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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?

3 Answers 3

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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.

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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.

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    This works over any field of characteristic $\ne 2$, and with a tiny modification also works for every field except $\mathbb F_2$.2015-03-19
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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 $A=(A-\alpha I)+\alpha I$.

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    @Edgar: My pleasure!2012-09-26