Show any matrix $A_{n\times n}$ can be written as sum of two nonsingular matrices

Question

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?

Answer 1

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 $A=(A-\alpha I)+\alpha I$.

Answer 2

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.