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I'm new to this subject so please bear with me. From my understanding, the dimension of some space is simply the length of the basis for that space. So the dimension for the kernel is the length of the basis for the set of vectors that map to the zero vector...

So, lets say we have a linear map $T=(a,b,c,d,e) = (a,b,0,0,0)$(which I saw from another example), and its said to have rank 2, nullity 3. I don't understand why this the case since all you need is two vectors {(1,0,0,0,0), (0,1,0,0,0)} to span the range, and I don't see why we would need 3 vectors to span the kernel being that the last three coordinates are always zero, so it all depends of the first two...

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  • $\begingroup$ If x = (a, b, c, d, e) is in the kernel then Tx = (0, 0, 0, 0, 0) so that a = 0 and b = 0. There is no restriction on c, d and e. $\endgroup$
    – Paul
    Nov 18, 2018 at 22:01
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    $\begingroup$ The kernel of a map is the set in the domain which is sent to zero. Thus the kernel of your map $T$ consists of the the elements of the form $(0, 0, c, d, e)$, since $$ T(0, 0, c, d, e) = (0, 0, 0, 0, 0) = 0. $$ This has dimension three. $\endgroup$
    – Xander Henderson
    Nov 18, 2018 at 22:01

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The kernel is the set of all vectors that are sent to $0$. A basis for the kernel here is $(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1)$. There are three of them, so it has dimension 3.

As far as visualising it: all linear maps (between finite dimensional spaces over a field $K$) are, up to change of basis in the domain and target separately, projections of $K^n$ onto the span of the first $m \leq n$ elements of the standard basis. That is: there is some number $m$ of directions that we don't do anything with, and some other number $n - m$ of directions that we squash down to nothing. Then the rank of the linear map is precisely the $m$ - the number of basis vectors that are left unchanged, and the nullity is precisely the $n - m$ - the number of basis vectors that are squashed.

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