I can't seem to wrap my head around this. I've read the book and looked up videos but it still doesn't make sense. It seems like the variables and the subscripts are throwing me off. Can't find an example similar to this one. Example: consider the transformations from $\mathbb{R}^3$ to $\mathbb{R}^3$. Is this transformation linear?


$y_1 = 2x_2$

$y_2 = x_2+2$

$y_3 = 2x_2$

  • $\begingroup$ Is this supposed to be a single transformation? (Because you wrote "these") $\endgroup$ – b00n heT Jul 8 '16 at 18:27
  • $\begingroup$ @b00nheT Yes, it is. I''ll edit my post. $\endgroup$ – Kevin R. Jul 8 '16 at 18:29
  • 1
    $\begingroup$ What is a bit peculiar here is that only $x_2$ appears yet not $x_1,x_3$. The $x_2$ is just the second coordinate of the input vector; the $y_1,y_2,y_3$ are the three coordinates of the image vector. You could write $y_1 = 0 x_1 + 2 x_2 + 0x_3$ and so on to make it more clear. $\endgroup$ – quid Jul 8 '16 at 18:38

A map $f$ is linear only if $f(u+v) = f(u) + f(v)$. For your map, note that \begin{align*} f((x_1,x_2,x_3) + (x_1',x_2',x_3')) &= (2(x_2+x_2'),\; x_2+x_2'+2,\; 2(x_2+x_2')) \\ & \neq (2x_2,\; x_2+2, \;2x_2) + (2x_2', \;x_2'+2,\; 2x_2') \\ & = f(x_1,x_2,x_3) + f(x_1', x_2', x_3') \end{align*} and thus $f$ cannot be linear.


Every linear transformation has $T(0) = 0$, so that the above transformation is not linear.


A linear operator, $L$, will satisfy the following conditions:

Let $\alpha$ be an element of the scalar field and $v, u$ both be vectors in the domain of $L$. Then:

  • $L(\alpha v) = \alpha L(v)$
  • $L(v+u)=L(v)+L(u)$

These imply

  • $L(0)=0$ for whatever concept of zero exists in the domain and codomain.

Ask yourself, if you try using the input of zero (in your case $(x_1,x_2,x_3)=(0,0,0)$) do you get the output of zero back?


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