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I am working on a problem where I am just given the characteristic polynomial of some 3x3 matrix $C$ and am asked to show that it is similar to an orthogonal matrix.

So I solved for the roots of the polynomial and found 3 distinct eigenvalues: 1 real, 2 complex.

So $C$ is similar to a diagonal matrix $D$, with the eigenvalues on the diagonal.

I computed $DD^*$, which happened to equal $D^*D$, which equaled to $I$. So, I know that $C$ is similar to a unitary matrix $D$.

However the problem statement asks to show the similarity to an orthogonal matrix, so I wonder whether the matrix $D$ suffices, or whether I have to now find another matrix, from knowing $D$.

Any ideas are welcome.

Thanks,

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I assume your matrix has real entries, in which case your two nonreal eigenvalues are complex conjugates, call them $a\pm bi$. Then $$\pmatrix{a&-b\\ b&a\\}$$ is the orthogonal matrix similar to the unitary matrix with $a\pm bi$ on the diagonal.

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  • $\begingroup$ Hi @GerryMyerson, thanks so much. I verified the computations. Is there a simple / deeper reason for why the matrix above works? It's easy to commit this matrix to memory, but I was just curious. Thanks, $\endgroup$
    – User001
    Dec 23 '15 at 4:11
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    $\begingroup$ Multiplication by $i$ is rotation through a right angle, counterclockwise; this is a linear transformation, represented by the matrix $$\pmatrix{0&-1\\1&0\\}$$ so we may call that matrix, $i$. Then the orthogonal matrix I gave is $a+bi$ (where I treat the identity matrix as 1). $\endgroup$ Dec 23 '15 at 4:18
  • $\begingroup$ Ah, ok. Can I ask one final question? So the first matrix you gave, can I treat it as a $2x2$ sub-matrix, and enlarge it to $3x3$, with the last column containing the real eigenvalue at the bottom right corner. I am guessing that this is the desired $3x3$ orthogonal matrix. So, is it true? That the eigenvalues for any sub-matrix will also be the eigenvalues for the full matrix? Is it a theorem? Thanks, @GerryMyerson, $\endgroup$
    – User001
    Dec 23 '15 at 4:26
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    $\begingroup$ Yes. So long as the matrix is what's called "block diagonal", that works. $\endgroup$ Dec 23 '15 at 20:20

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