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I am currently struggling on some linear algebra exercise. It says:

Proof that for any real $n \times n$-matrices $A,B$: $$\exists S\in GL_n(\mathbb C): SAS^{-1} =B \iff \exists T \in GL_n(\mathbb R): TAT^{-1}=B$$ Hint: Use the map $\phi :\mathbb C \to \mathbb C, z \mapsto \det(P+ zQ)$, while $S=P+iQ$ is the canonical decomposition into real- and imaginary part of $S$.

First i observed that the implication $"\implies"$ is rather trivial. So while thinking about the other direction i tried to use the hint. Since determinant is a polynomial function, the map $\phi$ is clearly continuous, hence defines a (compact) path from the real number $\det(P)$ to $\det(S)$. While the latter is $\neq 0$ i can find an environment of $i$, where $P+zQ$ is invertible, but what now? I still don't know how this matrix conjugates $A$ and the matrix is most likely not strictly real. I was thinking about using $P-Q$ instead because it kind of seemed possible to me, that the determinant could be equal to $\det (S)$. But even if that would be true i still don't know how this could lead the final result of "same conjugation". To be honest i don't understand the hint. This is some linear algebra class and i kind of doubt that they want us to use some analytic/approximation arguments with compactness, continuity etc.

I would be glad to hear any thoughts on this topic/any help so solve this exercise. Thanks in advance!

Tina
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    This is a similar result to your exercise ($A, B$ in this result have entries in the subfield instead): https://math.stackexchange.com/a/57257 Could you adapt this to your exercise? – Duong Ngo Nov 28 '24 at 09:08
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    @DuongNgo Yes thank you friend! That solves the exercise! – Tina Nov 28 '24 at 09:36
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    This is not true. E.g. we have $\pmatrix{1/2&1/2\ i/2&-i/2}\pmatrix{1\ &0}\pmatrix{1&-i\ 1&i}=\pmatrix{1/2&-i/2\ i/2&1/2}$, but there does not exist any real matrix $T$ such that $T\pmatrix{1\ &0}T^{-1}=\pmatrix{1/2&-i/2\ i/2&1/2}$, because the LHS is real but the RHS is not. Are $A$ and $B$ real matrices instead? – user1551 Nov 28 '24 at 13:35
  • @user1551 yes oops, you are right. Its fixed now. – Tina Nov 28 '24 at 14:01

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The answer is surprisingly simple:

$\phi$ gives rise to an $f \in \mathbb R [x]$, since determinant is polynomial. $f\neq 0$, since $\phi(i) \neq 0$, thus $\exists \lambda \in \mathbb R: \phi(\lambda)=f(\lambda) \neq 0$, because otherwise $f$ posseses infinetly many roots, forcing $f=0$, which is a contradiction. Define $T:= P+\lambda Q$, then:

$$TA=PA+\lambda QA= BP+\lambda BQ=BT$$ Because $$PA+iQA=SA=BS=BP+iBQ \implies PA=BP \wedge QA=BQ$$ Since $\det T \neq 0$, by definition of $T$, this is equivalent to: $$B=TAT^{-1}$$

Tina
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