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I'm studying Artin's Algebra right now, and I'm stuck from the beginning on his proof of the Jordan Decomposition Theorem. It goes somewhat like this:

Theorem: Let $T$ be a linear operator on a finite-dimensional complex vector space $V$. There is a basis B of $V$ such that the matrix of $T$ with respect to B has Jordan form.

Proof. Induction on the dimension of $V$ allows us to assume that the theorem is true for the restriction of $T$ to any proper invariant subspace. So if $V$ is the direct sum of proper $T$-invariant subspaces, say $V_1 \bigoplus \space . . .\space \bigoplus V_r$, with $r > 1$, then the theorem is true for $T$.

Artin then goes on to complete a (pretty long) proof of the theorem, but I can't even understand this first step; how does induction on the dimension of $V$ even allow us to assume anything? Can someone please clarify this for me? Thank you in advance!

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Artin says that if we can decompose $V$ as a direct sum of proper $T$-invariant subspaces, and you should be aware that $\dim V_i<\dim V$ with $T(V_i)\subset V_i$. That's where the induction works. And later on, you see that things will be straightforward in invariant subspaces.

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  • $\begingroup$ Sorry, I'm still a little confused. Which is the induction assumption, and which is the induction step? $\endgroup$ – DavidNiu Jul 23 at 3:23
  • $\begingroup$ @DavidNiu Artin latter shows the existence of Jordan generators, and he proves that we can reduce to Jordan form in each direct summand. $\endgroup$ – Bach Jul 23 at 3:32
  • $\begingroup$ Yes, I do understand that part, but how does that require induction on the dimension of $V$? Is it that for every new dimension, a new Jordan generator is essentially created? $\endgroup$ – DavidNiu Jul 23 at 3:38
  • $\begingroup$ @DavidNiu Maybe you want to read Step 3 where the induction assumption is used. $\endgroup$ – Bach Jul 23 at 3:45

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