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Given a linear transformation $T:B\to W$ defined on a subspace $B$ of $V$, I want to define a function $f: V\to W$ such that $f|_B = T$ and it’s also linear. I know that $B$ has a complement $M$ such that V = B + M (direct sum) and define $f$ to have the same values of $T$ on $B$ and value $0$ on $M$, but I don’t know how to prove that $f$ is linear. In particular, I’m stuck on showing for $v_1\in B, v_2\in M$ (none of two are zero, the cases with zeros is also ok). Any help?

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  • $\begingroup$ Write all the conditions and the conclusion using formulas, and it will be trivial. $\endgroup$ – W. mu Jan 27 '18 at 3:39
  • $\begingroup$ $f(v_1)+f(v_2)=f(v_1)=T(v_1)$ and $f(v_1+v_2)=0$. I can’t see why they must be equal. $\endgroup$ – AnalyticHarmony Jan 27 '18 at 3:51
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Since $V=M \oplus B$, every element in $V$ can be written uniquely as $v = m + b$, for $m \in M$ and $b \in B$.

Define the extension $\bar{f}$ of $f$ on $V$ as follows:

$\bar{f}(v):= f(b)$, if $v = m + b$.

We claim that this is linear. Given $v_1 = m_1 + b_1$ and $v_2 = m_2 + b_2$, we have

$\bar{f}(v_1 + v_2):= f(b_1 + b_2) = f(b_1) + f(b_2)=: \bar{f}(v_1)+\bar{f}(v_2)$,

where the second equality follows since $f$ is linear on $B$ by definition.

Now given $\alpha \in \Bbb R$ and $v = m + b \in V$, we have

$\bar{f}(\alpha v):=f(\alpha b) = \alpha f(b)=: \alpha \bar{f}(v)$

since $f$ is linear.

Since $v_1,v_2$ are arbitrary, $\bar{f}$ is linear and clearly extends $f$.

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  • $\begingroup$ Ok, but this is not the function I have defined, isn’t? The way I have defined will make the extension not linear? $\endgroup$ – AnalyticHarmony Jan 27 '18 at 3:58
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    $\begingroup$ @NoGoodAtMath This is exactly the function you have described in your post and it is linear. $\endgroup$ – 3-in-441 Jan 27 '18 at 3:59

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