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Suppose $x \in \operatorname{LS}(A,b)$ and $y \in \operatorname{N}(A)$. Show that $x+ty$ is in $\operatorname{LS}(A,b)$ for all $t \in \mathbb{C}$.

Edit: Here $LS(A,b)$ is the set of $x$ such that $Ax=b$, and $N(A)$ is the null space of $A$.

I believe this is considered to be in the category of null space... Because I think that by def. $y-w \in \operatorname{N}(A)$?

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    Ls=linear system , $A$= coefficients of the matrix and b is the vector of constants2012-01-30

1 Answers 1

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Here is a rewriting of the question, using only the explanations finally provided by the OP.

Assume that $A\cdot x=b$ and $A\cdot y=0$. Show that, for every $t$ in $\mathbb C$, $A\cdot(x+ty)=b$.

What this formulation shows, I believe, is the following:

  1. All the LS, N, w stuff in the original post is not needed and may be more an obstacle than a help to the comprehension of the question.
  2. Crucial hypotheses are implicit, which are that one is working with some vector spaces $V$ and $W$ (over $\mathbb C$), that $x$ and $y$ belong to $V$ and $b$ belongs to $W$, and that $A:V\to W$ is linear.
  3. The proof is direct once the definitions related to these hypotheses are recalled.

For example, to reduce the general case to the $t=1$ case, one could first show that $A\cdot(ty)=0$. Can you do that?