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annas

  • 2 years ago

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

  3. UnkleRhaukus
    • 2 years ago
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    what does ker mean

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

  5. UnkleRhaukus
    • 2 years ago
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    ?

  6. TuringTest
    • 2 years ago
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    kernel

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

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

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

  10. annas
    • 2 years ago
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    no problem

  11. TuringTest
    • 2 years ago
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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)

  12. annas
    • 2 years ago
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    ok

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

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

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

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

  17. annas
    • 2 years ago
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    ok

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

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

  20. annas
    • 2 years ago
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    ok

  21. TuringTest
    • 2 years ago
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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 \(A\vec x=\vec0\)

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

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

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