0
$\begingroup$

I'm reading a text for one of my classes, and I came across this theorem:

Theorem 4. Suppose $V$ is an $m$-dimensional vector space over $\mathbb F_p$. Suppose $T : V → W$ is a linear map. Write $n$ for the dimension of the null space of $T$, and $r$ for the dimension of the range. Then

a) The cardinality of $V$ is $|V | = p^m$.

b) The cardinality of the range of T is $p^r$.

c) The preimage of every vector in the range of $T$ has $p^n$ elements.

I don't understand why (c) is true. I can sort of see how given the inverse of an element in the range (call it $x$), one can get $p^n$ more elements mapping to said element by taking $T(x+u)$ where $u$ is in the null space. But then we'd have $q(1+p^n)$ elements in the preimage of every vector $w \in W$, where $q$ is the number of distinct elements of $(V - \text{null} \,V)$ mapping to $w$.

Could someone give me some intuition as to why (c) is true?

$\endgroup$
2
$\begingroup$

If $T$ is a linear map $V \to W$ then, for every $v\in V$, it is the case that: $$T^{-1}(\{Tv\})=\ker(T)+v.$$

Indeed, if $x \in \ker(T)$, then $T(x+v)=Tx+Tv=Tv$, so $x+v\in T^{-1}(\{Tv\})$. Viceversa, if $v' \in T^{-1}(\{Tv\})$, then $T(v'-v)=Tv'-Tv=Tv-Tv=0$ and $v'=(v'-v)+v\in\ker(T)+v$.

Moreover the map $\ker(T)\ni y \mapsto y+v \in\ker(T)+v=T^{-1}(\{Tv\})$ is obviously a bijection. Hence the preimage of every vector in the range of $T$ has the same cardinality of the kernel.

Hence we are done if we prove that the cardinality of the kernel is $p^n$. But this is true, because it is a linear space of dimension $n$ over a field of cardinality $p$. Indeed, if $\{e_1,...,e_n \}$ is a basis of $\ker(T)$, then every element is written uniquely as $a_1e_1+...+a_ne_n$ for $a_i \in \mathbb{F}_p$. Then the linear application: $$ \ker(T)\ni a_1e_1+...+a_ne_n \mapsto (a_1,...,a_n) \in\mathbb{F}_p^n$$ is bijective and then $|\ker(T)|=|\mathbb{F}_p^n|=p^n$.

$\endgroup$

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.