# Condition of tri-diagonal matrix A such that UAU^t=A for all orthogonal matrix U [closed]

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

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

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

• I don't get your point. Would you please elaborate the question more? May 6, 2019 at 8:20
• Are you calling the given matrix and its tri-diagonalisation both $A$? May 9, 2019 at 2:00
• 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. May 10, 2019 at 2:09

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.
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 $$A=\lambda I$$.