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I think I found a solution to question 16 from chapter 5.A of Axler's Linear Algebra book which states

Suppose V is a complex vector space and $T\in L(v)$, and the matrix of T with respect to some basis of V only has real entries show that if $\lambda$ is an eigenvalue then so is its complex conjugate.

and wanted to verify that the solution I attempted is indeed correct.

My Proof Attempt:

First I claim that for this matrix with the particular basis $C\circ T\circ C = T$ where $C$ is the complex conjugate map. Given any vector $v=\sum_{k=1}^{n}b_kv_k$, where $v_k$ is the particular basis, $T\circ C= \sum_{k=1}^{n}\overline{b_k}T(v_k)$. Hence $C\circ T\circ C= \sum_{k=1}^{n}b_k\overline{T(v_k)}$. But since the entries are real $T(v_k)=\overline{T(V_k)}$. Therefore the first claim is as such.

To prove the claim let $v$ be the given eigenvector of $T$ then $C\circ T\circ C (v)= \lambda v$, for some non-zero complex number $\lambda$. Since $C$ is its own inverse it then follows that $T\circ C(v) = C(\lambda v) $. By definition the map $C$ takes the coefficient of the vector and gives the vector with co-coefficients which are the complex conjugate of the orginal. Thus it is clear that $T\circ C(v) = \overline{\lambda}C(v)$ and the proof is complete.

I was wondering if this is correct and is clear enough, any critques and suggestions would be greatly appreciated, thank you!

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  • $\begingroup$ Seem OK to me. A alternative proof is to observe that the characteristic polynomial of the matrix has real coefficients and that if $\lambda$ is a root of a polynomial with real coefficients, then so is $\overline\lambda$. $\endgroup$ – amd Mar 28 '17 at 22:13
  • $\begingroup$ @amd True, and that proof is much more elegant but I wanted to give a determinant free proof in that they are not introduced until the very end of the text. $\endgroup$ – A.Riesen Mar 28 '17 at 22:22
  • $\begingroup$ If you’ve already established that the eigenvalues of $T$ are independent of basis, then I think you might be able to simplify the argument by working directly with the given matrix. Take the complex conjugate of both sides of the eigenvector equation and use the fact that conjugation distributes over addition and multiplication. $\endgroup$ – amd Mar 28 '17 at 23:28
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The idea of the proof is correct but there is one technical problem that should be addressed: A complex vector space $V$ has no natural conjugation map $C \colon V \rightarrow V$ so this map should be defined explicitly. Similarly, the expression $\overline{T(v_k)}$ makes no sense because.

Let me suggest how to fix that. If $\beta = (v_1,\dots,v_n)$ is the basis of $V$ with respect to which the matrix of $T$ is real, we have an isomorphism $\Phi \colon \mathbb{C}^n \rightarrow V$ given by

$$ \Phi \left( \sum_{i=1}^n a_i e_i \right) = \sum_{i=1}^n a_i v_i $$

where $(e_1,\dots,e_n)$ is the standard basis of $\mathbb{C}^n$ and $a_i \in \mathbb{C}$. Define the operator $T' \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ by $T' := \Phi^{-1} \circ T \circ \Phi$. Then $T$ and $T'$ have the same eigenvalues and the matrix representing $T'$ with respect to the standard basis $(e_1,\dots,e_n)$ of $\mathbb{C}^n$ is the same as the matrix representing $T$ with respect to the basis $(v_1,\dots,v_n)$ of $V$ so it has real entries.

Thus, by replacing $T$ with $T'$ and $V$ with $\mathbb{C}^n$ we need to solve the following question:

Let $T \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ be an operator such that $T(e_i)$ is a real linear combination of $(e_1,\dots,e_n)$ for all $1 \leq i \leq n$. Show that if $\lambda \in \mathbb{C}$ is an eigenvalue of $T$ then $\overline{\lambda}$ is an eigenvalue of $T$.

Now, in $\mathbb{C}^n$ a vector looks like $v = (z_1,\dots,z_n)$ so we can talk about $\overline{v} = (\overline{z}_1,\dots,\overline{z}_n)$ and your proof carries more or less through. Alternatively, it is actually easier to translate everything to matrices and solve it on the level of matrices.

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