Suppose $V=V_1 \oplus V_2$, and $\varphi$ is a linear transformation on $V$ satisfying $$\varphi(v_1+v_2)=v_1$$ where $v_i\in V_i$ for $i=1,2$. Show that $\varphi=\varphi^{2}$, and find Im $\varphi$ and Ker $\varphi$. If $V_1$ and $V_2$ have basis of $\{e_1, \cdots, e_r\}$ and $\{e_{r+1}, \cdots, e_n\}$, respectively, find the representation matrix of $\varphi$ on $V$.
I have proved the first statement, how do I solve the following subquestions?
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Hint: Let's do a concrete case. We can write $\mathbb{R}^2=\mathbb{R}\oplus \mathbb{R}$. Then $\varphi:\mathbb{R}^2\to \mathbb{R}$ is given by $\varphi(x,y)=x$. Indeed, $\varphi^2=\varphi$. Notice that every $x_0\in \mathbb{R}$ is hit by $\varphi$, because for any $y\in \mathbb{R}$, $(x_0,y)\mapsto x_0$. So, $\operatorname{im}(\varphi)=\mathbb{R}.$ Next, the kernel of $\varphi$ is $\varphi^{-1}(0)=\{(x,0): x\in \mathbb{R}\}$. That is the first direct summand above and is isomorphic to $\mathbb{R}$.
Finally, to calculate the matrix representation, we use as a basis for $\mathbb{R}\oplus \mathbb{R}$ the vectors $(1,0)=e_1$ and $(0,1)=e_2$. Then $\varphi(e_1)=1$ and $\varphi(e_2)=0$. So, the matrix is $$ \begin{bmatrix} 1&0 \end{bmatrix}.$$ Try to generalize this argument.
Since the sum is direct, the image is $V_1$ and the kernel is $V_2$. The representation matrix is composed of the unit matrix for $V_1$ and the zero matrix for $V_2$. Generally, if you take a basis $\{b_1,\ldots, b_k\}$ of $V$, then the transformation matrix is given by $A=(a_{ij})$ where $\phi(b_j) = \sum_{i} a_{ij}b_i$, i.e., the $j$th column of $A$ is given by $$A = \left(\begin{array}{ccc} \ldots & a_{ij} & \ldots\\ \vdots & \vdots & \vdots \\ \ldots & a_{kj}& \vdots \end{array}\right).$$