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I'm trying to solve some problems in Hoffman and Kunze and I'm kind of stuck on this one. This is 6.5.3 on Hoffman and Kunze.

Here is the question:

Let $T$ be a a linear operator on an $n$-dimensional space, and suppose that $T$ has $n$ distinct characteristic values. Prove that any linear operator $U$ which commutes with $T$ is a polynomial in $T$.

My work so far: Since $T$ has $n$ distinct characteristic values and since the space it acts on is also $n$ dimensional, $T$ must be diagonalisable. Now using the fact $UT=TU$ I can show that $U$ must also be diagonalizable. So these two operators must be simultaneously diagonalizable. Now I'm stuck with where this is going. I'm thinking of simultaneous diagonalizability because this is section 6.5 of Hoffman and Kunze and it deals with simultaneous diagonalizability. Am I on the right track with this?. Can anybody help?

Thanks so much for your time and your answers.

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  • $\begingroup$ So the claim is that the commutant of $T$ consists of only polynomials of $T$? That is not true in general. One can, for instance, take the operator-norm closure of the polynomials. E.g. take any spectral projection of $T$. $\endgroup$ – Michael Jun 17 '13 at 0:30
  • $\begingroup$ @Michael Apparently there are $n$ pairwise distinct eigenvalues, so indeed $T'=K[T]$. $\endgroup$ – Julien Jun 17 '13 at 0:33
  • $\begingroup$ @julien: Yes, U is simultaneously diagonalizable because the char. values are distinct. In other words because the char. spaces are one-dimensional. I used this fact to prove what I have. I don't understand what you mean by Lagrange interpolation. I've always been confused with Lagrange interpolation. Can you explain a bit?. Thanks for your comment. $\endgroup$ – minibuffer Jun 17 '13 at 0:38
  • $\begingroup$ @julien Ah OK. Thanks. $\endgroup$ – Michael Jun 17 '13 at 0:39
  • $\begingroup$ @julien, Why not turn that comment into an answer? $\endgroup$ – Stephen Jun 17 '13 at 0:46
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Note: we will prove that the commutant of $T$ is equal to $K[T]$. This property is actually equivalent to the fact that the characteristic and minimal polynomials are equal. That's also equivalent to similarity to a companion matrix.

Let $\lambda_j$ be the $n$ pairwise distinct eigenvalues of $T$. Such an operator must be diagonalizable, and its eigenspaces are all one-dimensional. Since $U$ commute with $T$, it leaves the eigenspaces of $T$ invariant. Since they are one-dimensional, this implies that every eigenvector of $T$ is an eigenvector for $U$. So $T$ and $U$ are simultaneously diagonalizable.

In a basis of simultaneous diagonalization, $T=\mbox{diag}(\lambda_1,\ldots,\lambda_n)$ and $U=\mbox{diag}(\mu_1,\ldots,\mu_n)$.

Now we will do Lagrange interpolation. Since the $\lambda_j$ are pairwise distinct, we can consider the degree $n$ polynomials $$ L_j(x)=\prod_{i\neq j}\frac{x-\lambda_i}{\lambda_j-\lambda_i}\qquad L_j(\lambda_i)=\delta_{ij}. $$ Now $$p(x)=\sum_{j=1}^n\mu_jL_j(x)\qquad\mbox{satisfies }\quad p(\lambda_j)=\mu_j \;\forall j. $$ Therefore $p(T)=\mbox{diag}(p(\lambda_1),\ldots,p(\lambda_n))=U$ belongs to $K[T]$. The converse is clear, so the commutant of $T$ is $$ \{T\}'=\{U\in L(V)\,;\,UT=TU\}=K[T] $$ the subagebra of all polynomials in $T$.

Note: for a less explicit argument, just compare the dimensions of $K[T]$ and the subspace of all diagonal operators in this basis. The latter is clearly $n$. The former is $n$ by minimal polynomial consideration and Euclidean division. And clearly $K[T]$ is contained in the commutant, so they must be equal. Oh...but that's TTS argument...

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  • $\begingroup$ Thanks for your answer. But I don't understand when you say, "since $U$ commute with $T$, it leaves the eigenspaces of $T$ invariant". Is this a Theorem or something that can be easily seen? $\endgroup$ – minibuffer Jun 17 '13 at 1:45
  • $\begingroup$ Yes. If $S$ and $T$ commute, then $S(\ker T)\subseteq \ker T$ and $S(\mbox{im} T)\subseteq \mbox{im} T$. Apply the former with $U$ and $T-\lambda Id$. $\endgroup$ – Julien Jun 17 '13 at 1:47
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It seems to me that you've found necessary and sufficient conditions on $U$. What do you think the dimension of the space of such $U$ is? You won't yet need to use that the eigenvalues are distinct.

Now, consider the space of polynomials in $T$. Elements of this space always commute with $T$. What's the dimension?

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  • $\begingroup$ It must be $n$ because the minimal polynomial of $T$ is degree $n$. $\endgroup$ – minibuffer Jun 17 '13 at 0:48
  • $\begingroup$ I don't quite get the "you won't yet need to use that the eigenvalues are distinct" sentence. Certainly that property was already used to show that $U$ commuting with $T$ must be diagonal when expressed on any basis that diagonalises$~T$. It is true that the condition will serve again later; this might raise the (academic) question of finding a proof that uses the condition just once. $\endgroup$ – Marc van Leeuwen Mar 19 '15 at 12:32

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