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I have the following question:

Given a certain transformation, determine if this transformation is linear or not. $T(1,0,0)=0,T(2,-1,0)=1,T(3,2,1)=2$

What I think needs to be done is to deduce a formula for this transformation and then determine if it linear or not (checking additive and homogeneous properties). I'm having hard time understanding how to deduce the formula. Thank you in advance.

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    How could you obtain a formula for $T$ from this information without knowing that $T$ is linear ? – 2017-01-04
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    Are you sure you need to deduce a formula **first**? You could check for additivity and homogeneity from the three points you have. If it holds there, you can assume $T$ is linear and only then derive a formula. If $T$ is not linear, 3 points are not enough to deduce $T$ uniquely. – 2017-01-04
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    I'll go with your idea, I might not understood the method of working completly. Let's assume that it's linea- derive a formula and then check whether the formula is linear or not. – 2017-01-04
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    To be more specific, your mapping seems to be ${\mathbb R}^3 \rightarrow {\mathbb R}^1$. Hence, if it is linear, it can be described by 3 coefficients àla $T = $, $t \in \mathbb{R}^3$. You are given 3 points which a linearly independent so you just have enough information to compute the coefficients (**knowing** its linear). To check for linearity you would need more. If you have a linearly dependent point $\bar{x}$ that you can write as $\bar{x} = \alpha_1 x_1+\alpha_2 x_2$ you can check whether $\bar{y} = T(\bar{x}) = T(\alpha_1 x_1+\alpha_2 x_2) = \alpha_1 T(x_1) + \alpha_2 T(x_2)$. – 2017-01-04
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    ok so if i try to sum it up: i have to find 2 points in the span (what you said as compute the coefficients) and then check if these two points are fulfilling additive property? – 2017-01-04

1 Answers 1

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$T(x,y,z)=4z-y$ is linear and satisfies the condition.

$T(x,y,z)=y-xy+6z$ isn't linear and also satisfies the condition.

So, we do not have information enough.

Edit. (response to the comment)

How did you calculate these formulas?

If $v_1,v_2,v_3$ are three linearly independent vectors in $\mathbb R^3$, and $\alpha_1,\alpha_2,\alpha_3$ are three arbitrary real numbers, then there is a unique linear transformation $T$ from $\mathbb R^3$ to $\mathbb R$ such that $T(v_1)=\alpha_1$, $T(v_2)=\alpha_2$, $T(v_3)=\alpha_3$. This is a particular case of a general result, which is stated and explained here. If you know how to prove it, you can calculate the first formula above. The general proof can be found here (Theorem 4.4). Motivated by this proof, we can deal with your particular case as follows.

Find $a_1,a_2,a_3$ (in terms of $x,y,z)$ such that $$(x,y,z)=a_1(1,0,0)+a_2(2,-1,0)+a_3(3,2,1).$$ (This can be done because $(1,0,0)$, $(2,-1,0)$, $(3,2,1)$ are linearly independent.) The answer is $a_1=x+2y-7z$, $a_2=2z-y$, $a_3=z$. So, the unique linear transformation from $\mathbb R^3$ to $\mathbb R$ that satisfies the condition given in the problem is given by $$\begin{align} T(x,y,z)&=T(a_1(1,0,0)+a_2(2,-1,0)+a_3(3,2,1))\\ &=T((x+2y-7z)(1,0,0)+(2z-y)(2,-1,0)+z(3,2,1))\\ &=(x+2y-7z)T(1,0,0)+(2z-y)T(2,-1,0)+zT(3,2,1)\\ &=(x+2y-7z)\cdot0+(2z-y)\cdot1+z\cdot2\\ &=4z-y \end{align}$$

For the second formula, I did not follow any specific rule. I just tried to find a formula with the term $xy$ such that:

  • For $x=1$, $y=0$ and $z=0$ the result is $0$.
  • For $x=2$, $y=-1$ and $z=0$ the result is $1$.
  • For $x=3$, $y=2$ and $z=1$ the result is $2$.

There are others. Can you find one?

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    how did you calculate these formulas? – 2017-01-04
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    @YaronScherf See my edit. – 2017-01-04