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

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

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

  • 3
    $\begingroup$ I don't get your point. Would you please elaborate the question more? $\endgroup$ May 6, 2019 at 8:20
  • $\begingroup$ Are you calling the given matrix and its tri-diagonalisation both $A$? $\endgroup$
    – LSpice
    May 9, 2019 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. $\endgroup$ May 10, 2019 at 2:09

1 Answer 1


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 $A=\lambda I$.

  • $\begingroup$ 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. $\endgroup$ May 10, 2019 at 2:11

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