# Issue with proof that restriction of a diagonalizable operator is also diagonalizable.

Suppose $$V$$ is a vector space over $$\mathbb{C}$$, and $$A$$ is a linear transformation on $$V$$ which is diagonalizable. I.e. there is a basis of $$V$$ consisting of eigenvectors of $$A$$. If $$W\subseteq V$$ is an invariant subspace of $$A$$ (so $$A(W)\subseteq W$$), show that $$A|_W$$ is also diagonalizable.

Below is a proof of this fact from This is a proof for Diagonalizable transformation restricted to an invariant subspace is diagonalizable.

Theorem. A linear transformation is diagonalizable if and only if its minimal polynomial splits and has no repeated factors.

Proof. This follows by examining the Jordan canonical form, since the largest power of $$(x-\lambda)$$ that divides the minimal polynomial is equal to the size of the largest block of corresponding to $$\lambda$$ of the Jordan canonical form of the linear transformation. (Use the fact that every irreducible factor of the characteristic polynomial divides the minimal polynomial, and that the characteristic polynomial must split for the linear transformation to be diagonalizable to argue that you can restrict yourself to linear transformations with Jordan canonical forms). QED

Theorem. Let $$A$$ be a linear transformation on $$V$$, and let $$W\subseteq V$$ be an $$A$$-invariant subspace. Then the minimal polynomial of the restriction of $$A$$ to $$W$$, $$A|_{W}$$, divides the minimal polynomial of $$A$$.

Proof. Let $$B=A|_{W}$$, and let $$\mu(x)$$ be the minimal polynomial of $$A$$. Since $$\mu(A)=0$$ on all of $$V$$, the restriction of $$\mu(A)$$ to $$W$$ is $$0$$; but $$\mu(A)|_{W} = \mu(A|_{W}) = \mu(B)$$. Since $$\mu(B)=0$$, then the minimal polynomial of $$B$$ divides $$\mu(x)$$. QED

Corollary. If $$A$$ is diagonalizable, and $$W$$ is $$A$$-invariant, then the restriction of $$A$$ to $$W$$ is diagonalizable.

Proof. The minimal polynomial of $$A$$ splits and has no repeated factors; since the minimal polynomial of $$A|_W$$ divides a polynomial that splits and has no repeated factors, it follows that it itself has no repeated factors and splits. Thus, the restriction of $$A$$ to $$W%$$ is also diagonalizable. QED

My question is about the corollary; how does the fact that the minimal polynomial $$A|_W$$ divides a polynomial that splits and has no repeated factors allow us to conclude that it also splits and has no repeated factors?

• I don't see how you can. Any scalar will divide $f$, but will not split into linear factors. Commented Apr 26, 2022 at 2:42
• I've edited it to be more specific to the issue I was having. Commented Apr 26, 2022 at 2:53

All polynomials split over an algebraically closed field such as $$\mathbb C$$. Over a non-algebraically closed field such as $$\mathbb R$$, splitting means that all of the roots are in that field. The roots of a factor of a polynomial are a subset (with multiplicity) of the roots of that polynomial, so it's trivially true that a polynomial splitting means that its factors split. Similarly for the factors being distinct: if the factor polynomial has repeated factors, then those factors must have been repeated in the original set of factors.