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Let $T=\begin{pmatrix}5 & 0 & 0 \\ 0 & 2 & i\\ 0 & -i & 2 \end{pmatrix}$.

I found the eigenvalues and eigenvectors already and they are $1,3,5$ and $\begin{pmatrix}0\\-i\\1\end{pmatrix}$,$\begin{pmatrix}0\\i\\1\end{pmatrix}$, and $\begin{pmatrix}1\\0\\0\end{pmatrix}$.

According to mathematica, their orthoganalized form will be $\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{array} \right)$

But I am getting wildly different answers when I try to find the normalized eigenvectors using the Gram-Schmidt process.

Julien
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emka
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  • Well, from you eigenvectors, it is clear what the orthogonal ones are. Are you using GS because you have to and are trying to validate that result? – Amzoti May 29 '13 at 14:33
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    You shouldn't be using GS at all. Just normalize each eigenvector individually. And keep in mind that even normalized eigenvectors are not unique---you can multiply them by any unit-magnitude complex scalar. – Michael Grant May 29 '13 at 14:50
  • You happen to have orthogonal eigenvectors. But if you hadn't, then making them orthonormal would most likely spoil the property of being eigenvectors (for instance it would be sure to spoil that property given that there are no repeated eigenvalues). – Marc van Leeuwen Nov 11 '13 at 17:03

2 Answers2

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You may be right. There is no unique orthonormal basis. Just check that whatever you are getting are eigenvectors, are orthogonal and have norm $1$ each.

Vishal Gupta
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The matrix that Mathematica gives is not the matrix formed by the unit eigenvectors, but the conjugate transpose of it. That is, it is a unitary matrix $U$ such that $T = U^\ast\pmatrix{5\\ &1\\ &&3}U$. You can easily see that the columns of $U^\ast$ are scalar multiples of your three eigenvectors.

user1551
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