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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?

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up vote 6 down vote accepted

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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First, I have seen $(A-\alpha*I)V\neq0$ and also $(\alpha*I-A)v\neq=0$ Is there a difference in values or should I view it like absolute values? Also, would using $\alpha*I$ count as a trivial case? –  Edgar Aroutiounian Sep 26 '12 at 1:28
    
@Edgar: The vectors $(A-\alpha I)v$ and $(\alpha I-A)v$ are additive inverses of each other, so in general they’re not the same, but if one is $0$, then so is the other. Whether $\alpha I$ counts as a trivial case depends on context: there’s no absolute definition of trivial here. (To get $\alpha I$, just leave a space between \alpha and I.) –  Brian M. Scott Sep 26 '12 at 1:31
    
thank you for your help! –  Edgar Aroutiounian Sep 26 '12 at 1:34
    
@Edgar: My pleasure! –  Brian M. Scott Sep 26 '12 at 1:36
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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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