, n are coprime: simply write Y=Xm (assuming Xn-c reducible), then Xmn - c = Yn - c is reducible. Am I doing something wrong here? Or just missing the obvious?
Thankyou very much for any responses, much appreciated indeed - 84.45.219.231 ( talk) 22:52, 20 August 2010 (UTC)
If you can't do Galois theory what hope do you have to suceed in math? So to help you, take the following advice: Compute as many Galois groups as possible. That should aid your disorder. No need to thank me. —Preceding unsigned comment added by 110.20.11.94 ( talk) 01:01, 21 August 2010 (UTC)
Mathematics desk | ||
---|---|---|
< August 19 | << Jul | August | Sep >> | August 21 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
115.178.29.142 (
talk) 03:25, 20 August 2010 (UTC)
Subtract twice the second row from the first row
Subtract multiples of the third row from the other rows
This matrix is in reduced row echelon form. Bo Jacoby ( talk) 06:42, 20 August 2010 (UTC).
Staecker ( talk) 11:47, 20 August 2010 (UTC)
That doesn't work since the (2,2)th entry x 2 + (1,2)th entry isn't 0; it's 2. [ personal attack removed ] —Preceding unsigned comment added by 110.20.55.15 ( talk) 12:00, 20 August 2010 (UTC)
I've always wanted to find "the" Linear algebra book. A book that would serve me for my entire math life. But I never could find it. Hence I ask the gurus at this reference desk.
My ideal linear algebra book would: (a) Cover all the fundamentals of linear algebra such as what's covered in Herstein's Topics in Algebra in a nice and elegant way (canonical forms, bilinear forms, hermitian matrices, you know advanced topics + the basics) (b) Be significantly more advanced than this and cover topics in linear algebra that pop up deeply in other branches of math (say tensor algebra) (c) Be a pure math textbook.
So is there such a book? I like the style of Herstein's topics in algebra so a book with this style would be desirable. My problem with Herstein's topics in algebra is: it's great and solid but I want a more advanced book. I like my book to give some calculations but also lots of nice theory in a elegant way. + the book should cover as many advanced topics as possible. This is important. It should be comprehensive. If a series of texts is required, that's fine. I don't mind if my book requires math background in general algebra, in fact I want a book like that. Thanks guys ... Oh and I forget to add: this book should be primarily on finite dimensional vector spaces. It's alright if other vector spaces are covered, but I want the emphasis on finite. Thanks guys again. .. —Preceding unsigned comment added by 114.72.248.27 ( talk) 07:29, 20 August 2010 (UTC)
Please answer my question? I request you as politely as I can. Why do people avoid me? What have I done? Please answer my question. I've made my question sound as polite as possible Thanks guys ... 114.72.244.249 ( talk) 08:37, 20 August 2010 (UTC)
Find the values of for which the simultaneous equations:
have solutions other than x=0, y=0
--
220.253.222.146 (
talk) 09:05, 20 August 2010 (UTC)
[ vulgarisms removed ]
For the record, the answer is to use matrixes. It's as imple as 1,2,3 ... [ personal attack removed ] Your system of equations is
k+1 -3
2 -k
and your goal is to determine when this matrix above kills something to 0. (like maps (x,y) to 0) But it does so when it's determinent is 0 and this is like when -k (k+1) = -6 or -k^2 - k + 6 = 0. But the quadratic has a solution = 1 +,- sqrt(1 + 24)/-2. So k = 3 or k = -2. I repeat
<<<<<k = -3 or k = 2>>>>>
Write that on your homework assignment with the advanced matrix stuff and 'all. Your teacher will be very impressed and you'll get an A+. If she's a hot lady, maybe she'll go a bit further than that for your own pleasure ...—Preceding unsigned comment added by 110.20.55.15 ( talk) 12:01, 20 August 2010
I'm just speaking my voice. Maybe I was vulgar in the 80's but nowadays the words I used (I won't state them cause clearly people have problems with it) are common. If you get out more Meni Rosenfeld, you'll see that. And yes, the guy who asked the question is a ******* if he wants his homework done for him. But ******** are ******** and so we might as well give the ******* the answers. I CAN'T see what I've done wrong. I CAN see why other people *THINK* I've done wrong. For some odd reason society frowns down upon telling the truth ... But why shouldn't I voice my opinions? BTW, now that you've told me I'll stop, but all I'm saying is that you guys are all against me. First some dude called Tango removes my question for no particular reason. Then some annonymous guy tries to block me, and then you tell me to "sign my post". Well, I complied. But now people don't answer my questions anymore. I mean what do I need to do to get some service around here? Don't you guys get paid for answering my questions? I'll report you guys to Wikipedia for not doing your job right ... I mean I don't want to, but I just lost my temper. I'm ignored. And some guy called Gandalf insults me above. So everyone hates me. I think I should just leave and save you all the miseries. Goodbye. Block me, I don't really care. But I'm not posting here anymore until you guys start treating me with some respect. —Preceding unsigned comment added by 110.20.55.15 ( talk) 13:28, 20 August 2010 (UTC)
Stop talking bullcrap 92.230.70.110. If you want to see bullshit in it's finest form look above LOL. —Preceding unsigned comment added by 110.20.11.94 ( talk) 00:57, 21 August 2010 (UTC)
In a Yoplait Yogurt commercial, the development of a woman's bone strength is represented as a curve that starts at (0,0) and rises quickly before dropping at a slightly slower rate which decreases, and then approaching the X-axis. Whether this is accurate or not I don't know, particularly since the curve breaks like a bone when it reaches the point where a woman's bones would break easily. But I recall seeing a function like this, which for values of x less than 0, F(-x) = -F(x). What might this function be? Vchimpanzee · talk · contributions · 21:36, 20 August 2010 (UTC)
Hi all!
I'm trying to solve the following:
Let K be a field and c ∈ K. If m, n ∈ Z>0 are coprime, show that Xmn−c is irreducible if and only if both Xm − c and Xn − c are irreducible. (Hint: Use the Tower Law.)
Problem is, I don't really understand how you can use the tower law when you don't know something is irreducible: surely you require irreducibility of f(x) for K[x]/(f(x)) to actually be a field? In which case, you can't really say anything about the degree of the relevant field extensions before you know they're irreducible? And obviously you can't say they're irreducible until you've proved the result, at which point it's moot.
For example, if Kmn, Km and Kn represent K[x]/(xmn-c), and so on, respectively, and we assume Km, Kn are irreducible, then [Kn:K][Km:K]=[Kmn:Km][Km:K]=mn I believe, but we can't then say '[Kmn:K]=mn' because we don't know that's a field necessarily, do we? I'd appreciate a nudge in the right direction, as I've been stuck on this for a while and I think it's probably a very short argument, just one I'm not getting - I believe I've proved Xmn−c is irreducible implies both Xm − c and Xn − c are irreducible via the contrapositive, but I didn't use the tower law, or the fact that m, n are coprime: simply write Y=Xm (assuming Xn-c reducible), then Xmn - c = Yn - c is reducible. Am I doing something wrong here? Or just missing the obvious?
Thankyou very much for any responses, much appreciated indeed - 84.45.219.231 ( talk) 22:52, 20 August 2010 (UTC)
If you can't do Galois theory what hope do you have to suceed in math? So to help you, take the following advice: Compute as many Galois groups as possible. That should aid your disorder. No need to thank me. —Preceding unsigned comment added by 110.20.11.94 ( talk) 01:01, 21 August 2010 (UTC)