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For a given lower triangular square matrix $T_1$ is there an orthogonal matrix $Q$ such that $QT_1$ is also lower triangular?

In the $2 \times 2$ case, I think the answer is no: $$ QT_1=\left[\begin{array}{cc} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right] \left[\begin{array}{cc} a_{11} & 0 \\ a_{21} & a_{22} \end{array}\right] = \left[\begin{array}{cc} a_{11}\cos(\theta)- a_{21}\sin(\theta) & - a_{22}\sin(\theta) \\ a_{11}\sin(\theta)+ a_{22}\sin(\theta) & a_{22}\cos(\theta) \end{array}\right] $$ so the only way $QT_1$ is lower triangular is if $\sin(\theta)=0$, which implies $\cos(\theta)=1$ and Q is the identity matrix. Is the answer no in the general $n \times n$ case?

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Assume that $T_1$ is nonsingular. If $T_1$ is lower triangular and $T_2:=QT_1$ is also lower triangular, we have $Q=T_2T_1^{-1}$. Since both $T_1$ and $T_2$ are lower triangular, it follows that $Q$ is lower triangular. But the inverse of a lower triangular matrix is lower triangular and the transpose is upper triangular. Since $Q^T=Q^{-1}$ we have that $Q$ is diagonal.

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For a given lower triangular square matrix A there always exists a orthogonal matrix Q such that QA is lower triangular. Take Q = I, the identity matrix which is orthogonal.

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  • $\begingroup$ +1. I do not know why the downvote(s). I think this answer is correct. The question does not say "other than the identity". $\endgroup$ – Luca Citi Jan 10 '19 at 23:13

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