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The problem

The goal is the show that two real matrices are similar over $\mathbb{R}$ if and only if they are similar over $\mathbb{C}$:

  1. Let $ T \in GL_{n}\mathbb{C}$, write $T = T_{1} + iT_{2}$ with $T_{1}, T_{2} \in Mat(n, \mathbb{R})$. Prove that there exists a real number $c \in \mathbb{R}$ such that $T_{1} + cT_{2}$ is invertible.
  2. Prove that two real matrices $A, B \in Mat(n, \mathbb{R})$ are similar over $\mathbb{R}$ if and only if they are similar over $\mathbb{C}$.

What I tried so far

  1. I tried a proof by contradiction:

Suppose $\nexists c \in \mathbb{R}$ such that $T_{1}+cT_{2}$ is invertible, but $T_{1} +iT_{2}$ is invertible. This is the same as saying $\forall c \in \mathbb{R}: det(T_{1} + cT_{2}) = 0$. Then for $c = 0$ we can see that $det(T_{1}) = 0$, so $T_{1}$ is not invertible. From here on I wanted to show that $T_{1} + iT_{2}$ cannot be invertible if $T_{1}$ is not invertible, but a simple counter example proves this false:

$ T_{1} = \left( \begin{array}{rrr} 0 & 0 \\ 1 & 1 \\ \end{array}\right), T_{2} = \left( \begin{array}{rrr} 2 & 3 \\ 1 & 2 \\ \end{array}\right)\\ $

$T_{1} + iT_{2}$ is invertible, but $T_{1} + cT_{2}$ is not for $c=1$. I am not sure how to go on from here.

  1. I tried this:

$"\Rightarrow"$ Is trivial, if $\exists S \in Mat(n,\mathbb{R}): B = S^{-1}AS$, then $S \in Mat(n, \mathbb{C})$. Following the definition of matrix similarity, $A, B$ are similar over $\mathbb{C}$.

$"\Leftarrow"$ There exists $S \in GL_{n}(\mathbb{C})$ such that $B = S^{-1}AS \Leftrightarrow SB = AS$. Write $S = S_{1} + iS_{2}$, with $S_{1},S_{2} \in Mat(n, \mathbb{R})$. Therefore $SB = S_{1}B + iS_{2}B = AS_{1} + iAS_{2} = AS$. Separating the real and imaginary parts of the equation gives us $S_{1}B = AS_{1}$ and $iS_{2}B = iAS_{2}$. We are only interested in the real part, which proves $A, B$ are similar over $\mathbb{R}$.

My question

First off, I am not sure wheter my work on 1. could lead somewhere. How would I prove this part? And second, is my proof for 2. correct?

user1551
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tobias ingold
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1 Answers1

1

You are almost there, but as it stands, your proof isn't correct because $S_1$ and $S_2$ can be both singular. Your argument can be salvaged by observing that some linear combination of $S_1$ and $S_2$ must be nonsingular. See this answer for more details.

From a higher perspective, two real matrices are similar over $\mathbb R$ iff they are similar over $\mathbb C$ because every matrix has a unique rational canonical form and the similarity transform $A\mapsto P^{-1}AP$ that brings a matrix $A$ to its rational canonical form can be carried out by using an invertible matrix $P$ whose entries are taken from the field generated by the entries of $A$. See this answer for more details.

user1551
  • 139,064