0
$\begingroup$

$A,U \in \mathbb{R}^{n\times n}$

$\exists A \in \text{{tri-diagonal}} \quad s.t \quad UAU^{t}=A \quad \forall U \in \text{{orthogonal}}$

I know it holds when A is a diagonal matrix, but have no idea when A is a tri-diagonal matrix..

$\endgroup$

closed as unclear what you're asking by user44191, Jan-Christoph Schlage-Puchta, Sean Lawton, Chris Godsil, LSpice May 9 at 2:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ I don't get your point. Would you please elaborate the question more? $\endgroup$ – Bullet51 May 6 at 8:20
  • $\begingroup$ Are you calling the given matrix and its tri-diagonalisation both $A$? $\endgroup$ – LSpice May 9 at 2:00
  • $\begingroup$ sorry for late comment. I wanna know the existence of that tridiagonal matrix A. I need to do SVD on matrix $ \sum {B \otimes \cdots A \cdots \otimes B}$ where A B are tridiagonal matrices. I hope that above condition holds, but it seems not true. $\endgroup$ – rae hyun kim May 10 at 2:09
1
$\begingroup$

In general, this doesn't hold for diagonal matrices $A$ in general.

Consider $$A=\begin{bmatrix} 1 & 0 &0 \\0 & 2 &0 \\ 0& 0& 3 \end{bmatrix}$$ and $$U = \begin{bmatrix} 0 &0 & 1 \\1 & 0 &0 \\ 0& 1&0 \end{bmatrix}$$ Then $$UAU^\mathrm{t}=\begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 &0 \\ 0 & 0 & 2 \\ \end{bmatrix} \neq A$$

Given any diagonal $A$ with 3 distinct eigenvalues, the permutation matrix $U$ (a matrix in the natural image of $S_3$) will act non-trivially by conjugation.

To better understand the case of a trivial action by conjugation. Consider the following.

If $UAU^\mathrm{t}=A$, then $UA=AU$; so $A$ commutes with all orthogonal matrices and must be a multiple of the identity matrix. To see this more clearly, suppose $v$ is an eigenvector of $A$ with eigenvalue $\lambda$. Then for all $u\in\mathbb{R}^3$ there is an orthogonal matrix $U$ such that $Uv=u$ (and thus $U^\mathrm{t}u=v$), and $$Au=UAU^\mathrm{t}u=UAv=\lambda Uv=\lambda u,$$ so every vector is an eigenvector of $A$ with eigenvalue $\lambda$. Thus $A=\lambda I$.

$\endgroup$
  • $\begingroup$ Thanks for answer. I totally misunderstood about the diagonal cases. It doesnt hold for even diagonal matrix A, so I need to find other way. $\endgroup$ – rae hyun kim May 10 at 2:11

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