# Show that $\phi: \mathbb{R^{n}}\to \mathbb{R^{m}}$ defined by $\phi(x)=Ax$ is a homomorphism.

Could someone let me know if my answer is correct?

Show that $\phi: \mathbb{R^{n}}\to \mathbb{R^{m}}$ defined by $\phi(x)=Ax$, $A\in \mathbb{M}_{m\times n}(\mathbb{R})$, is a homomorphism.

Since $\mathbb{R}^{n}$ is the external direct product of $\mathbb{R}$ with itself $n$ times, it is a group under addition. Similarly with $\mathbb{R}^{m}$.

Well-defined: If $x=y$ are equal $n$-vectors, then $\phi(x)=Ax=Ay=\phi(y)$.

Operation-preserving: $\phi(x+y)=A(x+y)=Ax+Ay=\phi(x)+\phi(y)$ (because matrix multiplication distributes over addition)

$$A(\lambda X)= \lambda AX ?$$
• You proved that $\phi$ is a homomorphism of groups, not a "homomorphism of vector spaces" (read "linear transformation"). – Ivo Terek Mar 19 '18 at 16:41