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Can someone please show me how to diagonalize a matrix such as the one below using an orthogonal similarity transformation? $$ \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ \end{bmatrix} $$

I have been looking everywhere online to find an example of orthogonal similarity transformations but I can't find any. Am I searching for the wrong thing? Is there another name for it, because similarity transformations seem awfully close to Jordan canonical form?

Please help. Thank you in advance.

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  • $\begingroup$ Is there some difference between an orthogonal similarity transformation and a regular similarity transformation? $\endgroup$ – user137731 Nov 21 '15 at 1:40
  • $\begingroup$ Here's the usual process for diagonalizing a matrix. $\endgroup$ – user137731 Nov 21 '15 at 1:41
  • $\begingroup$ I wish I knew the answer to your question..I was hoping someone would explain that to me. Appreciate the feedback. $\endgroup$ – thepillsbury Nov 21 '15 at 1:49
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For stuff like that WolframAlpha is a great help: $$ \pmatrix{2&1&1\\1&2&1\\1&1&2}=SDS^{-1}= \pmatrix{ -1&-1&1\\ 0&1&1\\ 1&0&1 } \pmatrix{ 1&0&0\\ 0&1&0\\ 0&0&4 } \frac13 \pmatrix{ -1&-1&2\\ -1&2&-1\\ 1&1&1 } $$

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  • $\begingroup$ This is not an orthogonal transformation since the inverse of S does not equal the transpose of S. I figured out a way to do it using the Gramm-Schmidt process. $\endgroup$ – thepillsbury Nov 24 '15 at 21:30
  • $\begingroup$ @thepillsbury yes I know, but I thought you'll figure out the details yourself, whic was correct... $\endgroup$ – draks ... Nov 26 '15 at 20:29

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