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While attempting to answer a question here (namely, the finite dimensional case of the title question: Prove that if $\lambda$ is an eigenvalue of $T$, a linear transformation whose matrix representation has all real entries, then $\overline{\lambda}$ is an eigenvalue of $T$), I noticed the asker did not specify a finite dimensional vector space. Though the person who asked the question was satisfied with a finite-dimensional response, I was wondering if the analogue was true for infinite dimensional vector spaces.

I have seen several proofs of this fact relying on $V$ being finite dimensional. One proof utilizes the roots of the characteristic polynomial; if the coefficients are real then the roots come in conjugate pairs.

The second notable proof I've seen goes something like:

$$(T-\lambda I)v = 0$$ $$\overline{(T-\lambda I)v} = 0$$ $$(\overline{T} - \overline{\lambda I})\overline{v} = 0$$ $$(T- \overline{\lambda}I)\overline{v} = 0$$

where we define $\overline{T}$ as taking the conjugate of each element of the matrix representation of $T$, and we define $\overline{v}$ as conjugating each entry in the n-tuple representation of $v$ with respect to a basis. Going backwards will give you that, given the conditions set earlier, $\lambda$ is an eigenvalue if and only if $\overline{\lambda}$ is an eigenvalue, and also, $v$ is an eigenvector with eigenvalue $\lambda$ if and only if $\overline{v}$ is an eigenvector with eigenvalue $\overline{\lambda}$.

The first thing we would have to do is have some notion that is similar to the matrix representation of $T$ having all real entries. What exactly would that be? Would we have to work with infinite matrices, or (assuming the axiom of choice) could we define $T$ such that it takes basis vectors to linear combinations of basis vectors with real coefficients and that would suffice?

If we assume the axiom of choice and take a basis of $V$, I am under the impression that the second proof I provided for the finite dimensional case could extend to the infinite dimensional case. Is it necessary to use the axiom of choice for a proof, though?

Overall, my question is: First, is there an analog of $T$ having all real matrix entries in an infinite dimensional case? Denote this property, if it exists, $P$.

Second: Does anyone have a proof or counterexample of the following?: Let $V$ be an infinite dimensional complex vector space, and let $T$ be a linear transformation with $P$. $\lambda$ is an eigenvalue of $T$ if and only if $\overline{\lambda}$ is an eigenvalue of $T$.

Can we also add: $v$ is an eigenvector with eigenvalue $\lambda$ if and only if $\overline{v}$ is an eigenvector with eigenvalue $\overline{\lambda}$? Whatever $\overline{v}$ may happen to mean in this case.

If we can do this without infinite matrices, infinite basis, or assuming the axiom of choice, I would much prefer that! But I understand it may be necessary.

Christian
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  • Even in infinite-dimensional vector spaces, linear combinations are finite by definition. For your 2nd question, it is already false in finite-dim. case (simply take $\lambda.\mathrm{Id}$ where $\lambda\in\mathbb C\setminus \mathbb R$). – paf Jul 27 '16 at 15:51
  • You may want to look at http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/complexification.pdf – paf Jul 27 '16 at 15:54
  • To be honest, I'm not quite sure I see what you mean. The finite dimensional analog of $P$ is that $T$ has all real entries with respect to some basis. So the statement is: Let $V$ be a complex, finite dimensional vector space and let $T$ be a linear transformation whose matrix representation has all real entries with respect to some basis. Then $\lambda$ is an eigenvalue of $T$ if and only if $\overline{\lambda}$ is an eigenvalue of $T$. I am confident this is true (I provided two sketches of proofs in the question). Certainly if we do not require $T$ to have real entries then it is false. – Christian Jul 27 '16 at 16:11
  • With the linear combinations, we do define them to be finite, and my idea of using "infinite linear combinations" is likely very wrong. But I suppose my point of needing infinite "linear combinations" (or some notion like it) is that I thought not all vectors in infinite dimensional vector spaces can be represented as a finite linear combination. In some spaces, like $\mathbb{R}[x]$, they can, but consider $\mathbb{R}^{\mathbb{R}}$. I find it very unlikely every vector can be represented as a finite linear combination of some "basis". If I am wrong though, I would love to be enlightened! – Christian Jul 27 '16 at 16:16
  • @Christian you are wrong, so long as you accept the axiom of choice (or equivalently Zorn's lemma). Every vector space has a basis which, by the definition of a basis, requires only finite linear combinations. – Ben Grossmann Jul 27 '16 at 19:04
  • @Christian there are contexts, however, where infinite linear combinations are important to consider, but note that such linear combination don't necessarily "converge". You may want to look into the difference between a "Hamel basis" and "Schauder basis". – Ben Grossmann Jul 27 '16 at 19:07
  • @Omnomnomnom Thank you for pointing out my incorrect intuition! This is something quite astonishing and amazing to me, especially when considering vector spaces requiring an uncountable basis, even when assuming choice! I have adjusted my question to avoid any talk of infinite linear combinations, as I do not want to muddle the original question with my incorrect ideas, and I now do not think they are valuable to solving this problem. – Christian Jul 27 '16 at 19:22

1 Answers1

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Let $V$ be a vector space over $\Bbb R$. As elaborated in the link in the comment, let $V_{\Bbb C} = V \oplus V$ denote the complexification of $V$, in which $v + iw = v \oplus w = (v,w)$. We define the conjugation map by $$ J(v + iw) = v - iw $$ For any $v,w \in V$. Note that $J$ is $\Bbb R$-linear and that for any $\lambda = a+bi$, $x = v + iw$, we have: $$ J(\lambda x) = J[\lambda(v+iw)] = \overline{\lambda}(v - iw) = \overline{\lambda}J(v + iw) = \overline{\lambda}J(x) $$ which I will let you verify. In other words, $J$ is antilinear.

Now, if $T:V \to V$ is linear, then the unique $\Bbb C$-linear extension to $V_{\Bbb C}$ is given by $$ \tilde T(v + iw) = T(v) + iT(w) $$ It follows that $$ \tilde TJ(v + iw) = T(v - iw) = T(v) - iT(w) = J\tilde T(v + iw) $$ That is, if $\tilde T:V_{\Bbb C} \to V_{\Bbb C}$ is the $\Bbb C$-linear extension of an $\Bbb R$-linear map, then $\tilde TJ = J\tilde T$. With that, we may proceed:

Theorem: Suppose that $\tilde T:V_{\Bbb C} \to V_{\Bbb C}$ is the $\Bbb C$-linear extension of an $\Bbb R$-linear map on $V$. If $\lambda$ is an eigenvalue of $\tilde T$ with eigenvector $v \in \Bbb V_{\Bbb C}$, then $\overline{\lambda}$ is an eigenvalue of $\tilde T$ with eigenvector $\overline{v} = Jv$.

Proof: Note that $$ \tilde T\overline{v} = \tilde T(Jv) = J(\tilde Tv) = J(\lambda v) = \overline{\lambda}J(v) = \overline{\lambda} \overline{v} $$ as desired.

Or, to more closely mirror your referenced proof: $$ (T - \lambda I)v = 0 \implies\\ J(T - \lambda I)v = 0 \implies\\ (JT - J(\lambda I))v = 0\implies\\ (TJ - \overline{\lambda}IJ)v = 0 \implies\\ (T - \overline{\lambda} I)(Jv) = 0 \implies\\ (T - \overline{\lambda} I)\overline v = 0 $$

Ben Grossmann
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