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Let $f, g : U\rightarrow V$ be linear maps and $\lambda\in F$. Then the maps $f + g : U\rightarrow V$ and $\lambda f : U \rightarrow V$ are linear.

My attempt at the proof for the first statement is as follows:

Let $u,z\in U$ and $a\in F$, using a linearity check

by definition of $f + g$ $$(f + g)(au + z) = f(au + z) + g(au + z)$$ by linearity of $f$ and $g$

$$= (af(u) + f(z)) + (ag(u) + g(z)) $$ by basic properties of vector spaces

$$= af(u) + ag(u) + f(z) + g(z)$$ by an axiom of vector spaces

$$= a(f(u) + g(u)) + (f(z) + g(z))$$ by definition of $f + g$.

$$= a(f + g)(u) + (f + g)(z)$$ Hence, $f + g$ is linear

Is this the correct approach. What is the proof of $\lambda f : U \rightarrow V$ to be linear?

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  • $\begingroup$ Yes, you are correct. Do the same for $(\lambda f)(x)=\lambda f(x)$. $\endgroup$ – Emilio Novati May 7 '16 at 15:54
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You have $$ (\lambda f)(u+av)=\lambda f(u+av)=\lambda (f(u)+af(v))=\lambda f(u)+a\lambda f(v) =(\lambda f)(u)+a(\lambda f)(v). %Comment added into order to correct one character typo. $$

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