# 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)

## 1 Answer

So far so good.

You still need to show the scalar multiplication is preserved.

$$A(\lambda X)= \lambda AX ?$$

• Thank you for taking the time to look over my solution. Why do I have to show that? – numericalorange Mar 19 '18 at 16:30
• Aren't we talking about vector spaces? If so, the scalar product should be also preserved under homomorphism. – Mohammad Riazi-Kermani Mar 19 '18 at 16:32
• 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
• If you're working in the category of groups, then you're done. – Ivo Terek Mar 19 '18 at 16:43
• @IvoTerek Okay! Thanks for clearing that up. I apologize for not being specific about which homomorphism I was proving. – numericalorange Mar 19 '18 at 16:44