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I'm in a matrix theory class, and today we started talking about linear transformations. My professor noted that the range and domain of a linear transformation must be vector spaces over the same field. This made sense to me at first because of the fields are different, then something like $T(cx)=cT(x)$ doesn't make any sense when c is only in one of the vector spaces.

But then I thought that maybe some fields might be compatible, like for example, $\mathbb{Z}_5$ and $\mathbb{Z}_7$

Are there any linear transformations for two spaces over different fields? or Are there functions that have all the same properties as linear transformations except that the spaces involved use different fields?

If the answer is no, can you prove it?

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  • $\begingroup$ What is mean "compatible"? $\endgroup$ Sep 21, 2016 at 15:12
  • $\begingroup$ As in, $T(cx)$ and $cT(x)$ are both defined and also $T(cx)=cT(x)$ $\endgroup$
    – Zachary F
    Sep 21, 2016 at 15:14
  • $\begingroup$ I haven't checked if $Z_5$ and $Z_5$ have a T such that $T(cx)=cT(x)$ but I mentioned them as candidates because $T(cx)$ and $cT(x) are both defined. $\endgroup$
    – Zachary F
    Sep 21, 2016 at 15:17
  • $\begingroup$ I still don't understand how it can be the same $c$ if fields are different. Explain me please. $\endgroup$ Sep 21, 2016 at 15:21
  • $\begingroup$ @Weaam: do you understand that $5$ in $\mathbb Z_7$ and $5$ in $\mathbb Z_5$ are different things? Just ignore the fact they notated the same way. We can identify elements only in some rare cases (like isomorphism or embedding). $\endgroup$ Sep 21, 2016 at 15:29

2 Answers 2

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If you have an injective field endomorphism $\varphi$ from field $Fa$ to field $Fb$ - said differently:

$$\forall (x, y) \in Fa \times Fa, \varphi(a+b) = \varphi(a) + \varphi(b) \space \mathrm{and} \space \varphi(a.b) = \varphi(a).\varphi(b)$$

then $\varphi(Fa)$ is a sub field of $Fb$ isomorphic to $Fa$

If $Vb$ is a vector space over $Fb$ of finite dimension, you can exhibit a base from $Vb$. You can then build a sub set (not a sub vector space) of $Vb$ from the base and the field $\varphi(Fa)$. This sub set (let us call it $Wb$) is a vector space over $\varphi(Fa)$ (*).

By identifying $Fa$ and $\varphi(Fa)$, you can indeed build a linear transformations (say $\Lambda$) from a vector space over $Fa$ (say $Va$) and $Wb$.

But you cannot call it a linear transformations from $Va$ to $Vb$ because $\Lambda(Va)$ is not a sub vector space of $Vb$ (except for the trivial $\Lambda(Va) = \{ 0_{Vb} \}$ see below). It is only a sub vector space of $Wb$.


Well to be honest, there is one trivial case where you could exhibit a pseudo linear transformation for $Va$ to $Vb$:

$$\begin{align}\Lambda: &Va \rightarrow Vb \\ &x \mapsto 0_{Vb}\end{align}$$

It respects all the properties for a linear transformation, and $\Lambda(Va) = \{ 0_{Vb} \}$ is indeed a sub vector space but it is not really interesting, and on a strict point of view, as the fields are different, it is not a true linear transformation.

(*) I suppose it could be extended to vector space of infinite dimensions, but I do not know how to prove it

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Let's suppose $V$ is a vector space over a field $K$ (i.e., an abelian group onto which $K$ acts). If $L$ is another field, and $K$ can act on the addtive structure of $L$ (i.e., the additive abelian group of $L$ is a $K$ vector space), then we can extend $V$ to a vector space over $L$ using tensor products. This construction is known as the extension of scalars. We take the tensor product of $V$ and $L$ over $K$:

$$V \otimes_K L$$

It is naturally both a $K$ vector space (by the tensor product constrution) and a $L$ vector space (by multiplication on the $L$ component). There is a natural function from $\varphi: V \to V\otimes_K L$, which is additive, and compatible with multiplication by scalars on $K$:

$$\varphi(x + y) = (x+y)\otimes 1 = (x\otimes 1) + (y\otimes 1)$$ $$\varphi(kx) = (kx\otimes 1) = (x\otimes 1_k) = k(x\otimes 1)$$

Here, $1_k$ detones the image of the action of $k$ on the unit of $L$.

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