# A problem on invariant subspaces.

Let $$W$$ be an invariant subspace for $$T$$.Prove that the minimal polynomial for the restriction operator $$T_W$$ divides the minimal polynomial for $$T$$,without referring to matrices.

Attempt Suppose $$f$$ is the minimal polynomial for $$T_W$$ where $$f(T_W)=0$$. If $$f$$ is the minimal polynomial polynomial for $$T$$ then $$f(T)\alpha=0$$ , for every $$\alpha \in V$$. If $$f(T_W)$$ doesn't divide $$f(T)$$ then there exist $$g(T_W)$$ and $$r(T_W)$$ such that $$f(T)=g(T_W)f(T_W)+r(T_W)$$ where either $$r(T_W)=0$$ or $$deg(r(T_W)). If $$r(T_W)=0$$ then we're done. Otherwise we'll get a contradiction because $$f(T_W)$$ is minimal polynomial.

This is second way to approach this problem:

Given $$T (W)\subseteq W$$. Let $$f$$ be the minimal polynomial of $$T_W$$ then $$f(T_W)=0$$. The minimal polynomial of $$T_W$$ divides any polynomial $$f(t)$$. If $$f$$ is the minimal polynomial of $$T$$ then $$f(T)v=0$$, where $$v\neq 0$$. Also we have $$f(T_W)w=0$$ for every $$w\in W$$.

From here how to show that $$f(T_W)|f (T)$$?

Is my first proof correct?

Any help or suggestion will be appreciable.

Thanks!

There are some stylistic/notational suggestions that I have, but in terms of ideas, both work. In fact, both are really just the same idea. You use the following theorem:

If $$p$$ is the minimal polynomial of some $$T$$, and $$h$$ is a polynomial such that $$h(T) = 0$$, then $$p$$ divides $$h$$.

The proof of this is exactly what you laid out: apply the division algorithm to get a partial quotient $$q$$ and remainder $$r$$, which satisfy $$h = p \cdot q + r$$, where $$\deg(r) < \deg(p)$$. But then $$r(T) = 0$$, with degree less than the minimal polynomial, which implies $$r = 0$$.

So, with that in mind, there's not a lot of differences in your approach. In the first approach, you simply spell out the steps in the theorem explicitly, rather than referring to them implicitly through invoking the theorem.

Now for the nitpicks:

• "If $$f$$ is the minimal polynomial polynomial for $$T$$ then $$f(T)\alpha=0$$, for every non zero $$\alpha\in V$$"
It's also true (as it will be for any polynomial) when $$\alpha = 0$$. I don't see the value in excluding $$\alpha = 0$$ here.
• "If $$f(T_W)$$ doesn't divide $$f(T)$$..."
I very much disapprove of using $$f$$ to denote both the minimal polynomial of $$T$$ and the minimal polynomial of $$T_W$$. The notations $$f(T_W)$$ and $$f(T)$$ suggest a single polynomial $$f$$, but with two different operators $$T$$ and $$T_W$$ substituted in. So, $$f$$ is a single polynomial, $$f(T)$$ is an operator on the domain of $$T$$, and $$f(T_W)$$ is an operator on $$W$$.
More generally, I think you should learn to distinguish between a polynomial $$f$$ and $$f(x)$$ or $$f(T)$$. Writing $$f$$ means you are just referring to the polynomial. If you write $$f(x)$$ where $$x$$ lies in the domain of $$f$$ (e.g. complex numbers), then $$f(x)$$ is the image of the point $$x$$ under $$f$$, and lies in the codomain of $$f$$. Similarly, $$f(T)$$ means to substitute $$T$$ into the polynomial (even though the domain doesn't include operators!), producing a single operator.