|delta||ihope, ok. a, A, d and D are arbitrary?|
|ihope||Yes, arbitrary real numbers.|
|delta||ihope, did you try something like p(X) = bX^2 + (a-d)X - c, etc?|
|ihope||What would that do?|
|delta||ihope, the discriminant of this trinom is the first term. Being < 0 meaning there's no real root.|
ihope, same for the second. So, the third one is the discriminant of p + q.
|ihope||The third is the discriminant of p+q?|
(b+B)^2 - 4(a+A)(c+C) = b^2 - 4ac + B^2 - 4AC, isn't that?
(p + q)(X) = (b+B)X^2 + (a - d + A - D)X - (c + C).
|delta||So, if bB > 0, you're done.|
Investigate others cases maybe.
|ihope||x^2 + (a+d)x + ad-bc is the polynomial I got the stuff from.|
At least, I think it is.
Also, the existence of a nice definition of e^x is nice, no?
|mike8901||can someone guide me to the solution of this linear algebra question? The question is: If the vector (a,b) is a multiple of the vector(c,d) show that the vector(a,c) is a multiple of (b,d). The hint given(a very vague one) is that if A=[a,b;c,d] then AB=I and BA=I. However, I thought the matrix wouldn't be invertable.|
*wouldn't have an inverse
|seb--||mike8901: what do you mean 'multiple of the vector' ?|
mike8901: (a,b) = k (c,d) for some integer k?
|mike8901||(a,b)=s*(c,d) where s is a scalar multiplier|
|FatalError||mike8901, consider [a,b] = [c,d] and since [c,d] is a multiple of [a,b] we can rewrite it k[a,b] = [ka, kb]|
|mike8901||FatalError: I got that far :P|
then ka=c kb=d
|FatalError||well then you have the vectors [a, ka] = j*[b,kb] -> a*[1,k] = j*b*[1,k]|
|seb--||yes..i see it now too|
|seb--||mike8901: so are you happy?|
|mike8901||not really :/|
[a, ka] = j*[b,kb] how did you get there?
|seb--||mike8901: he worked backward from your goal|
mike8901: do that
mike8901: ASSUME what you are trying to prove and massage it into something obvious
|doyley|uk||DCC SEND MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM|