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A matrix $A \in R^{n\times n}$ is said to be orthogonally equivalent to $B\in R^{n\times n}$ if there is an orthogonal matrix $U\in R^{n\times n}$, $U^T U=I$, such that $A=U^T B U$. My question is what kind of matrices are orthogonally equivalent to themselves? i.e., $A=U^T A U$

A similar interesting question is: if $$U^T \Lambda U=\Lambda $$ where $\Lambda$ is a diagonal matrix and $U$ is a orthogonal matrix, are the diagonal entries of $\Lambda$ equal? That is whether $\Lambda=kI$.

Look forward to your opinion. Thank you very much.

Shiyu

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    do you mean that $A=U^T A U$ for every orthogonal matrix U? Some specific U? because you can always take $U=I$2011-02-18
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    @Prometheus: if $U=I$ of course all matrices satisfy $A=IAI$. I mean $A=U^T A U$ for some non-identity orthogonal matrices but not for all orthogonal matrices.2011-02-18
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    The converse is not right. Take a 3x3 diagonal matrix with entries 2,2,1 and call it A. Use as the orthogonal matrix U a block 2x2 rotation with a 1 in the bottom right corner.2011-02-18
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    The converse doesn't seem right. Take A=I, then any orthogonal matrix will do2011-02-18
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    @Thomas: What I mean is $ k$ can be a scalar to be determined. If $ A=I$, $k=1$, $A=kI$ still holds.2011-02-18
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    @ Jason: Thank you! You are right! But from your example, we can see 1) A is a diagonal matrix. 2) $A=diag(2I,1)$ which is still very similar to $A=kI$. If $A=kI$ does not hold can we say $A$ must be diagonal? Moreover, the orthogonal matrix you chose is very special. It is a rotation matrix with rotation axis as $z-axis$. What will happen for a generic rotation matrix? Anyway, I think I should add some constraints to the orthogonal matrix.2011-02-18
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    Even if A is not diagonal there are still many counterexamples. Use a 4 x 4 matrix A with A_11 = 2, A_22 = 2, A_33, A_34, A_43, and A_44 arbitrary and all other elements 0. For U, use a block 2 x 2 rotation and 1s on the rest of the diagonal.2011-02-18
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    @Jason: Thanks. Yeah, your counterexample is right again. But as I mentioned, the rotation matrix is very special. And can you choose $A_{11}\ne A_{22}$ in the counterexample? I'm afraid not. So for the parts that really times a generic rotation matrix instead of the identity, it seems they should be like $kI$.2011-02-18

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The family of matrices $U^{T}BU$, where $B$ is a fixed, positive definite matrix $\mathbb{R}^{n\times n}$, and $U$ varies over the orthogonal group $O(n)$, is obtaining by rigidly rotating and reflecting the eigenvectors of $B$. The matrix $B$ is invariant under such a transformation iff its eigenspaces are preserved. Even if there are $n$ distinct eigenvalues (so that all eigenspaces are $1$-dimensional), there are $2^n$ discrete choices for $U$ that preserve $B$: namely, reflections of any subset of the eigenvectors. Note that these form a discrete subgroup of $O(n)$ under matrix multiplication: it can be represented as $O(1)^n$. When eigenvalues are degenerate, then additional orthogonal transformations of the higher-dimensional eigenspaces will preserve the matrix $B$. In general, if the eigenspaces of $B$ associated with eigenvalues $\lambda_1 < \lambda_2 < ... < \lambda_k$ have dimensions $d_1,d_2,...d_k$, with $d_1+d_2+...+d_k=n$, then the subgroup of $O(n)$ that preserves $B$ is isomorphic to $O(d_1)\times O(d_2) \times ... \times O(d_k)$.

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    I'm not sure I understand the bit "..., is obtaining by rigidly rotating and reflecting the eigenvectors of B". Could you please elaborate it? Thank you.2011-03-07
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    Can you prove "The matrix B is invariant under such a transformation iff its eigenspaces are preserved"? You analysis is based on this proposition. But it is not so obvious to understand.2011-03-25