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anonymous
 3 years ago
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anonymous
 3 years ago
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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let 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
 3 years ago
Best ResponseYou've already chosen the best response.0just explain the procedure how to do it. @lalaly

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0what does ker mean

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i guess its just L . ker is a printing mistake sorry for that uncle

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe i m sure about kernel its only written ker L

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i just need the way to solve it

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2I'm looking over my notes, I always forget this stuff

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2oh, 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)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2that is, solve\[ \left[\begin{matrix}1 & 2 \\ 2 & 4\end{matrix}\right] \left(\begin{matrix}a \\ b\end{matrix}\right)=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then we'll get two eqs right ???

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2yes, from which we will get the two vectors that will make up the basis of our kernel

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2at least I think that is what they want

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0about what about part B ???

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2consulting notes again...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2hm... 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok thank you i got it :) i just need some practice now

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2yeah, me too :) welcome, good luck
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