Show that $Ax=0$.

Question

I need a hint to help me get started with this problem:

Given the sequence of homomorphisms, $\mathbb{Z}^{m}\to \mathbb{Z}^{n} \to M\to 0$, where $M=\mathbb{Z}^{n}/K$ and $K=im(\phi_{A})\subseteq \mathbb{Z}^{n}$ and $\phi_{A}:\mathbb{Z}^{m}\to \mathbb{Z}^{n}$ with $\phi_{A}(\mathbf{x})=\mathbf{x}A$, show that for $\mathbf{m}=(m_{1},m_{2},...,m_{n})$ (images in M of the standard basis vectors in $\mathbb{Z}^{n}$, $A\mathbf{m}=0$ in $M$.

Any idea on how to prove this?

Answer 1

This is a response to @numericalorange 's comment but posted as an answer due to its length.

@numericalorange , $d_{n+1}d_{n}$ is a composition of two morphisms in a sequence.

For example, consider the sequence $$A \to B \to C \to 0,$$ and define morphisms $$d_1 : A \to B, d_2 : B \to C.$$ Clearly, when composed, we have $d_2 d_1$. Now, if we define $\text{Im}(d_1) = \text{ker}(d_2)$, we have $d_2 d_1 = 0$.

As to your question, hint:

Recall that any homomorphism between vector spaces $U$ and $V$, $\psi: U \to V$, can be represented as $\psi(x) = Ax$, where $A$ is a matrix.