# If T is diagonalizable then is restriction operator diagonalizable?

• I
CGandC
TL;DR Summary
Does minimal polynomial zero out the linear operator restricted to any subspace?
The usual theorem is talking about the linear operator being restricted to an invariant subspace:
Let ##T## be a diagonalizable linear operator on the ##n##-dimensional vector space ##V##, and let ##W## be a subspace of ##V## which is invariant under ##T##. Prove that the restriction operator ##T_W## is diagonalizable.​
I had no problem understanding its proof, it appears here for example: https://math.stackexchange.com/ques...-t-w-is-diagonalizable-if-t-is-diagonalizable

However, I had difficulty understanding why we needed the assumption that ## W ## is ##T##-invariant, I mean - If ## m_T(x) ## is the minimal polynomial of ##T## so ## m_T(T)=0 ## and thus for any subspace ## W \subseteq V ## ( not necessarily ## T##-invariant ) ## m_T(T_W) =0 ##; so why in the above theorem it was necessary for ## W \subseteq V ## to be ## T ##-invariant?

Homework Helper
2022 Award
If $W$ is not $T$-invariant, then the matrix representation of $T_W$ is not square: it must include additional rows to account for the part of $T_W(W)$ which is not in $W$. In what sense is this non-square matrix "diagonalizable"?

This theory is defined for linear maps $T: V \to V$ where the codomain is the same as the domain, rather than some different space. A restriction of $T: V \to V$ to a subspace $W \subset V$ will only qualify as such a map if we have $T_W: W \to W$, ie. $W$ is $T$-invariant.

• WWGD, CGandC and topsquark
CGandC
If $W$ is not $T$-invariant, then the matrix representation of $T_W$ is not square: it must include additional rows to account for the part of $T_W(W)$ which is not in $W$. In what sense is this non-square matrix "diagonalizable"?

This theory is defined for linear maps $T: V \to V$ where the codomain is the same as the domain, rather than some different space. A restriction of $T: V \to V$ to a subspace $W \subset V$ will only qualify as such a map if we have $T_W: W \to W$, ie. $W$ is $T$-invariant.
Although it makes sense that the matrix representation of a diagonalizable operator should be square matrix, I still don't see how this knowledge is necessary for proving the theorem for non-invariant subspace since a linear operator is also called diagonalizable iff its minimal polynomial decomposes to distinct linear factors of multiplicity 1 - this theorem shows the definition of diagonalizability is independent of a matrix representation, thus the minimal polynomial of T restricted to some arbitrary subspace will still divide the minimal polynomial of T itself which is composed of different linear factors of multiplicity 1 thus the minimal polynomial of the restriction will also be composed of different linear factors of multiplicity 1, hence T restricted to some subspace will still be diagonal ( regardless of the subspace's invariance ); so I still don't see how the knowledge that the subspace should be invariant is required to prove the above theorem.

CGandC
Ok, I think I understand fully now what you have said.
The minimal polynomial is defined as the minimal polynomial which zeros out a square matrix.
The minimal polynomial is also defined for linear transformations whose domain is the same as the co-domain ( i.e. ## T: V \to V ## ) as the minimal polynomial which zeros such linear transformation.

So although it is true that ## m_T(T_W) =0 ## for arbitrary subspace ## W \subseteq V ##, it is undefined to talk about a minimal polynomial of ## T_W ## if ## W## is not ##T##-invariant since it isn't true that the domain is equal to the co-domain.

Am I correct?