Is there ever a case where a mapping is not preserved/closed under scalar multiplication? Every problem I have encountered has always been closed under scalar multiplication. However, closed under addition is easier to see if a mapping is linear or not. What mappings are there that prove otherwise?
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A "linear mapping" by definition preserves both scalar multiplication and vector addition.
If you allow an arbitrary mapping, then anything goes. For example, let $V=\mathbb{R}^2$, we can define $f:V\to V$ via $f(x,y)=(1,2)$. This mapping sends everything to the same vector, namely $(1,2)$. This one is not linear, it preserves neither scalar multiplication nor vector addition.
vadim123
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Or perhaps, for something that may feel less like a cheat, $f(x,y)=f(x+1,y)$. – hmakholm left over Monica Dec 04 '14 at 18:36