I am learning linear algebra and bought the book from Gilbert Strange: Introduction to linear algebra and trying to understand the four fundamental subspaces. I know that a system is solvable if b is in the column space of A. A system is not solvable if there are numbers that multiply the equations in A so they add up to $0 = 1$
As he says, this is called Fredholm's law:
A system of linear equations is solvable if one of these problems has a solution: $Ax = b$ or $A^Ty = 0$ with $y^Tb = 1$
I don't really get the last part of this one. Lets assume I have a system of linear equations like the following: $$ A = \begin{bmatrix} 1&-1&0\\ 0&1&-1\\ 1&0&-1\\ \end{bmatrix}, b = \begin{bmatrix} 1\\1\\1\\ \end{bmatrix} $$ According to Fredholm's law the system is solvable if b is in $C(A)$ which is not the case in this example. But what about the second part? If I choose the following values for $y = \begin{bmatrix}1\\1\\-1\\\end{bmatrix}$ then $A^Ty = 0$ and $y^Tb = 1$. But this system is not solvable. What am I understanding wrong here?
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$\begingroup$As you already stated, the system has no solution, since $b\notin C(A)$. According to Wikipedia, exactly one must hold:
$1)$ Either $Ax = b$ has a solution or
$2)$ The equation $A^T y = 0$ has a non - trivial solution $y$, with $y^T b \neq 0$.
In the example you described, we are in the second case, which means the proposition $1)$ cannot happen.
Equivalently, if we read the next line, the equation $A x = b$ has a solution iff for any $y$ s.t $A^T y = 0$, also holds $y^T b=0$.
As far as the book is concerned, could you provide the chapter which includes Fredholm's alternative?
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