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junior college 1 | H1 Maths
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10b,11b,k=8,help me,ty
Date Posted: 6 years ago
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P2 needs 3 independent polynomials to form a basis; 2x2 matrices needs 4 independent matrices.
Date Posted:
6 years ago
Please show me how to do it, do not show only results
It's explained in the comments.
Why is not p0, p1, p2 or 1 2 3 but p0, p1, p4?
Part b has no relation to the numbers in the matrix in question? Please make it clearer, thanks
Regarding 10b), since we are certain that P0, P1 & P4 form a basis, P2 & P3 are dependant to this basis, so they are no necessary. On how to spot the simplest basis, it's to look for the simplest, namely, one scalar without t, one term with t only and one term with t^2 only. P0 & P1 meet the first 2 requirements; P4 has an extra scalar term, but it can be easily eliminated with a combination with P0. Therefore, P0, P2 & P4 form a basis.
Regarding 11b), 2x2 matrix has 4 variables, so we need 2 more to form a basis, notice the 2 given matrices are linearly independent? If they are dependent to each other, we will need 3 more. On what matrices should these other 2 be? Choose the simplest. The two given by me are the simplest.
Must get the definition of basis of vector space first. Then it's easy to solve these questions. Basis is like the minimum elements in a recipe, short of one, we can't make the soup.
oh,I tried to do it in a formula and failed but now I understand,thank you very much
In lesson 11, k = 8, I wrote in the description
You are welcome. Btw, are you studying in Singapore?
For 10b, other combinations can form a basis of P2, but the one we chose is the simplest. P0, P1 & P2 is a basis too. So is P0, P1 & P3.
Sorry, not P0, P1 & P3, because P3 has no t^2 term.
I am studying in vietnam
The answer is p0, p1, p4?
I can not find a site in vietnam great as this site :)
P0, P1& P4 is just one basis. They are other combinations which can be basis too. Question is asking for ONE basis, not ALL basises.
Glad you find it useful. Enjoy math!
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