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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.
just explain the procedure how to do it. @lalaly
what does ker mean

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

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