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$A=\begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$ has eigenvalue $\lambda=1$. The eigenvector is (1, 1).
To make triangular matrix,
$U^{-1}AU=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \end{pmatrix}$ $\begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$ $\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \end{pmatrix}$=$\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}=T$

In here, I understand that the first column of $U$ is (1,1) because it's the eigenvector, and dividing it by $\sqrt{2}$ is to make it unit length.
But how can I derive the second column (1,-1)? I know it is orthogonal to column one, and also has unit vector by dividing $\sqrt{2}$. Is there any formula I can use to calculate the second column? (Sine there is only one eigenvector, we have to make second column.)

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In general, what you need are generalized eigenvectors, but in your particular case, actually any nonzero vector that is not parallel to $(1,1)^T$ can serve as the second column, so the second column does not need to be orthogonal to the first one. – user1551 Nov 25 '12 at 16:49
@user1551 I suspect that thistime wants a Schur decomposition, so an orthogonal matrix is actually needed. – Phira Nov 27 '12 at 15:21
@Phira You're right, of course. I overlooked the "Schur" in the title. – user1551 Nov 27 '12 at 15:25

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