Consider a $2\times 2$ matrix, A, with real entries. The goal is to show that we can write $A$ in the following form (Schur form):
$$\tag{1}
A=QTQ^T,\quad T=\begin{bmatrix}B_1 &\times &\dots &\times\\
& B_2 & &\times\\
\ \ \ \ \ O & & \ddots\\
& & & B_j\end{bmatrix}
$$
Where the $B_i$'s are either $1\times 1$ or $2\times 2$ matrices (each $2\times 2$ block will correspond to a pair of complex conjugate eigenvalues). Also, $O$ denotes the zero matrix
The case where the eigenvalues of $A$ are real can be shown by letting $Q$ be constructed from a unit eigenvector and a unit vector orthogonal to that unit eigenvector.
If the eigenvalues of $A$ are complex the proof I am looking t states "we simply set $T=A$ and $Q=I$
Since, when $A$ has complex eigenvalues, we can just set $T=A$, then $A=T$ must be of the form specified in (1).
QUESTION: (Given the previous sentence) Does this mean $A$ is upper triangular? If so, why? Or does A not need to be upper triangular since $B$ will be a $2\times 2$ block and $A$ is a $2\times 2$ matrix?
(It seems to me that we would need $A$ to be upper triangular though because the form of $T$ in (1) has the zero matrix $O$ below the diagonal.)
Best Answer
What about $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$?