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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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    Why negative vote?2012-01-30
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    For starters, maybe because none of LS, N, A, b or w were defined. (Did not downvote, yet.)2012-01-30
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    In adding $\LaTeX$ to your question I’ve also added my guesses about the unexplained notations; please make sure that I’ve preserved the intended meaning. Also, what is $w$?2012-01-30
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    If @Brian is right in his interpretation/rewriting of the question, here is a hint: compute $A\cdot(x+ty)$.2012-01-30
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    Ls=linear system , A= coefficients of the matrix and b is the vector of constants2012-01-30

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