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We are given $A,B$ are orthogonal $4$ by $4$ matrices with real values only. We are given $\det(A) = \det(B) = 1$ and $\mathrm{trace}(A) = \mathrm{trace}(B)$. Is $A$ similar to $B$?

My solution:

I think that it is, but a friend told me that she managed to contradict it.

I think they are similar because, the determinant is the product of the eigenvalues. and since $A,B$ are orthogonal, we know all the eigenvalues are either $1$ or $-1$.

So the set of eigenvalues is any of the following:

$1,1,1,1$ or $1,1,-1,-1$ or $-1,-1,-1,-1$

Since $1+1+1+1 \neq 1+1-1-1 \neq -1-1-1-1$ we get that $A$ and $B$ must have the same set of eigenvalues. And so they are similar to the same diagonal matrix, and so they are similar to each other.

Where is the problem in this argument?

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  • $\begingroup$ I think the argument's fine. $\endgroup$ – DonAntonio Jan 4 '14 at 17:47
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    $\begingroup$ Yes, I edited that 1/2 second before your comment, @Git...thanks. $\endgroup$ – DonAntonio Jan 4 '14 at 17:47
  • $\begingroup$ You mean all eigenvalues are real, or all entries are real? $\endgroup$ – Andreas Caranti Jan 4 '14 at 17:49
  • $\begingroup$ @DonAntonio I take back my thanks and undelete my answer :P $\endgroup$ – Git Gud Jan 4 '14 at 17:50
  • $\begingroup$ Well, until the OP addresses the doubts in the comments I shall not add anything more to this... $\endgroup$ – DonAntonio Jan 4 '14 at 17:53
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The problem is this step:

I think they are similar because, the determinant is the product of the eigenvalues. and since $A,B$ are orthogonal, we know all the eigenvalues are either $1$ or $−1$.

There are (real) orthogonal matrices which don't have real eigenvalues.

To look for simple counter examples, it might be useful to add the restriction $\text{trace}(A)=0=\text{trace}(B)$.

One of the simplest orthogonal matrices with non-real eigenvalues that comes to mind is $\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$, so it's sensible to consider the block diagonal matrix $$\color{grey}{A:=}\begin{bmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\0 & 0 &-1 & 0\end{bmatrix}.$$

A matrix $B$ with real eigenvalues only which satisfies the conditions is easy is to find.

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  • $\begingroup$ And have zero trace. $\endgroup$ – egreg Jan 4 '14 at 17:33
  • $\begingroup$ The OP though says the matrices only have real eigenvalues... $\endgroup$ – DonAntonio Jan 4 '14 at 17:44

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