Let $A = [a_{ij}] \in \mathbb{K}^{n \times n}$, where $\mathbb{K} = \mathbb{R}$ or $\mathbb{K} = \mathbb{C}$.
In the case $\mathbb{K} = \mathbb{C}$, for example, via the Jordan decomposition theorem, there exists an invertible matrix $B$ with coefficients from $\mathbb{K} = \mathbb{C}$ such that $C := B^{-1}AB$ is an upper triangular matrix, i.e., $C=[c_{ij}]$ with $c_{ij}=0$ for $i>j$.
Is it true that for every matrix $A$ of type $n \times n$ with real coefficients, which characteristic polynomial has $n$ real roots not necessarily different, there exists an invertible matrix $B$ of the same type with real coefficients such that $B^{-1}AB$ is upper triangular?
Thanks.
Best Answer
The fact that the coefficients are different are different or not in $\chi_A(x)$ makes no change. The only thing that matters is whether all the roots are real or not. It is true that every real matrix with real eigenvalues (roots of $\chi_A(x)$) is triangulable. To prove this, you need the following two facts:
If $\lambda$ is an eigenvalue of $A$ then there exists $0\neq v\in\mathbb{R}^n$ such that $A\cdot v=\lambda v$.
If $W\lneqq \mathbb{R}^n$ and all roots of $\chi_A(x)$ are real, then there exists $v\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}$ such that $v\notin W$ and $Av-\lambda v\in W$
Using those two claims, you can construct a basis of $\mathbb{R}^n$, $b=(v_1,...,v_n)$ such that for all $1\leq i\leq n$ $Av_i=a_{i,1}v_1+...+a_{i,i}v_i$. Now let $B$ be a matrix whose columns are $(v_1,...,v_n)$. It's easily checked that $B^{-1}AB$ is upper-triangular.