# 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.

• Define "orthogonally similar." – David G. Stork Jun 28 '18 at 1:18
• @DavidG.Stork If A and B are orthogonally similar, then P is an orthogonal matrix such that, $P^TAP = B$ – kronos Jun 28 '18 at 1:23
• I don't know what you really mean with "orthogonally similar" but I think you can use the result with the upper triangular matrix with the matrix $\textbf A^T$ and therefore you will have the desired result. – Gonzalo Benavides Jun 28 '18 at 1:24
• So you can prove that $A$ is orthogonally similar to an upper triangular matrix, and you want to be able to prove $A$ is orthogonally similar to a lower triangular matrix, right? – Robert Lewis Jun 28 '18 at 2:06
• @RobertLewis yes, what I'm thinking is that if an arbitrary matrix B (such that its transpose is A) is orthogonally similar to an upper triangular matrix. Then $P^TBP = T^T$ or $(P^TBP)^T = (T^T)^T$ or $P^TAP = T$ – kronos Jun 28 '18 at 2:20

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 $O$.

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.