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Dec 22, 2014 at 18:33 vote accept Scentless Apprentice
Dec 20, 2014 at 4:41 comment added Scentless Apprentice That's what I thought. Thanks!
Dec 20, 2014 at 2:19 comment added Nick Matteo @ScentlessApprentice: Suppose $w$ and $v$ are eigenvectors for the eigenvalue $\lambda$, so $T w = \lambda w$ and $T v = \lambda v$. Then $T(w + v) = T w + T v = \lambda w + \lambda v = \lambda(w + v)$.
Dec 19, 2014 at 23:18 comment added Scentless Apprentice Terrific explication, @Kundor. Very much appreciated. One last request for clarification. You say that "because $u_{i}$ is a linear combination of eigenvectors for $\lambda_{i}$, it is also an eigenvector for $\lambda_{i}$." This wasn't obvious to me at all. Is this because null($T - \lambda_{i}$) is a subspace of $V$?
Dec 19, 2014 at 20:01 history answered Nick Matteo CC BY-SA 3.0