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Our lecturer gave us the following proof for the Solutions of Linear Equations but I don't understand the point of what he does in the second part. What is the point of the second part where he has $T(z - x_0) = 0$, so $z - x_0$ is an element of Ker(T). I don't see what this is proving.

It seems to me that the statement is proved in the first part that ends with 'so x is a solution'. As he has shown T(x) = y is satisfied by the particular solution $x_0$ along with the general solution of the homogeneous equation u. As in T(u) = 0.

Anyone able to explain this solution to me?

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3 Answers 3

The first part proves that all vectors of the form $x_0 + u$ with $u \in \ker T$ are solutions. This provides a sufficient condition for a vector to be a solution. But note that this does not mean that all solutions are of this form. This is what is proved in the second part.

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T(z-x0)=T(z)-T(x0)=y-y=0. He proves that evry solution is of the form of the statement in the second part, and in the first he proves that all vector of the form of the statement (x0+u) is also a solution.

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$z= x_0 + (z-x_0) $
$T(z)= T(x_0) + T(z-x_0) $
$T(z)= 0 + T(z-x_0) $

z is solution $ \iff $ $z-x_0 \in Ker(T)$

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