# Product of lower triangular matrix and orthogonal matrix

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?

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.