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Give an example of two commuting operators , on a finite-dimensional real vector space such that + has a eigenvalue that does not equal an eigenvalue of plus an eigenvalue of and has a eigenvalue that does not equal an eigenvalue of times an eigenvalue of .

I tried constructing diagonal matrices yet it seems the theorem still holds. What are examples of S and T to demonstrate this property doesn't hold over real vector spaces.

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    Well, in $\mathbb R^2$ already we can construct examples where $S$ and $T$ don't have (real) eigenvalues, but $S+T$ and $ST$ do... – Misha Lavrov Mar 25 '24 at 03:59
  • @Misha, but two nonreal numbers can add up to (and multiply to) a real number. – Gerry Myerson Mar 25 '24 at 04:17
  • Duplicate of https://math.stackexchange.com/questions/19464/spectrum-of-the-sum-of-two-commuting-matrices ? – Gerry Myerson Mar 25 '24 at 04:25
  • Diagonal matrices are almost certainly not going to help here: if $\lambda_i$ and $\alpha_i$ are the $i$th diagonal entries on $S$ and $T$ respectively, then $S + T$ and $ST$ will also be diagonal matrices. The $i$th entry of each will be $\lambda_i + \alpha_i$ and $\lambda_i \alpha_i$ respectively. So diagonal matrices (in the usual sense of the word) won't actually help here unfortunately. They're great for identifying eigenvalues, but are a little too nice to give a counterexample. – Kyle Mar 25 '24 at 05:40
  • So do no counterexamples exist where both S and T have real eigenvalues? Because I couldn't find any after hours of trying to construct one. –  Mar 25 '24 at 12:24
  • If all we want is for $S$ and $T$ to have some real eigenvalues, we can embed the accepted answer in $\mathbb R^n$ for $n \ge3$ and have $n-2$ dimensions where $S$ and $T$ can be well-behaved. But the question linked to in the comments above shows that we need some real/non-real trickery to happen here; there's no example over $\mathbb C$. – Misha Lavrov Mar 25 '24 at 16:47

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Consider the matrix \begin{bmatrix}0&-1\\1&0\end{bmatrix} whose minimal polynomial is $x^2 + 1$, which has no solution in $\mathbb{R}$, so this matrix has no real eigenvalues, since the zeros of the minimal polynomial are the eigenvalues. Let $S$ be the matrix above and $T$ be the additive inverse of $S$: $$S=\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \qquad T=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$$ hence the addition and the multiplication of $S$ and $T$ are: $$S+T=\begin{bmatrix}0&0\\0&0\end{bmatrix}, \qquad ST=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ As the multiplication equals the identity matrix, $S$ and $T$ are inverses of each other, so they commute. Moreover, addition and multiplication have eigenvalues $0$ and $1$, respectively - which are not the sum or product of an eigenvalue of $S$ and an eigenvalue of $T$, simply because $S$ has no eigenvalues.

Frank
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  • No real eigenvalues, but it does have eigenvalues, $\pm i$. And $0=i+(-i)$, and $1=(i)(-i)$. – Gerry Myerson Mar 25 '24 at 07:56
  • @GerryMyerson, yes, that is obvious. But that is not what the question asks, here the operators are over a real vector space, so the question is: Give an example of two commuting operators , on a finite-dimensional real vector space such that + has a real eigenvalue that does not equal a real eigenvalue of plus a real eigenvalue of and has a real eigenvalue that does not equal a real eigenvalue of times a real eigenvalue of . – Frank Mar 25 '24 at 16:29
  • Think about it this way, let $V$ be a nonzero vector space over some field $\mathbb F$, and $T$ be an operator on $V$. Because $V$ is non-zero, the minimal polynomial of $T$ is nonconstant. But any nonconstant polynomial over $\mathbb F$, has a zero in the algebraic closure $\bar{\mathbb F}$, let $\mu$ be a zero of the minimal polynomial of $T$, and also let's assume $\mu \in \bar{\mathbb F} \setminus \mathbb F $, as in our example where $\pm i \in {\mathbb R}(i)\setminus {\mathbb R}$. – Frank Mar 25 '24 at 17:50
  • So, do we now say that $T$ has an eigenvalue $\mu$, in other words, do every operator on a nonzero vector space have an eigenvalue?
    No, we don't say that, because $\mu v$ in the equation $Tv=\mu v$ does not make any sense, what we only have is the action of $\mathbb F$ over $V$, and extending it to an action of $\bar{\mathbb F}$ would make $V$ no longer a vector space over $\mathbb F$, assuming $\bar{\mathbb F} \neq \mathbb F$.
    In short, when we say an operator doesn't have any eigenvalues, we mean that it doesn't have any eigenvalues in the underlying field.     – Frank Mar 25 '24 at 17:51
  • Well, Frank, maybe that's what you mean, but it's not what I mean. If I wanted to mean that, I would say the operator has no eigenvalues in $F$. I wouldn't assume that, just because it's an operator on an $F$-vector space, that people are only interested in eigenvalues in $F$. When you solve systems of linear constant-coefficient differential equations, even if everything is explicitly stated as being real, you still may need to get at the nonreal eigenvalues of the system to write the solutions in terms of real sines and cosines. It's no good just saying "no eigenvalues, so no solutions." – Gerry Myerson Mar 26 '24 at 02:51
  • @GerryMyerson, I attempted to explain what is meant by the question, if we consider the complex eigenvalues of these real operators then the question does not make any sense. If $V$ is a finite-dimensional complex vector space and $S$, $T$ are commuting operators on $V$. Then every eigenvalue of $S+T$ is an eigenvalue of $S$ plus an eigenvalue of $T$, and every eigenvalue of $ST$ is an eigenvalue of $S$ times an eigenvalue of $T$. The whole purpose of this question is to show how this fails when the underlying field is $\mathbb R$ and not $\mathbb C$. – Frank Mar 26 '24 at 10:26
  • Although I agree with you that saying "has no eigenvalues in $\mathbb F$" is more precise and prevents vaguenesses like this one. – Frank Mar 26 '24 at 10:30
  • I'm not sure how you know what OP meant by the question. It's not clear to me that OP knew the result about complex vector spaces. OP certainly didn't say anything like, "I know how this works over the complex numbers, I want to know why it's different over the reals." – Gerry Myerson Mar 26 '24 at 23:21
  • @GerryMyerson, it's Exercise $10$ of Section $5E$ in Axler's Linear Algebra Done Right. The exercise, after stating the question, comments on what I said earlier, that's how I knew what the question meant and also that OP would know about it. – Frank Mar 27 '24 at 00:37
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    OK. I wish OP had come clean and cited the source of the problem, and the context. (Of course, it's possible that the question was set as an exercise by a teacher who wasn't using Axler as the textbook, but was getting problems from it. When I was teaching, I often harvested assignment questions from texts other than the ones the students were meant to acquire.) – Gerry Myerson Mar 27 '24 at 01:35