2
$\begingroup$

Theorem: A linear transformation is injective if and only if its nullity is trivial

Let $V$ be vector space and $T:V\to \mathbb{R}$

Let $x\in V$

$T(x) = | x |$

$T(x) = 0$ if $x = 0$

By the theorem, $T$ is injective. How?

$T(x) = 2$ if $x = 2, -2 $

(I may have messed up the conditions, for example, maybe it was $T:V\to W$. I don't remember the exact question, sorry.)

$\endgroup$
2
  • 1
    $\begingroup$ Why is $|x|\in V$, what is $|x|$? How would define it if $V$ is the vectorspace of polynomials over $\mathbb{R}$? $\endgroup$
    – Mathematician 42
    Dec 11, 2016 at 2:29
  • 4
    $\begingroup$ Why is $T$ linear? $\endgroup$
    – Asaf Karagila ♦
    Dec 11, 2016 at 5:32

2 Answers 2

Reset to default
9
$\begingroup$

The theorem is valid for linear transformations, the absolute value function (also known as norm) may not be a linear transformation.

$\endgroup$
14
  • 1
    $\begingroup$ Did not downvote, but why 'may'? When is the absolute value linear? $\endgroup$
    – Theoretical Economist
    Dec 11, 2016 at 2:34
  • 6
    $\begingroup$ If $V$ is trivial ;-) $\endgroup$
    – Lukas Betz
    Dec 11, 2016 at 2:37
  • 1
    $\begingroup$ Good question. I am going to look for an answer.@TheoreticalEconomist $\endgroup$
    – Arthur
    Dec 11, 2016 at 2:50
  • 1
    $\begingroup$ @user29418 due to $n(-v) = -n(v)$ for any linear $n$, but $\|-v\| = |-1|\cdot\|v\| = \|v\|$, the norm can only be linear if it is zero for any vector, i.e. only on the trivial zero-dimensional vector space. $\endgroup$
    – leftaroundabout
    Dec 11, 2016 at 14:09
  • 1
    $\begingroup$ @Arthur interesting idea, but you can't have a $V\to\mathbb{R}$ norm on such a space, because $k + (p-k) \equiv 0$, so $\|k + (p-k)\| = 0$, hence <del>you could only fulfill the triangle inequality with $\|k\|=\|p-k\|=0$, which contradicts $\|k\|\neq 0 \ \forall k \neq \vec{0}$.</del> You could perhaps have a $(\mathbb{Z}_p)^d \to \mathbb{Z}_p$ “norm”, but that would also have its problems. $\endgroup$
    – leftaroundabout
    Dec 11, 2016 at 15:11
8
$\begingroup$

You have chosen your "linear transformation" (in the statement of the theorem) to be the map $x\mapsto\lvert x\rvert,$ but you haven't checked that this is actually a linear transformation.

Indeed, if $T\colon\mathbb{R}\to\mathbb{R}$ is intended, then $x\mapsto\lvert x\rvert$ is not a linear map (for example, $\lvert -1 + 2\rvert = 1 \neq 3 = \lvert -1\rvert +\lvert 2\rvert$), so, although the theorem is true, it simply doesn't apply to your choice of $T,$ because your choice of $T$ doesn't satisfy the given conditions.

[If, as others have astutely pointed out, you take $T\colon\{0\}\to\mathbb{R}$ instead, then you do get a linear map, but this map is obviously injective.]

Try something like $T_{1}\colon\mathbb{R}\to\mathbb{R},x\mapsto2x$ or $T_{2}\colon\mathbb{R}^{2}\to\mathbb{R}^{2},\begin{pmatrix}x \\ y\end{pmatrix}\mapsto\begin{pmatrix}2x \\ x+2y\end{pmatrix}$ instead. In general, if you want examples of linear maps, you should think about multiplication by matrices.

In the case of $T_{2},$ I picked the matrix $\begin{pmatrix} 2 & 0 \\ 1 & 2\end{pmatrix}$ (written under the standard choices of bases). This matrix has determinant $4,$ which is non-zero, so we see that $T_{2}$ must therefore be invertible, and hence injective. I'll leave it to you to check that its kernel is trivial, as the theorem says it should be.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.