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$f:\Bbb Z^2\to\Bbb Z^2:{n\brack m}\mapsto\begin{bmatrix}3&2\\4&3\end{bmatrix}{n\brack m}$

it is onto because:

$\left\{\begin{align*}&m = 9d - 4s\\ &n = 6d + 3s\end{align*}\right.$ (after solving the system of equations)

onto because m and n are integers

one-to-one because:

$\left\{\begin{align*}&3n + 2m = 3p + 2q\\ &4n + 3m = 4p + 3q\end{align*}\right.$

which is equal to $L_2 - L_1$ and $L_1 - 2L_2$:

$\left\{\begin{align*}&n = p\\ &n + m = p + q\end{align*}\right.$

equal to:

$\left\{\begin{align*}&n = p\\ &m = q\end{align*}\right.$

I have to solve the system of equations for both (to prove it is onto and one-to-one) if I understood correctly?

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    There's a section [in this tutorial](http://meta.math.stacke$x$ch$a$nge.com/questions/5020/mathja$x$-basic-tutorial-and-quick-reference) about how to do matrices.2012-11-07

2 Answers 2

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Yours can be interpreted as linear transformation/homomorphism, from $\Bbb Z^2$ to itself.

Injectivity and surjectivity then boil down to the matrix being invertible or not, and the value of it's determinant. We have $\begin{vmatrix}3&2\\4&3\end{vmatrix}=3\cdot 3 -4\cdot 2 =9-8=1$

Since the determinant is $1$, the function is invertible (which means it is one-one and onto), and it's inverse matrix $\left(\begin{matrix}3&-4\\-2&3\end{matrix}\right)$ will also send integers to integers, which is indeed what we want.

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    @GladstoneAsder As far as I can tell, your solution is OK.2012-11-07
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Let $f \colon X \to Y$, then:

  1. $f$ is onto (surjective) if and only if for every $y \in Y$, there exists some $x \in X$ such that $y = f(x)$.

  2. $f$ is one-to-one (injective) if and only if $f(a) \neq f(b) \implies a \neq b$ for every $a, b \in X$

Claim: $f$ is surjective.

Proof:

Observe that we have a matrix with $\textrm{Det}(M)=1$, then we know (from linear algebra) $M$ is invertible.

Let $ M = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}.$

Then $ M^{-1} = \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix}.$

Since $f(n,m) = (3n + 2m, 4n + 3m)$ and $f^{-1}(r,s) = (3r -2s, -4r + 3s)$, we know the map is surjective since for every $(r,s) \in \mathbb{Z}^{2}$, it is always true that $(r,s) = f(f^{-1}(r,s))$.

Lemma: Let $f \colon X \to Y$ such that $f^{-1}(f(x)) = x$ for every $x \in X$, then $f$ is injective.

Claim: $f$ is injective.

Proof:

Follows immediately from the above lemma, since it is easily verified that $f$ is left-invertible.

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    By "not the right approach" I simply mean "not the easiest approach". The proof of $f$ being a bijection follows directly from the matrix $M$ being invertible.2012-11-07