Prove that an $n \times n$ matrix ${\bf A}$ is orthogonally similar to a lower triangular matrix Prove that if ${\bf A}$ is an $n \times n$ matrix with real eigenvalues, then ${\bf A}$ is orthogonally similar to a lower triangular matrix ${\bf T}$.
I can prove that ${\bf A}$ is similar to an upper triangular matrix (using induction), but I can't find a way to prove it similar to a lower triangular matrix.
 A: We are given the 
Statement: If $A$ is a real square matrix with all real eigenvalues, then $A$ is orthogonally similar to an upper triangular matrix.
Now let $A$ be as given above.
Consider $A^T$; it too is a real square matrix, and since its eigenvalues are the same as those of $A$, we see they are all real, and thus there is an orthogonal matrix $O$ and an upper triangular matrix $\Delta$ such that
$A^T = O\Delta O^T; \tag 1$
then
$A = (A^T)^T = (O \Delta O^T)^T = (O^T)^T \Delta^T O^T = O \Delta^T O^T; \tag 2$
we thus see the matrix $A$ is similar to the lower triangular matrix $\Delta^T$ by the same orthogonal transformation $O$.
A: If $U$ is upper triangular, let $P$ be the matrix with ones on the antidiagonal and zeros elsewhere. Then $P$ is orthogonal (actually $P = P^T = P^{-1}$), and $P^T UP$ is lower triangular (it is the matrix obtained from $U$ by rotating the entries 180 degrees). So any upper triangular matrix is orthogonally similar to a lower triangular matrix. Since any matrix is similar to an upper triangular matrix, it is therefore also similar to a lower triangular matrix.
