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?

  • $\begingroup$ 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, 2014 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. $\endgroup$ Oct 15, 2014 at 1:16

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Oct 15, 2014 at 1:39
  • 1
    $\begingroup$ 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, 2014 at 1:42

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 a 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 wrt the transformation applied to $\vec{0}$ is not also $\vec{0}$.

  • $\begingroup$ 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 $\endgroup$ Jun 9, 2017 at 15:10

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