# What kind of matrices are orthogonally equivalent to themselves?

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 $U=I$ –  Prometheus Feb 18 '11 at 13:39
@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. –  Shiyu Feb 18 '11 at 13:41
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. –  Jason DeVito Feb 18 '11 at 13:47
The converse doesn't seem right. Take A=I, then any orthogonal matrix will do –  Thomas Rot Feb 18 '11 at 13:48
@Thomas: What I mean is $k$ can be a scalar to be determined. If $A=I$, $k=1$, $A=kI$ still holds. –  Shiyu Feb 18 '11 at 13:56

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 $O(d_1)\times O(d_2) \times ... \times O(d_k)$.