Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
    
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. –  Weltschmerz Mar 7 '11 at 13:56
    
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. –  Shiyu Mar 25 '11 at 15:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.