Show any matrix $A_{n\times n}$ can be written as sum of two nonsingular matrices A nonsingular matrix is one which is invertible, and hence the determinant is not equal to $0$. So at first I thought about having $\det(A_{1})+\det(A_{2})=\det(A_{1}+A_{2})$ and then the resulting sum being a invertible matrix, but this is not generally the case. 
Then I thought about eigenvalues, as the determinant is the product of the eigenvalues and that using that I could showing that the spectrum of the sum is equal to the sum of the spectrums. 
Is that a better way? 
 A: Write $A=L+U$ where $L$ is lower triangular, $U$ is upper triangular. $L$ and $U$ have the same diagonal, equal to half the diagonal of $A$, except where zeros appear. In this case, use $1$ in $L$ and $-1$ in $U$. This makes the diagonals of $U$ and $L$ have no zero entries and so $L$ and $U$ are nonsingular.
A: Suppose that $\alpha\ne 0$ is not an eigenvalue of $A$. Then for any non-zero $v$ we have $(A-\alpha I)v\ne 0$, so $A-\alpha I$ is non-singular. So is $\alpha I$, and $A=(A-\alpha I)+\alpha I$.
A: The purpose of this answer is to elaborate Erick Wong's commentary and say something about the $\mathbb{F}_2$ case:
Suppose $F$ is a field with at least $3$ distinct elements $0,1,\alpha$. Let $A\in$Mat$_n(F)$ and $d_i=a_{ii}$ be its diagonal elements. If $d_i\neq1$, put $e_i:=1$; if $d_i=1$, put $e_i:=\alpha$. Then $$A=L+U,$$
where $L$ is lower triangular with diagonal $e_i$ and $U$ is upper triangular with diagonal $d_i-e_i$. By construction, the diagonal entries of both $L$ and $U$ are all nonzero; since the determinant of a triangular matrix is the product of its diagonal entries, both $L$ and $U$ are nonsingular.
When $F:=\mathbb{F}_2$ is the field with $2$ elements, we can get every matrix in Mat$_n(F)$ as a sum of at most $n+2$ nonsingular matrices (I don't know if this number is sharp; this is just what I came up with). Here is how:
Denote by $G_i$ the matrix full of ones except for the diagonal, which is $0$ except for the $i$th entry, which is $1$ (i.e., $G_i=1-I+E_i$, where $1$ is the full-of-ones matrix, $I$ is the identity and $E_i$ is the diagonal $i$th elementary matrix). Observe that $G_i$ has a row $R$ full of ones, while the other rows have just one $0$, each in a different position. Since the characteristic is $2$, by sequentially adding $R$ to the other rows and after that sequentially adding the other rows to $R$, we get $\det(G_i)=\det(I)=1$. Therefore the matrices $G_i$ are nonsingular.
Now write
$$A=L+U+\sum_{i=1}^n d_iG_i,$$
where $L$ is lower triangular with diagonal full of ones and $U$ is upper triangular with diagonal full of ones.
A: The works over any field. Suppose $A$ is a matrix whose size is at least $2\times2$. By using elementary row and column operations or by using Smith normal form, we can write a square matrix $A$ in the form of $PDQ$ where $P,Q$ are invertible and $D$ is a diagonal matrix. Let $C\ne I$ be any circulant permutation matrix. Then $CD$ has a zero diagonal. Hence $CD=L+U$, where $L$ is strictly lower triangular and $U$ is strictly upper triangular. Now
$$
A=PC^T(L+I)Q+PC^T(U-I)Q
$$
is the sum of two invertible matrices.
The case where $A$ is $1\times1$ is a bit different. If the underlying field $\mathbb F$ has at least three elements, we have $A=(A-B)+B$ where $B$ is any element in $\mathbb F\setminus\{0,A\}$. However, if the field has only two elements (i.e. if it is isomorphic to $GF(2)$), the element $1$ is not the sum of two invertible elements in $\mathbb F$.
