0
$\begingroup$

I'm trying Linear Algebra Done Right 5.A 35.

  1. Suppose $V$ is finite-dimensional, $T \in \mathcal L(V) $, and $U$ is invariant under $T$. Prove that each eigenvalue of $T/U$ is an eigenvalue of $T$.

What I'm confused is whether $T/U(v+U)=\lambda(v+U)$, where $\lambda$ is eigenvalue of $T/U$ ($v \notin U$), equals $\lambda v + \lambda U$ or $\lambda v +U$ and how the right equality makes sense.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Scalar multiplication on $V/U$ is defined by $$\lambda(v+U)=\lambda v+U$$

Have a look at page $96$.

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .