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I am able to prove the above-asked using the fact that a linear map preserves linear dependence. I also vaguely suspect a connection from group theory (homomorphisms?). But I'm having trouble getting an intuition for that fact that a linear transformation cannot increase the dimension of a vector space. Help?

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You can think of a linear transformation as being determined by its action on a basis. If the basis elements are mapped to linearly independent vectors, then their images span the same dimensional space (say finite dimensional). Of course, since the transformation is linear, they will span the entire image as well. Otherwise, some basis vector gets mapped into the linear span of the others, and the dimension will drop. – Andrew Aug 2 '12 at 14:23
I'd like to say that all first the 4 answers I've received (incl. the above comment-- and all within 10 mins!) are superb and all helpful to me, you guys rock! – Ryan Aug 2 '12 at 14:36
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My intuition would be this: linear maps preserve lines (that's the intuition behind them, anyway). If they are degenerate, the lines might collapse to points, but points which are colinear remain such after the transformation.

Dimension of a space is, intuitively, the number of independent lines you can draw in a space. Because linear transformation preserves not just lines, but also linear subspaces of higher dimensions (so coplanar points remain coplanar etc.), it can't "split" a line into more of them, even if it can join some of them, and it can't turn lines which weren't independent into ones which are (because that would be "splitting" some higher-dimensional subspace; it's a little circular, but we're talking about the intuition here), so the dimension can't go up.

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This is because linear transformations are closed in a vector space. There may be smaller subspaces inside, but there is no way to go into a larger one.

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Is this true? Couldn't you have a linear transformation from $\mathbf{R}^4$ to the space of 2x2 matrices? – kennysong Oct 12 '13 at 13:17
@jellyksong, you're right. I was only thinking of linear transformations as matrices. I had no though at all about something like $\mathbb{R}^ \to \mathbb{C}$. – Karolis Juodelė Oct 13 '13 at 7:37

One possible intuition could be that, say, a 3-dimensional space contains "many more" points that a 2-dimensional one does, so there are not enough points in $K^2$ to hit every point in $K^3$.

Strictly speaking this is of course rubbish -- when the scalar field $K$ is infinite we can prove that the cardinalities of $K$, $K^2$, $K^3$, etc. are all the same. But the requirement that the map must be linear forces it to be "well-behaved" enough to respect the naive intuition about the sizes of sets anyway.

(The argument does work when $K$ is a finite field, however).

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