# Proof: are restriction to operators diagnolizable if the operator is?

"Let $$V$$ be a vector space of finite dimension $$n\ge1\ \ T:V\to V$$ a linear operator and $$S$$ a T-invariant space of $$V$$. Consider the restriction $$T|_S:S\to S$$ and $$(T|_S)(w):=T(w)$$ If $$T$$ is a diagonal operator in $$V$$ prove that $$T|_S$$ will be diagolizable in $$S$$"

Ok so, as in matrix form the diagonal operator can be divided in "blocks" of same eigenvalues, each block related to its eigenspace I thought about these blocks as T-invariant subspaces, and, as the matrix is diagonal the blocks must be diagonal too so the operator restricted to this eigenspaces would be diagonalizable. My problem is basically how to work this ideas into the proof(Are them right?) Thanks in advance.

• Do you know that/how diagonalizability can be characterised using the minimal polynomial? Commented Jun 5, 2015 at 1:22
• $\Pi_1^k(\mu_i-\lambda)^n$ (n at the power has a index i also), is it? Commented Jun 5, 2015 at 1:37

One way to turn that idea into a proof is to work in terms of a specific choice of eigenvectors which act as bases of the eigenspaces you mention. To get there, I would use the characterization of a diagonalizable matrix as (taken from Diagonalizable Matrix)

A matrix is diagonalizable if there exists a basis of V consisting of
eigenvectors of T. With respect to such a basis, T will be represented by a
diagonal matrix.


Now you can appeal to the definition of an eigenvector as well: $$Tv = \lambda v$$ to make statements about whether those eigenvectors are still eigenvectors of the restricted operator.

Ok, I came up with this trial now, and I'd like to know if it's acceptable in your opinion - my teacher is really strict when it comes to prove something:

Let $w\in S$ and $S=V(\lambda_k)$ a eigenspace of $T$ and $B=${$v_1,v_2,...,v_n$} a bases of eigenvectors of $V$ where $[T]_B$ is diagonal

As $S$ is a subspace of $V$ let $B^S=${$v_1,...,v_k$} be a bases of $S$ $k\le n$.

By the definition of a eigenvector we have $$T(v)=\lambda v$$ And by the defiinition of a application of a restrict operator $$(T|_S)(w):=T(w)$$

If we take $w=\alpha_1v_1+...+\alpha_k v_k \rightarrow$

$T(w)=\alpha_1T(v_1)+...+\alpha_k T(v_k)$ and as all this $v_i$ are associated with the same eigenvalue $\lambda_k$ thus we have:

$T(w)=\lambda_k(\alpha_1v_1+...+\alpha_k v_k$) which in fact is in the subspace $S$ holding the definition of the strict operator

So $(T|_S)(w)=\lambda_k(\alpha_1v_1+...+\alpha_k v_k$)

Now at least, for a convenient choice $w=v_i(1\le i \le k)$ as the vectors of the bases $B^S$ we can see that: $$(T|_S)(v_i)=\lambda_kv_i$$ definig eigenvectors of $T|_S$, hence it must be diagonalizable since it holds for every choice of $1\le i \le k$, the whole bases of $S$.

I think I may be missing something about algebraic-geometric multiplicity but at the same time I have the sensation that if $T(v)=\lambda v$ holds already itd be kinda redundant.

• Your proof looks fine to me. I would point out in the beginning that B is a basis of eigenvectors. You implicitly use that later, but that is the important thing about diagonalizable operators. Commented Jun 5, 2015 at 18:39
• You start with the basis of eigenvectors for $V$ and then take a subset of this basis as a basis for $S$. It is not clear why you can do this. Commented May 4 at 10:12

Yes, $$T$$ is diagonalizable iff for each eigenvalue $$\lambda$$ of $$T$$,$$N(T-\lambda I)=N((T-\lambda I)^2)$$ ($$N(T)$$ stands for null space of $$T$$).

And in any T-invariant subspace $$W$$,$$N(T_W-\lambda I)=N((T_W-\lambda I)^2)$$ still holds, so $$T_W$$ is diagonalizable.