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How can I verify that the following function is linear?

L : $R^2 \Rightarrow R^3$ such that $L(v_1,v_2) = (v_2,4v_1+v_2,0)$

I know that the theorem states that:

  1. for all $v \in R^m$ and all $\alpha \in R^m$, we have $L(\alpha v) = \alpha L(v)$.

  2. for all $v,w \in R^m$, we have $L(v + w) = L(v) + L(w).$

But I need some help starting with the arithmetic of the proof, how do I do this when they are in different dimensions?

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  • $\begingroup$ $L(\alpha (v_1,v_2))=L(\alpha v_1,\alpha v_2)$. Compare with $\alpha (L(v_1,v_2))=\alpha (v_2,4v_1+v_2,0)$. $\endgroup$ – Mauro ALLEGRANZA Sep 11 at 14:28
  • $\begingroup$ The same for 2. $\endgroup$ – Mauro ALLEGRANZA Sep 11 at 14:33
  • $\begingroup$ Quibble: "how can I show the following linear transformation is linear" is silly. If it is a linear transformation, then by definition it is linear. You mean, how can you show the given function is in fact a linear transformation. $\endgroup$ – Arturo Magidin Sep 11 at 14:37
  • $\begingroup$ That’s not a “theorem.” That’s the definition of linearity. $\endgroup$ – amd Sep 11 at 16:57
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$L(\alpha v) = (\alpha v_2, 4 \alpha v_1+\alpha v_2, 0)$

$\alpha L(v) = (\alpha v_2, 4 \alpha v_1+\alpha v_2, 0)$

Are these the same for all $v$?

Now try the other.

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  • $\begingroup$ If i try to solve for $L(\alpha v)$ using the individual componets of $v_1$ and $v_2$, what do you suggest I name the components of $v_1$? For example if I was using $x$ and $y$, I would use ($x_1,x_2$) and ($y_1,y_2$).. $\endgroup$ – atn Sep 11 at 15:10
  • $\begingroup$ You are not solving anything as such, just checking equivalene. You can use whatever symbols you want. However, checking that the two formular are the same will be easier if you use consistent symbols. $\endgroup$ – copper.hat Sep 11 at 15:16
  • $\begingroup$ I wanted to solve for both and then check equivalency. Like what they do here: yutsumura.com/are-these-linear-transformations/ I was just asking how I would define those components to do so without it being confusing $\endgroup$ – atn Sep 11 at 15:17
  • $\begingroup$ I think you are making this more complicated than necessary. You just need to notice that the above two expressions the same, for example. $\endgroup$ – copper.hat Sep 11 at 15:23

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