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annas Group Title

question:

  • 2 years ago
  • 2 years ago

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  1. annas Group Title
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    Let L: \[R ^{2} -> R ^{2} \] be the linear transformation defined by \[L[\left(\begin{matrix}a \\ b\end{matrix}\right)]= \left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right] \left(\begin{matrix}a \\ b\end{matrix}\right)\] a)find ker L b) find a set of vectors spanning range L.

    • 2 years ago
  2. annas Group Title
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    just explain the procedure how to do it. @lalaly

    • 2 years ago
  3. UnkleRhaukus Group Title
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    what does ker mean

    • 2 years ago
  4. annas Group Title
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    i guess its just L . ker is a printing mistake sorry for that uncle

    • 2 years ago
  5. UnkleRhaukus Group Title
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    ?

    • 2 years ago
  6. TuringTest Group Title
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    kernel

    • 2 years ago
  7. annas Group Title
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    maybe i m sure about kernel its only written ker L

    • 2 years ago
  8. annas Group Title
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    i just need the way to solve it

    • 2 years ago
  9. TuringTest Group Title
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    I'm looking over my notes, I always forget this stuff

    • 2 years ago
  10. annas Group Title
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    no problem

    • 2 years ago
  11. TuringTest Group Title
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    oh, kernel is the same thing as null space for a linear transformation i.e. the set of all \(\vec x\) in \(A\vec x=\vec 0\) is the kernel, so just solve that matrix to find part a)

    • 2 years ago
  12. annas Group Title
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    ok

    • 2 years ago
  13. TuringTest Group Title
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    that is, solve\[ \left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right] \left(\begin{matrix}a \\ b\end{matrix}\right)=0\]

    • 2 years ago
  14. annas Group Title
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    then we'll get two eqs right ???

    • 2 years ago
  15. TuringTest Group Title
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    yes, from which we will get the two vectors that will make up the basis of our kernel

    • 2 years ago
  16. TuringTest Group Title
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    at least I think that is what they want

    • 2 years ago
  17. annas Group Title
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    ok

    • 2 years ago
  18. annas Group Title
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    about what about part B ???

    • 2 years ago
  19. TuringTest Group Title
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    consulting notes again...

    • 2 years ago
  20. annas Group Title
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    ok

    • 2 years ago
  21. TuringTest Group Title
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    hm... I'm perhaps a bit confused because my notes don't seem to agree with wikipedia. I think the dimension of the null space, i.e. the nullity is the kernal in that case, the answer to part a) would be the number of vectors that make up the basis for the null space of your transformation matrix the answer to part b) must then be the basis itself I guess, which is the set of vectors that span \(A\vec x=\vec0\)

    • 2 years ago
  22. annas Group Title
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    ok thank you i got it :) i just need some practice now

    • 2 years ago
  23. TuringTest Group Title
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    yeah, me too :) welcome, good luck

    • 2 years ago
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