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Let $f:\mathbb{Z}_{5}^{3} \rightarrow \mathbb{Z}_{7}^{2}$ be a transformation such that $$f\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right):=\begin{bmatrix}x+2z\\x+y-z\end{bmatrix}$$

My homework first asks me to show this is a linear transformation and decide whether or not it's surjective. This $f$ transformation is then used in several questions which suggest its linearity (finding its Kernel, finding its transformation matrix, etc.).

My confusion is that the question doesn't specify what field these vector spaces are on. So the question would be if, say, I decide to pick $\mathbb{Z}_{5}$ as the field I work with, is there any way to make sense of this? after all, the operations in my two vector spaces are defined differently (namely, using their respective mod equivalence classes).

If it helps, the context of this homework is diagonalization and triangularization of transformation matrices.

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  • $\begingroup$ It does. $\mathbb Z$ is german for Zahl, integers. You write $f : \mathbb Z_5^3 \to \mathbb Z_7^2$, right? $\endgroup$ Jun 17, 2019 at 4:59
  • $\begingroup$ Do you really want $\Bbb Z_7^2$? $\endgroup$ Jun 17, 2019 at 5:00
  • $\begingroup$ That's what my sheet says, hence my confusion, do you think it could just be a typo from the person who made it? $\endgroup$
    – gr8astard
    Jun 17, 2019 at 5:04
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    $\begingroup$ Is this well-defined? $\endgroup$
    – Seewoo Lee
    Jun 17, 2019 at 5:23
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    $\begingroup$ In order for a map to be linear, it must be a function between two vector spaces over the same field. It is provably impossible to define addition and scalar multiplication on $\Bbb{Z}_5^3$ and $\Bbb{Z}_7^2$ over a common field in such a way that both are vector spaces, so my guess is that it's a typo. $\endgroup$ Jun 17, 2019 at 5:52

2 Answers 2

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I think there is probably a typo, as I don't see how to talk about a linear transformation when the codomain is a vector space over a different field than the domain.

Let's say $f:\Bbb Z_5^3\to\Bbb Z_5^2$.

The matrix is $\begin{pmatrix}1&0&2\\1&1&-1\end{pmatrix}$.

Row-reduce: $\to\begin {pmatrix}1&0&2\\0&1&2\end {pmatrix}$. It has rank $2$.

Thus it's surjective.

The kernel is $\{\begin{pmatrix}-2t\\-2t\\t\end{pmatrix}\,,t\in\Bbb Z_5\}$.

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  • $\begingroup$ I think it's a typo too, and this interpretation makes sense considering the questions that come after. I'll run with this, thank you. $\endgroup$
    – gr8astard
    Jun 17, 2019 at 6:16
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Starting point:

So let $u,v\in \mathbb{Z}^3_5$ (where $u=(u_1,u_2,u_3)^T$) then you need to show $$ f((u+v)\mod5) = (f(u) + f(v)) \mod 7 $$ as clearly $f(u), f(v) \in \mathbb{Z}^2_7$, where each element of the vectors $u$ and $v$ are mapped according to your function $f$ defined above (where the right hand side summation should be interpreted as $\mod 7$); and for $\alpha \in \mathbb{Z}$ then $$ f(\alpha u \mod 5) = \alpha f(u) \mod 7 $$

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  • $\begingroup$ This makes sense. I will try it and report back in a bit! $\endgroup$
    – gr8astard
    Jun 17, 2019 at 5:43

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