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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.

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  • $\begingroup$ 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 $\lambda\in\mathbb C\setminus \mathbb R$). $\endgroup$ – paf Jul 27 '16 at 15:51
  • $\begingroup$ You may want to look at math.uconn.edu/~kconrad/blurbs/linmultialg/complexification.pdf $\endgroup$ – paf Jul 27 '16 at 15:54
  • $\begingroup$ 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. $\endgroup$ – Christian Jul 27 '16 at 16:11
  • $\begingroup$ 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! $\endgroup$ – Christian Jul 27 '16 at 16:16
  • $\begingroup$ @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. $\endgroup$ – Ben Grossmann Jul 27 '16 at 19:04
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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: $$ (T - \lambda I)v = 0 \implies\\ J(T - \lambda I)v = 0 \implies\\ (JT - J(\lambda I))v = 0\implies\\ (TJ - \overline{\lambda}IJ)v = 0 \implies\\ (T - \overline{\lambda} I)(Jv) = 0 \implies\\ (T - \overline{\lambda} I)\overline v = 0 $$

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