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Let the function $T$ be a mapping from $\mathbb{R^3} \rightarrow \mathbb{R^4}$ given by

$$T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix}1\\2\\-1\\-2\end{bmatrix}$$

Is this function a linear transformation?

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  • $\begingroup$ It would help to go back to the definition of a linear transformation. This function does not fulfill the necessary criteria; can you see why? $\endgroup$ – Michael M Oct 15 '14 at 1:14
  • $\begingroup$ @amras1 I know that there are two necessary conditions i. T(u + v) = T(u) + T(v) ii. T(cv) = cT(v) I'm just not sure how to apply them. $\endgroup$ – KawaiiExpert Oct 15 '14 at 1:16
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Every linear transformation applies $$T(0)=0$$ The given transformation : $$T(0,0,0)=(1,2,-1,-2) \neq (0,0,0,0)$$ Hence, Not a linear transformation.

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  • $\begingroup$ Oh so is the function saying that no matter what $x1$, $x2$, $x3$ we have, it must always give us $[1,2,-1,-2]$? Because if so, then I can see how your answer is correct. $\endgroup$ – KawaiiExpert Oct 15 '14 at 1:39
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    $\begingroup$ Yes, Since it not dependent on $x1,x2,x3$, For any $x1,x2,x3 \rightarrow T(x1,x2,x3)=(1,2,-1,-2)$. $\endgroup$ – JaVaPG Oct 15 '14 at 1:42
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It might also be helpful to consider the proof of why T(0) must be 0 from the axioms of linearity

0 in a field with (second operation multiplication, which is what we consider here in this scalar multiplication) is defined to be the element at which $\forall f \in F$, we have $0 + f = f + 0 = f$ (note that I made this definition two-sided, because in any ring [all fields are rings] the additive operation forms an abelian group). It can be shown using the distributive axiom of rings that $0 * f = f* 0 = 0$. This is also true for rings in general, not just fields, but we use fields with vector spaces for more useful axioms. https://en.wikipedia.org/wiki/Proofs_of_elementary_ring_properties may also be helpful for general ring/field properties.

By linearity we have $T(0 * v) = 0 * T(v)$. This is not finished, but that tangent on fields was to really show that vector spaces have very similar structure, except that scalar multiplication is NOT binary operation. However, we can use the distributive property in similar manner as in Workaholic's answer. Vector spaces - Multiplying by zero scalar yields zero vector

Then we can see that this transformation will not be linear we the transformation applied to $\vec{0}$ is not also $\vec{0}$.

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  • $\begingroup$ not sure why I got downvoted for this. Suppose that I should edit. It is not an axiom that 0*f = 0 $\forall f \in F$. However, it can be proven by the ring axioms, and every field is a ring. We use mainly distributive axiom $\endgroup$ – Charlie Tian Jun 9 '17 at 15:10

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