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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.)

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

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

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    $\begingroup$ Did not downvote, but why 'may'? When is the absolute value linear? $\endgroup$ Dec 11, 2016 at 2:34
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    $\begingroup$ If $V$ is trivial ;-) $\endgroup$
    – Lukas Betz
    Dec 11, 2016 at 2:37
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    $\begingroup$ Good question. I am going to look for an answer.@TheoreticalEconomist $\endgroup$
    – Arthur
    Dec 11, 2016 at 2:50
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    $\begingroup$ @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. $\endgroup$ Dec 11, 2016 at 14:09
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    $\begingroup$ @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. $\endgroup$ Dec 11, 2016 at 15:11
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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.

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