anonymous
  • anonymous
question:
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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.
anonymous
  • anonymous
just explain the procedure how to do it. @lalaly
UnkleRhaukus
  • UnkleRhaukus
what does ker mean

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anonymous
  • anonymous
i guess its just L . ker is a printing mistake sorry for that uncle
UnkleRhaukus
  • UnkleRhaukus
?
TuringTest
  • TuringTest
kernel
anonymous
  • anonymous
maybe i m sure about kernel its only written ker L
anonymous
  • anonymous
i just need the way to solve it
TuringTest
  • TuringTest
I'm looking over my notes, I always forget this stuff
anonymous
  • anonymous
no problem
TuringTest
  • TuringTest
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)
anonymous
  • anonymous
ok
TuringTest
  • TuringTest
that is, solve\[ \left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right] \left(\begin{matrix}a \\ b\end{matrix}\right)=0\]
anonymous
  • anonymous
then we'll get two eqs right ???
TuringTest
  • TuringTest
yes, from which we will get the two vectors that will make up the basis of our kernel
TuringTest
  • TuringTest
at least I think that is what they want
anonymous
  • anonymous
ok
anonymous
  • anonymous
about what about part B ???
TuringTest
  • TuringTest
consulting notes again...
anonymous
  • anonymous
ok
TuringTest
  • TuringTest
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 \(A\vec x=\vec0\)
anonymous
  • anonymous
ok thank you i got it :) i just need some practice now
TuringTest
  • TuringTest
yeah, me too :) welcome, good luck

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