Question: I understand that if $A$ is unitary, then $A^{-1} = A^\ast$ are also unitary matrices.

But if $A$ is a unitary upper triangular matrix, then how are $A^{-1}$ and $A^\ast$ also upper triangular? This was from a particular class.

Then the statement goes on to say that $A$ will be a diagonal matrix with diagonal entries having absolute value $1$.

Answer: 
For something closer to an analytic approach (that hints at a very useful inequality from Schur) just compute the squared Frobenius norm.

You know all the eigenvalues of a unitary matrix all have modulus 1 and you know they are on the diagonal of a triangular matrix. Computing the squared Frobenius norm two different ways:

*(i.)* $\Big \Vert A\Big \Vert\_F^2 = \text{trace}\Big(A^\*A\Big)=\text{trace}\Big(I\_n\Big)=n$

*(ii.)* $\Big \Vert A\Big \Vert\_F^2=\Big(\sum\_{i=1}^n \vert a\_{i,i}\vert^2\Big)+\Big(\sum\_{i=1}^{n-1}\sum\_{j=i+1}^n \vert a\_{i,j}\vert^2\Big)=n+\Big(\sum\_{i=1}^{n-1}\sum\_{j=i+1}^n \vert a\_{i,j}\vert^2\Big)$

$\implies \Big(\sum\_{i=1}^{n-1}\sum\_{j=i+1}^n \vert a\_{i,j}\vert^2\Big) = 0$  

i.e. all entries above the diagonal of your upper triangular matrix are zero, so $A$ is diagonal.
Upper triangular matrices are closed under multiplication.

If you consider the elementary matrices corresponding to row reduction, all the upper right entries concern changing the upper right entries, so changing them back would use matrices of the same shape.

$A^\*$ is a notation that I have seen mean different things in different contexts. Does it mean complex conjugation here? If so, no elements get swapped so of course it would still be upper triangular.