Determining if L is a linear map, and checking if its injective or surjective I was working through some examples from my textbook and got stuck on this question, was wondering if someone could help me understand how to do it.

$L: M_{2\times 2}(\mathbb R) \to M_{2 \times 2}(\mathbb R)$ is defined by $L(M) = M + M^T$. Determine if $L$ is a linear map or not and if it is surjective or injective.

From what I can gather, I need to prove the following to confirm if it is a linear map:
$L_1: L(x+y) = L(x) + L(y)$
$L_2: L(sx) = sL(x)$
Im not sure how to approach solving these two in $M_{2 \times 2}(\mathbb R)$ and I do not know where to start for checking for injectivity and surjectivity. Any help would be appreciated!
 A: This should get you started: Write out $x$ and $y$ as general $2 \times 2$ matrices:
$$
x =
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix},
y = \begin{pmatrix}
e & f \\
g & h \\
\end{pmatrix}
$$
Now compute, what is $L(x)$? What is $L(y)$? What is $L(x) + L(y)$? What is $L(x+y)$? Are these last two equal?
Next, if $s \in \mathbb R$, what is $sx$? What is $L(sx)$? What is $s L(x)$? Are these last two equal?
A: $L : M_2(\Bbb{R}) \to M_2(\Bbb{R}) $ defined by $$L(M) =M+M^T$$
$•\forall M, N\in M_2(\Bbb{R}) $ and $\lambda \in \Bbb{R}$
$\begin{align}L(M+\lambda N)&=(M+\lambda N) +(M+\lambda N)^T\\&= (M+\lambda N)+(M^T+\lambda N^T)\\&=(M+M^T)+\lambda (N+N^T)\\&=L(M)+\lambda L(N)\end{align}$
Hence $L$ is linear.
$\begin{align}•\ker(L)& =\{M\in M_2(\Bbb{R}): L(A) =\textbf{O}\}\\&=\{M\in M_2(\Bbb{R}):M^T=-M\}\\&=Skew M_2(\Bbb{R})\end{align}$
A linear map  is injective iff kernel is trivial.
$•$ Suppose for $A\in  M_2(\Bbb{R}) $ $\exists M\in M_2(\Bbb{R})$ such that $L(M)=A$ .
$A=M+M^T$
$A^T=(M+M^T) ^T= M^T+M=A$
Hence  $A\in M_2(\Bbb{R}) $ is symmetric matrix.
Hence $A\in Skew M_2(\Bbb{R}) $ has no preimage.
Hence $L$ is not surjective.
A: $L(A+B)=(A+B)+(A+B)^{T}=A+B+A^{T}+B^{T}=(A+A^T)+(B+B^T)=L(A)+L(B)$. Now check the second property. To check injectivity, let $A$ be a matrix such that $A=-A^{T}$ . Must $A$ be the zero matrix? For surjectivity, let $B$ be a $2x2$ matrix. Must there exist a $2x2$ matrix  $A$ such that $A+A^{T}=B$?
