Identity Matrix = Orthogonal?

Question

Why is the identity matrix the only n × n matrix that is orthogonal, upper triangular and has positive entries on the diagonal? Aren't there other examples where this is the case?

Answer 1

Since it's upper triangular, the first column has a nonzero entry only in the first row, and it must be $\pm 1.$ Since the diagonal entries are positive, it must be $1$. Then all the other elements of the first row must be $0$, so the second column has a nonzero entry only on the diagonal ...

Answer 2

The fact that the transpose $Q^t$ of such a matrix $Q$ is its own inverse implies that all of the dot products $q_i\cdot q^j = \delta_i^j$ where $q_i$ is row $i$ and $q^j$ is column $j$ of $Q$. This implies that all of the off-diagonal entries of $Q$ must be $0$. The only thing left is the diagonal entries, which must be $1$ for $\|Qe_i\| = 1$ for each basis vector $e_i$.

Answer 3

Suppose a matrix, $A$, satisfies orthogonal, upper triangular, and has positive entries.

Then $A^TA=I$, that is $A^T$ is the inverse of $A$, but inverse of upper triangular matrix is upper triangular but $A^T$ is lower triangular, hence $A^T$ must be a diagonal matrix. Hence $A$ is a diagonal matrix.

The eigenvalues of a diagonal matrix are the diagonal entries and we know that the absolute value is $1$ due to orthogonality. We are also told that it has positive entries, hence those diagonal entries are $1$. Hence, $A$ must be the identity matrix.