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annasBest ResponseYou've already chosen the best response.1
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

annasBest ResponseYou've already chosen the best response.1
just explain the procedure how to do it. @lalaly
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
what does ker mean
 one year ago

annasBest ResponseYou've already chosen the best response.1
i guess its just L . ker is a printing mistake sorry for that uncle
 one year ago

annasBest ResponseYou've already chosen the best response.1
maybe i m sure about kernel its only written ker L
 one year ago

annasBest ResponseYou've already chosen the best response.1
i just need the way to solve it
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
I'm looking over my notes, I always forget this stuff
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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

TuringTestBest ResponseYou've already chosen the best response.2
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

annasBest ResponseYou've already chosen the best response.1
then we'll get two eqs right ???
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
yes, from which we will get the two vectors that will make up the basis of our kernel
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
at least I think that is what they want
 one year ago

annasBest ResponseYou've already chosen the best response.1
about what about part B ???
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
consulting notes again...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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

annasBest ResponseYou've already chosen the best response.1
ok thank you i got it :) i just need some practice now
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
yeah, me too :) welcome, good luck
 one year ago
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