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I am just checking different analogous of $\alpha:V \longrightarrow W$ being affine. I have problems with this one:

$\alpha:V \longrightarrow W$ affine $\iff$ $G_\alpha=\{(v,\alpha(v): v\in V)\}$ is a affine subapce of $V\times W$.

Intuitively ($\Leftarrow$) should be clear, because any linear combination of $(v,\alpha(v))$, $\sum_{i=1}^{k}\lambda_i (v,\alpha(v))$ with $\sum_{i=1}^{k}\lambda_i=1$ is still in A but how is possible to formalize it in a nice way?

($\Rightarrow$) is not so clear to me

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up vote 1 down vote accepted

If $\alpha$ is afine, let $(v_i, \alpha(v_i)$ be arbitrary tuple of vectors in $G_\alpha$. You want to show that $\sum_{i=1}^k \lambda_i (v_i, \alpha(v_i))$ is in $G_\alpha$ for any $\lambda_i$ with $\sum \lambda_i=1$. But that is just $(\sum_{i=1}^k \lambda_i v_i, \sum_{i=1}^k \lambda_i \alpha(v_i))$. Now you can use the assumption that $\alpha$ is affine to finish.

Conversely, suppose $G_\alpha$ is affine. You want to see that for any tuple $v_i$ in $V$ the map $\alpha$ takes $\sum_{i=1}^k \lambda_i v_i$ to $\sum_{i=1}^k \lambda_i \alpha (v_i)$. That is just saying that $(\sum_{i=1}^k \lambda_i v_i, \sum_{i=1}^k \lambda_i \alpha (v_i))$ is in $G_\alpha$. Now use that $(v_i, \alpha(v_i))$ are in $G_\alpha$ and $G_\alpha$ is affine.

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