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I learned from 3Blue1Brown's Linear Algebra videos that a 2-D transformation is linear if it follows these rules:

  • lines remain lines without getting curved
  • the origin remains fixed in place
  • grid lines remain parallel and evenly spaced

I'm now going through linear algebra from a textbook, which lays out this definition of a linear transformation:

  • T(u+v) = T(u) + T(v)
  • T(cu) = cT(u)

I'm wondering, is there a connection between these two ways of thinking of linear transformations? Do the visual ways of seeing 2-D linear transformations correspond to the formal definition when in 2-D?

Neil Aronson
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  • I recently asked a question in this spirit with some insightful answers (though I asked in the more abstract setting of vector spaces over arbitrary fields) http://math.stackexchange.com/questions/2035208/is-a-map-that-preserves-lines-and-fixes-the-origin-necessarily-linear – eepperly16 Dec 06 '16 at 03:06
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    I've read your question and have to say I don't fully understand it or the answers as I'm just starting out in linear algebra, but thank you for the link. – Neil Aronson Dec 06 '16 at 04:02

1 Answers1

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If we make reasonable algebraic definitions for the three points above, then the answer is yes: there is a correspondence between the geometric perspective and the algebraic perspective.

Let $f$ be a map that fixes the origin ($f(\vec{0}) = \vec{0}$) and sends lines of evenly spaced points to lines of evenly spaced points ($f(\vec{u} + c\vec{v}) = \vec{u}' + c\vec{v}'$). Then we have

$$ f(c\vec{v}) = c\vec{v}' $$

If we take $c=1$ then we get $f(\vec{v}) = \vec{v}'$. So $f(c\vec{v}) = cf(\vec{v})$.

Now consider

$$ f(c_1\vec{u} + c_2\vec{v}) = \vec{u}' + c_2\vec{v}' $$

If we set $c_2 = 0$, we get

$$ f(c_1\vec{u}) = c_1 f(\vec{u}) = \vec{u}' $$

So we have

$$ f(c_1 \vec{u} + c_2 \vec{v}) = c_1f(\vec{u}) +c_2\vec{v}' $$

But if $c_1 = 0$, we then have

$$ f(c_2 \vec{v}) = c_2 f(\vec{v}) = c_2 \vec{v}' $$

So $f(c_1 \vec{u} + c_2 \vec{v}) = c_1 f(\vec{u}) + c_2f(\vec{v})$. So $f$ is linear. Therefore, if $f$ fixes the origin and sends lines of evenly spaced points to evenly spaced points, then $f$ is linear.

Can you show the converse ?

eepperly16
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  • Ok, I think the key insight for me is where you say $f(\vec{u} + c\vec{v}) = \vec{u}' + c\vec{v}'$. Is it correct that this is the mathematical way of saying that f sends lines of evenly spaced points to lines of evenly spaced points because they are moved apart by a constant c? – Neil Aronson Dec 06 '16 at 05:28
  • I thought of $\vec u + c \vec v$ as a line that touches the point $\vec u$ oriented in the direction $\vec v$ since we are adding multiples of $\vec v$ to $\vec u$. So if $\vec u' + c \vec v'$ is parallel then $\vec v'$ needs to be a multiple of $\vec v$ right? I don't see why $f(c\vec{v}) = c\vec{v}'$ needs to be true. – Aditya P Apr 23 '19 at 16:11
  • @Aditya A linear transformation sends a collection of parallel lines in the input space to a collection of parallel lines in the output space. Note the lines in the input space need not to be parallel to the lines in the output space. $f(c\vec{v})=c\vec{v}’$ must be true because $f$ must send an equally spaced set of points along the line spanned by $\vec{v}$ to an equally spaced set of points spanned by another line, which we define to be $\vec{v}’$ – eepperly16 Apr 23 '19 at 16:49
  • Now that you point it out, it's laughable that I was thinking v was parallel to v'. One is in input space, the other in output space. Ok so I think I get you, you are considering the set of all equally spaced points that lie on a line, that would be line itself. Since the line oriented along v passing through u becomes a line oriented along v' and passing through u'. Here you just take u to be the zero vector and then that is true. – Aditya P Apr 23 '19 at 18:08