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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.)

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    $\begingroup$ The eignevalues of a triangular matrix will correspond to the entries on the main diagonal. If the main diagonal is real, the eigenvalues are real. If the eigenvalues are complex, the matrix in this form will have complex entries. $\endgroup$ – Doug M Sep 27 '18 at 23:49
  • $\begingroup$ @DougM That is indeed true, and it is straightforward. Thank you. $\endgroup$ – user106860 Sep 27 '18 at 23:52
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If $A$ is a $2\times2$ matrix, then there are only two possibilities for $T$: it will either be upper-triangular, corresponding to two $1\times1$ blocks along the main diagonal, or it will consist entirely of a $2\times2$ block. The zeros below the main diagonal in the generic template for $T$ simply indicate that all entries below the blocks along the diagonal are zero, not that there must be a zero below the main diagonal.

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  • $\begingroup$ Your edit made things clear, so I am happy. Thank you. A follow up question: If I wanted to indicate a matrix of the form of (1) but with the requirement that there are 0's below the main diagonal, how would this be done? Would I just use an additional sentence? I guess this condition would require that the blocks are all upper triangular, as well... so perhaps this is too tangential a question. $\endgroup$ – user106860 Sep 27 '18 at 23:59
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    $\begingroup$ @user106860 Yes, you’d need to add something along those lines, but unless that block structure is somehow significant it would be simpler just to say that the matrix is upper-triangular and be done with it. $\endgroup$ – amd Sep 28 '18 at 0:03
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What about $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$?

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  • $\begingroup$ Has eigenvalues of $-i$ and $i$, so a $2\times 2$ matrix with complex eigenvalues need not be upper triangular. Thank you, that helps a bit. But then... if $T$ is $2\times 2$ and is of the form specified in (1), $T$ does not need to be upper triangular even though the form in (1) has the zero matrix below the main diagonal? Sorry that this is probably just me having an issue with notation. $\endgroup$ – user106860 Sep 27 '18 at 23:50
  • $\begingroup$ Apparently. It looks like $B$ could be this matrix, or some of the $B_i$ could. Surely there are other examples. $\endgroup$ – Chris Custer Sep 27 '18 at 23:54
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    $\begingroup$ I guess the form you gave isn't necessarily upper triangular. $\endgroup$ – Chris Custer Sep 27 '18 at 23:57

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