Proof: Space of real matrices is the direct sum of spaces of symmetric and skew-symmetric matrices This is the problem I am trying to solve (from Artin's Algebra, Chapter 3: Vector Spaces, Section 5). I did some searching around here, and I found similar problems but no problems phrased in exactly this way:

Prove that the space of real  $n \times n$ matrices is the direct sum of the spaces of symmetric matrices ($A^t = A$) and skew-symmetric matrices ($A^t = -A$)

This is my solution so far:

Let $Y$ be the space of symmetric real nxn matrices and $Z$ be the space of skew-symmetric real nxn matrices.
We will prove that (I based my strategy on the direct sum definition provided in the text):


*

*$Y$ and $Z$ are independent, and $Y+Z = \mathbb{R}^{nxn}$

*The bases of $Y$ and $Z$, $B_y$ and $B_z$ respectively, can be used to construct a basis $B = B_y \cup B_z$ of $B$



Part 1:
Let $y \in Y$, $z \in Z$. Let $(y)_{ij}$ denote the i,j-th element of y and let $(z)_{ij}$ denote the i,j-th element of z. 
Consider the i,j-th element of their sum, $(y+z)_{ij}$. $(y+z)_{ij} = 0 \implies (y)_{ij} = -(z)_{ij} \implies y = -z$
Since both Y and Z are closed under scalar multiplication and addition, $y = -z$ implies that $z = -y$ must be an element of Z, and therefore, $z = y$ must be an element of Z. So, we have
$$ y = -z \implies y = -y \implies y = \bar{0}$$
We have proven that $Y$ and $Z$ are independent spaces. 
It seems trivial that $Y+Z = \mathbb{R}^{nxn}$, and I'm struggling to prove that bit.

Part 2: Construct $B_y$ as follows: Let there be $\frac{n^2-n}{2}$ $b_{ij}$ matrices, populated with zeroes, except for the i,j-th and j,i-th entries, which equal 1.
Construct $B_z$ as follows: Let there be $\frac{n^2-n}{2}$ $b_{ij}$ matrices, populated with zeroes, except for the i,j-th entry, which equals 1, and the j,i-th entry, which equals -1.
It seems trivial that these are bases for their respective spaces, so I think I can skip that. However, how do I prove that their union spans $\mathbb{R}$? It seems really obvious and intuitive but I am getting frustrated trying to prove this.

You do not need to work element-wisely to get the proof done. In addition, while there may be multiple equivalent conditions to show a sum of two subspaces $V_1 + V_2$ is a direct sum, to verify that $V_1 \cap V_2$ is $\{0\}$ is the most expedient for your question. Below is a complete proof.
Let $M$ denote the space of real $n \times n$ matrices, $S$ denote the space of real $n \times n$ symmetric matrices and $K$ denote the space of real $n \times n$ skew-symmetric matrices. I assume that you have verified $M$ is a vector space and $S$, $K$ are subspaces of $M$. To show $M$ is a direct sum of $S$ and $K$, it suffices to show 


*

*For any $A$ in $M$, $A = B + C$, where $B \in S$ and $C \in K$.

*$S \cap K = \{0\}$. 
Per amd's comment, 1. is true by taking $B = (A + A')/2$ and $C = (A - A')/2$. It is readily to check $B = B'$ and $C = -C'$. Hence $B \in S$ and $C \in K$.
To prove 2., suppose $D \in S \cap K$, then $D = D'$ and $D = -D'$, which implies $D = -D$, i.e., $D = 0$, therefore 2. holds. This completes the proof.
