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annas

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  • one year ago
  • one year ago

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  1. annas
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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.

    • one year ago
  2. annas
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    just explain the procedure how to do it. @lalaly

    • one year ago
  3. UnkleRhaukus
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    what does ker mean

    • one year ago
  4. annas
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    i guess its just L . ker is a printing mistake sorry for that uncle

    • one year ago
  5. UnkleRhaukus
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    ?

    • one year ago
  6. TuringTest
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    kernel

    • one year ago
  7. annas
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    maybe i m sure about kernel its only written ker L

    • one year ago
  8. annas
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    i just need the way to solve it

    • one year ago
  9. TuringTest
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    I'm looking over my notes, I always forget this stuff

    • one year ago
  10. annas
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    no problem

    • one year ago
  11. TuringTest
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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)

    • one year ago
  12. annas
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    ok

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

    • one year ago
  14. annas
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    then we'll get two eqs right ???

    • one year ago
  15. TuringTest
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    yes, from which we will get the two vectors that will make up the basis of our kernel

    • one year ago
  16. TuringTest
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    at least I think that is what they want

    • one year ago
  17. annas
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    ok

    • one year ago
  18. annas
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    about what about part B ???

    • one year ago
  19. TuringTest
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    consulting notes again...

    • one year ago
  20. annas
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    ok

    • one year ago
  21. TuringTest
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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 \(A\vec x=\vec0\)

    • one year ago
  22. annas
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    ok thank you i got it :) i just need some practice now

    • one year ago
  23. TuringTest
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    yeah, me too :) welcome, good luck

    • one year ago
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