Direct sum of V1 = {A ∈ V : At = A} and V2 = {A ∈ V : At = −A} equals V = Mn(R)

Question

Be:

• $V=M_n(ℝ)$
• $V_1$ = {A ∈ V : $A^t$ = A}
• $V_2$ = {A ∈ V : $A^t$ = -A}

Knowing that $V_1$ and $V_2$ are subspaces of V, prove that $V = V_1 ⊕ V_2$.

So, first of all, I already know how to prove that $V_1$ ⊕ $V_2$ is a direct sum.

To prove that V = $V_1$ ⊕ $V_2$, I need to prove that:

1. $V_1 ⊕ V_2 ⊂ V$, and
2. V ⊂ $V_1$ ⊕ $V_2$

Thanks to subspace properties, proving 1) is easy enough. But I couldn't prove 2). My Linear Algebra professor published his answer, which is something like:

Take $A ∈ V$, and

$B=\frac{1}{2}(A+A^t)$, and

$C=\frac{1}{2}(A-A^t)$

Therefore:

$B^t=\frac{1}{2}(A+A^t)^t$

$B^t=\frac{1}{2}(A^t+(A^t)^t)$

$B^t=\frac{1}{2}(A^t+A)$

$B^t=\frac{1}{2}(A+A^t)$, thus $B^t = B$ and B ∈ $v_1$

Similarly, for C:

$C^t=\frac{1}{2}(A-A^t)^t$

$C^t=\frac{1}{2}(A^t-(A^t)^t)$

$C^t=\frac{1}{2}(A^t-A)$

$C^t=\frac{1}{2}(A-A^t)$, thus $C^t = -C$ and C ∈ $v_2$

Thus: $B + C ∈ V_1 + V_2$, and $B + C = A$. Proving that $V_1 ⊕ V_2 ⊂ V$.

I kind of get the proof. But I can't muster the thought process which led him to come up with this:

Take $A ∈ V$, and

$B=\frac{1}{2}(A+A^t)$, and

$C=\frac{1}{2}(A-A^t)$

And what that thought process was, is precisely what I'd like to know. (Different ways to prove what the question asks are welcome too).