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Assume $A$ is a $n\times n$ matrix where $A=B+E$, $E$ is an identity matrix. Prove:

If $B$ is a positive definite matrix, then $Ax=F$ has unique solution.

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$x,F$ are vectors, right? Also, is there an assumption that $F \neq 0$? Otherwise $x$ is not unique unless $A$ is full rank (i.e., $x = 0$). –  user2468 Mar 13 '12 at 5:44
    
If $A=B+I$, $Ax=f = (B+I)f = Bf + If$. $If$ has one solution for obvious reasons. $Bf$ also has a unique solution since $B$ is Positive Definite. (Thus meaning all eigenvalues are $>0$ and hence the matrix has trivial nullspace). Thus proving what you need. –  Inquest Mar 13 '12 at 6:18

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Suppose there are two solutions $x$ and $y$ such that $Ax=Ay=F$, then take the difference: $A(x-y)=0$, but $A$ is positively definite, hence, $x=y$.

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