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REMAINDER

  • 2 years ago

find the polynomial in P2 whose coordinate matrix with respect to the basis

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  1. REMAINDER
    • 2 years ago
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    is|dw:1349421000654:dw|

  2. REMAINDER
    • 2 years ago
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    \[B=\left\{ x+x ^{2} ,1+x\right\}\] this is the basis

  3. REMAINDER
    • 2 years ago
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    is it gonna be B=(0,0,0) ;(0,1,1);(0,1,0) |dw:1349421351446:dw| |dw:1349421421254:dw|

  4. Coolsector
    • 2 years ago
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    we need a matrix to make the change between the two basis

  5. REMAINDER
    • 2 years ago
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    how to find that matrix

  6. Coolsector
    • 2 years ago
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    well i have some problem here .. how can B be a basis for some R3 if it has only two vectors

  7. REMAINDER
    • 2 years ago
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    i think the firt vector is zero

  8. Coolsector
    • 2 years ago
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    it's given ?

  9. Coolsector
    • 2 years ago
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    but still i dont think it is a basis in this case

  10. REMAINDER
    • 2 years ago
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    no

  11. Coolsector
    • 2 years ago
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    i might dont understand the question but B isnt a basis for P2

  12. Coolsector
    • 2 years ago
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    we cant generate for example "1" which is a valid P2

  13. REMAINDER
    • 2 years ago
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    yes it is

  14. Coolsector
    • 2 years ago
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    how come ?

  15. Coolsector
    • 2 years ago
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    here i just showed you that it doesnt span P2

  16. REMAINDER
    • 2 years ago
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    we have been told

  17. Coolsector
    • 2 years ago
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    so i dont know .. i dont understand im sorry

  18. REMAINDER
    • 2 years ago
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    ok in general how to anser this kind of questions

  19. Coolsector
    • 2 years ago
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    you have to find a matrix that converts between the two basis and then you only have to multiply

  20. REMAINDER
    • 2 years ago
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    eg like wat i did

  21. Coolsector
    • 2 years ago
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    for example moving from S = { (1,2) , (3,5) } basis into the standard R2 basis the matrix is 1 3 2 5 now if you multiply this matrix with S vectors you get them in the standard R2 basis

  22. Coolsector
    • 2 years ago
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    im sorry i couldnt help more but i really dont know how

  23. REMAINDER
    • 2 years ago
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    ok thnx

  24. SWAG
    • 2 years ago
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    hmm

  25. REMAINDER
    • 2 years ago
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    @Coolsector i din't find the answer ,bul i'll try consult to my lecture

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