Construct an isomorphism between $V$ and $\Bbb F^2$ and justify Consider the subspace of $M_2(\Bbb F):V = \biggl\{\begin{pmatrix}a+b&a\\0&b\end{pmatrix}\Biggm\vert  a,b\in \Bbb F\biggr\}$
Construct an isomorphism between $V$ and $\Bbb F^2$ and justify
 A: I assume $F$ is a general field and $M_2(F)$ is the set og $2 \times 2$ matrices under that field. An isomorphism $T: F^2 \rightarrow V$ could simply be the function that maps a point $(a,b) \in F^2$ to the associated matrix in $V$, i.e.
$$T(a,b) = \begin{bmatrix} 
a+b & a \\
0 & b \end{bmatrix}.$$
It is clear that $T$ is a linear tranformation: Let $(a,b), (\alpha, \beta) \in F^2$ and $x, y \in F$. We have
$$T(x(a,b)+y(\alpha, \beta)) = \begin{bmatrix} 
xa + y\alpha + xb + y\beta & xa+ y\alpha \\
0 & xb + y\beta \end{bmatrix} = xT(a,b)+yT(\alpha, \beta)$$.
We can also show that $T$ is injective by assuming $T(a,b) = T(\alpha, \beta)$. We get
$$\begin{bmatrix} 
a+b & a \\
0 & b \end{bmatrix} = \begin{bmatrix} 
\alpha+\beta & \alpha \\
0 & \beta \end{bmatrix} \implies a=\alpha \wedge b=\beta$$
It is also clear that $T$ is surjective by construction. Thus, $T$ is an isomorphism between $V$ and $F^2$.
I may have misinterpreted the question. I hope this could be of some help anyway
A: Consider $\phi:V\to F^2$ defined as $\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}=(a,b)$.
Then
$\phi\Big(\begin{pmatrix}a+b&a\\0&b\end{pmatrix}+\begin{pmatrix}c+d&c\\0&d\end{pmatrix}\Big)=\phi\begin{pmatrix}a+b+c+d&a+c\\0&b+d\end{pmatrix}=(a+c,b+d)=(a,b)+(c,d)=\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}+\phi\begin{pmatrix}c+d&c\\0&d\end{pmatrix}$
$\phi\Big(\lambda\begin{pmatrix}a+b&a\\0&b\end{pmatrix}\Big)=\phi\begin{pmatrix}\lambda a+\lambda b&\lambda a\\0&\lambda b\end{pmatrix}=(\lambda a,\lambda b)=\lambda(a,c)=\lambda\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}$
So $\phi$ is linear.
If $\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}=\phi\begin{pmatrix}c+d&c\\0&d\end{pmatrix}$ then $(a,b)=(c,d)$, so $\begin{pmatrix}a+b&a\\0&b\end{pmatrix}=\begin{pmatrix}c+d&c\\0&d\end{pmatrix}$, so $\phi$ is injective.
Take some $(r,s)\in F^2$, then we can consider $\begin{pmatrix}r+s&r\\0&s\end{pmatrix}$ and thus $\phi\begin{pmatrix}r+s&r\\0&s\end{pmatrix}=(r,s)$, so $\phi$ is surjective.
We conclude $\phi$ is an isomorphism, so $V\simeq F^2$
A: Construct a map
\begin{align*}
\Phi: V &\to \mathbb{F}^2
\\ \boldsymbol{\{}a \sim b\boldsymbol{\}} &\mapsto (a, b)
\end{align*}
Where $\boldsymbol{\{}a\sim b\boldsymbol{\}} \overset{\operatorname{def.}}{=} \begin{bmatrix} a +  b & b \\ 0 & a \end{bmatrix}$. It's easy to see that $\Phi$ is a linear map.
Now, note that $\Phi$ is a surjection, as the pre-image of $(a,b)$ is in $V$ for every $a, b \in \mathbb{F}$. Thus $\Phi(V) = \mathbb{F}^2$.
Let's consider $ \boldsymbol{\{}c \sim d\boldsymbol{\}} \in \ker(\Phi)$. Then we have that:
$$ \Phi(\boldsymbol{\{}c \sim d\boldsymbol{\}}) = (c, d) = \boldsymbol{0}_{\mathbb{F}^2} = (0, 0) \implies  c = d =0 $$
Hence, $\boldsymbol{\{} c \sim d\boldsymbol{\}} = \boldsymbol{\{} 0 \sim 0\boldsymbol{\}} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = \boldsymbol{0}_V $. i.e., $\ker(\Phi) = \{\boldsymbol{0}_V \}$. Thus $\Phi$ is injective.
Hence, $\Phi$ is an isomorphism. $V \cong \mathbb{F}^2$
