Anonymous at Tue, 5 Mar 2024 22:53:11 UTC No. 16058921
>>16058912
99% of people who can solve this hate themselves.
Anonymous at Wed, 6 Mar 2024 00:04:38 UTC No. 16059005
>>16058912
i am the basis for i am BASED
Anonymous at Wed, 6 Mar 2024 01:36:55 UTC No. 16059099
Choose one of the four variables and write it in terms of the other, let's say that a=2b-c+d. Then substitute that in the general form of the polynomials in M. This means that every polynomial in M is of the form:
2b+bx-c+cx^2+d+dx^3
Then, group them in the form ax_by+cz. Namely, every element is of the form:
b(x+2)+c(x^2+1)+d(x^3+1).
This means that M is actually the subspace spanned by the polynomials x+2. x^2+1 and x^3+1.
If you prove that they are linearly independent, the set given above is a basis, if not remove linearly dependent vectors until you get a basis. Can you take it from here? Remember that polynomials are equal to zero if and only if every single one of their coefficients is zero.
Anonymous at Wed, 6 Mar 2024 01:38:47 UTC No. 16059102
>homework thread
Anonymous at Wed, 6 Mar 2024 01:38:58 UTC No. 16059103
>>16059099
small correction, the second polynomial is x^2-1, the procedure is still the same though
Anonymous at Wed, 6 Mar 2024 03:19:57 UTC No. 16059180
>>16058912
2+x, x+2x^2, x^2+x^3
I just chose one to make a - 2b = 0
one to make -2b + c = 0
and one to make c - d = 0
Anonymous at Wed, 6 Mar 2024 04:01:39 UTC No. 16059239
>>16059103
thanks man
Anonymous at Wed, 6 Mar 2024 04:04:35 UTC No. 16059248
>>16059102
yeah whatever buddy linear algebras hard and homework threads should be allowed anyway.
Anonymous at Wed, 6 Mar 2024 05:02:42 UTC No. 16059328
>>16058912
Ok so you're in a polynomial space of size 3 with the condition that a-2b+c-d = 0
Now let's just show that this is a vector space:
1. we can multiply by a constant and still be in P3
Thus it complete under scale
2. we can add two elements and still be in P3
This is it complete under addition
Now let's look up the definition of a basis, I don't know what you textbook says but here's what google says
>it is a a finite space that is linearly independent and spans the enclosing vector space
Ok I think I remember this from my time in college
So uhhh I guess choose a,b,c, and d to satisfy the condition.
1, x, x^2, x^3 are linearly independent and span P3
and uhhhhhhhhhh
so.... the basis of M is a linear independent combination of these that satisfies the condition...
Ok I got frustrated and just asked Claude ai
This makes sense if you think about the equivalence relation of equality, but this method was not obvious to me. I think it has something to do with if two vectors are linearly independent than adding another basis to it is also linearly independent.