# Jordan Decomposition Theorem Proof with Induction

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!

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.
• 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? – DavidNiu Jul 23 at 3:38