I If T is diagonalizable then is restriction operator diagonalizable?

[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?

 

[pasmith]

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.

 

[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?