Product of minimal polynomials of restriction to invariant subspaces equals the minimal polynomial of the vector space

100 Views Asked by At

I have been reading Schaum's outlines Linear Algebra and this was a motivating result to prove the primary decomposition theorem. Statement of Theorem 10.8: enter image description here

Its proof as in my textbook: enter image description here

I got stuck in the fourth line which says

$g(T_1)=0$ and $h(T_2)=0$ because $U=Ker[g(T)]$ and $W=Ker[h(T)]$

I don't know how this argument works. An explanation would be helpful.

1

There are 1 best solutions below

1
On BEST ANSWER

Use that $T_i$ is the restriction of $T$:
If $u\in U=\ker g(T)$, then $g(T_1)u=g(T|_U)u=g(T)u=0$.