Let $A$ be a $3 \times 3$ real matrix not similar to any upper-triangular matrix on $\Bbb R$, prove $A$ is similar to a diagonal matrix on $\Bbb C$.

Question

Suppose a $3 \times 3$ real matrix $A$ is not similar to any upper-triangular matrix on the real field $\mathbb{R}$, that is, there is no $3 \times 3$ invertible real matrix $P$, such that $P^{-1}AP$ is an upper-triangular matrix. Prove that $A$ is similar to a diagonal matrix on the complex field $\mathbb{C}$.

My Solution

I need to cite the following result:

Two real matrices $A$ and $B$ are similar on $\mathbb{C}$ if and only if $A$ and $B$ are similar on $\mathbb{R}$. In detail, if there exists an invertible $P \in \mathbb{C}^{n \times n}$ such that $B = P^{-1}AP$, then there exists a $Q \in \mathbb{R}^{n \times n}$ such that $B = Q^{-1}AQ$.

I then prove the conclusion by contradiction. If $A$ is not similar to a diagonal matrix on $\mathbb{C}$, $A$ would have at least two identical eigenvalues. Because $n = 3$ and the imaginary roots appear in pair, this means all $A$'s eigenvalues are real numbers. There are two cases.

• All eigenvalues equal to $\lambda_0$. Using the Jordan canonical form theory, $A$ is similar to one of the following three matrices: \begin{equation*} \begin{pmatrix} \lambda_0 & 0 & 0 \\ 0 & \lambda_0 & 0 \\ 0 & 0 & \lambda_0 \end{pmatrix}, \quad \begin{pmatrix} \lambda_0 & 1 & 0 \\ 0 & \lambda_0 & 0 \\ 0 & 0 & \lambda_0 \end{pmatrix}, \quad \begin{pmatrix} \lambda_0 & 1 & 0 \\ 0 & \lambda_0 & 1 \\ 0 & 0 & \lambda_0 \end{pmatrix}. \end{equation*} In either case, $A$ would be similar to an upper-triangular matrix on $\mathbb{R}$ (using the result stated at the very beginning), contradiction.

• Two eigenvalues equal to $\lambda_1$, the remaining eigenvalue equals to $\lambda_2 \neq \lambda_1$. Under this case, $A$ is similar to one of the following two matrices: \begin{equation*} \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_2 \end{pmatrix}, \quad \begin{pmatrix} \lambda_1 & 1 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_2 \end{pmatrix}. \end{equation*} In either case, $A$ would be similar to an upper-triangular matrix on $\mathbb{R}$, contradiction.

Therefore, $A$ has three different eigenvalues, hence it's diagonalizable on $\mathbb{C}$.

My Question

Is there any more direct solution that possibly avoids using Jordan form and the result I cited? Because this is an exercise appeared before the Jordan form chapter is introduced.

Answer 1

$A$ must have an eigenvalue with a non zero imaginary part (otherwise all eigenvalues are real and hence would be similar to an upper triangular matrix, cf. the real Schur form).

Since $A$ is real the non real eigenvalues must be a complex conjugate pair. Since all the eigenvalues are distinct it is diagonalisable (over $\mathbb{C}$).