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A subspace $W$ of $V$ is called invariant under the linear transformation $T:V\to V$ if $T(W)\subseteq W$. Prove that $S: V/W\to V/W$ defined by $v + W\mapsto T(v) + W$ is a linear transformation.

Tried to compute $S(\alpha_1v_1 + W + \alpha_2v_2 + W)$, but couldn't get it to work out. Any direction would be helpful!

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  • $\begingroup$ As is often the case with quotient spaces, the important point is showing that $S$ is well-defined. This is where $T(W)\subset W$ is needed. $\endgroup$ – carmichael561 Dec 20 '16 at 1:27
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Use the fact that $W + W = W$ and that $\alpha W = W$ for all nonzero scalars $\alpha$. So, $$ \begin{array}{lll} & & S(\alpha_1v_1 + W + \alpha_2v_2 + W) \\ & = & S(\alpha_1v_1 + \alpha_2v_2 + W)\\ & = & T(\alpha_1v_1 + \alpha_2v_2) + W\\ & = & \alpha_1 T(v_1) + \alpha_2 T(v_2) + W \\ & = & \alpha_1 T(v_1) + W + \alpha_2 T(v_2) + W\\ & = & \alpha_1 S(v_1 + W) + \alpha_2 S(v_2 + W)\\ \end{array} $$

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  • $\begingroup$ So where did you use the fact that $W$ is invariant under $T$? $\endgroup$ – Nate Stemen Dec 20 '16 at 1:40
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Recall that: $\alpha_1 v_1 + W + \alpha_2 v_2 + W = (\alpha_1 v_1 + \alpha_2 v_2) +W$. So,

$$S(\alpha_1 v_1 + W + \alpha_2 v_2 + W) = S((\alpha_1 v_1 + \alpha_2 v_2) +W) = T(\alpha_1 v_1 + \alpha_2 v_2) + W.$$

Now, using linearity of $T$, can you finish?

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This is well explained in Chapter 3 of Axler's Linear Algebra Done Right.

First of all, one needs

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where the "affine subsets" are given by the following definition: enter image description here

Then

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