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I'm trying to gasp the visual representation of $V/\ker f$ and in continuation the first isomorphism theorem in linear algebra.

My understanding is that, given the representation of a vector space $V,$ $V/\ker f$ is created by "erasing" all the dimensions $f$ maps to $0.$ That's where I get stuck. Any help would be appreciated

P.s The way I have linear algebra in mind is how it is presented in 3blue1brown's channel, if this is of any importance https://www.youtube.com/watch?v=XkY2DOUCWMU&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=5

Michael Hardy
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Fotis Rokanis
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  • So, if my understanding is correct, if V a finite dimension vector space and W a subspace, V/W is the set of all vector spaces with dimension dimV - dimW that are parallel to W.For example $R^3/R$ is the set of all parallel planes – Fotis Rokanis Jun 03 '18 at 00:09
  • $R$ is not a subspace of $R^3$. – Christian Sykes Jun 03 '18 at 00:14
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    That is very well said indeed, even though your final example would need to be more fleshed out. Now if you have a complement to $W$, each of these spaces parallel to $W$ intersects the complement exactly once, and you can "think" of an element of $V/W$ as a vector in that complement. But it is not canonical since there are many possible complements that you could use for that. – Arnaud Mortier Jun 03 '18 at 00:16
  • So, to tie it to the first isomorphism theorem. Let's take $f:R^2->R$ such that $f(x,y) = x -y$. $R^2/kerf$ is the set of all lines parallel to the bisector of the 2 axis that passes through the first and third quarter of the plane. Every line is mapped to 1 and only one element of imf, since if 2 elements of $R$ differ by l they will be projected onto the same element, and the function is obviously onto, thus the 2 vector spaces are isomorphic – Fotis Rokanis Jun 03 '18 at 00:32

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