1. ## homomorphism

I know that for it to be homomorphism;

A homomorphism from two rings is a function such that

F(a+b)=F(a)+F(b) and F(a*b)= F(a)*f(b)

but how do i apply it to this example thanks.

2. $\displaystyle f\left( \begin{bmatrix}{a}&{b}\\{0}&{d}\end{bmatrix}+ \begin{bmatrix}{x}&{y}\\{0}&{z}\end{bmatrix}\right )=f\left( \begin{bmatrix}{a+x}&{b+y}\\{0}&{d+z}\end{bmatrix} \right)=a+x=f\left( \begin{bmatrix}{a}&{b}\\{0}&{d}\end{bmatrix} \right)+f\left( \begin{bmatrix}{x}&{y}\\{0}&{z}\end{bmatrix}\right )$

Try proving the other condition. To compute the kernel: $\displaystyle f\left(\begin{bmatrix}{a}&{b}\\{0}&{d}\end{bmatrix } \right)=0 \Leftrightarrow a=0 \implies ker(f)=\hdots$. To compute the image, note that the parameters of each matrix are in $\displaystyle \mathbb{R}$, then, the parameter $\displaystyle a$ can be any number in $\displaystyle \mathbb{R}$. Now you can conclude.

3. Originally Posted by felper
$\displaystyle f\left( \begin{bmatrix}{a}&{b}\\{0}&{d}\end{bmatrix}+ \begin{bmatrix}{x}&{y}\\{0}&{z}\end{bmatrix}\right )=f\left( \begin{bmatrix}{a+x}&{b+y}\\{0}&{d+z}\end{bmatrix} \right)=a+x=f\left( \begin{bmatrix}{a}&{b}\\{0}&{d}\end{bmatrix} \right)+f\left( \begin{bmatrix}{x}&{y}\\{0}&{z}\end{bmatrix}\right )$

Try proving the other condition. To compute the kernel: $\displaystyle f\left(\begin{bmatrix}{a}&{b}\\{0}&{d}\end{bmatrix } \right)=0 \Leftrightarrow a=0 \implies ker(f)=\hdots$. To compute the image, note that the parameters of each matrix are in $\displaystyle \mathbb{R}$, then, the parameter $\displaystyle a$ can be any number in $\displaystyle \mathbb{R}$. Now you can conclude.

Thus the image is $\displaystyle \mathbb{R}$?

4. Originally Posted by SubZero
Thus the image is $\displaystyle \mathbb{R}$?
Exactly.