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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 $1$.

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  • $\begingroup$ Easy to prove that the inverse of an invertible upper triangular matrix is still upper triangular. $\endgroup$
    – xbh
    Commented Apr 29, 2021 at 16:50
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    $\begingroup$ I can't think of any non-diagonal upper triangular unitary matrices. $\endgroup$ Commented Apr 29, 2021 at 16:51

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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.

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If $A$ is an upper triangular matrix, then $A^{-1}$ is an upper triangular matrix, and the conjugate transpose $A^*$ is a lower triangular matrix.

If $A$ is upper triangular and unitary, i.e. $A^{-1} = A^*$, then $A$ is simultaneously upper triangular and lower triangular. The only matrices that are both upper and lower triangular are diagonal matrices.

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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.

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    $\begingroup$ In order to get $A^{-1} = A^*$, we must take $A^*$ to mean the conjugate transpose. So if we know $A$ is unitary and upper triangular, then $A^{-1}$ is upper triangular and $A^*$ is lower triangular, so we conclude $A^{-1}$ is diagonal and $A$ is diagonal. $\endgroup$
    – GEdgar
    Commented Apr 29, 2021 at 16:58
  • $\begingroup$ $A^*$ means conjugate transpose according to what has been done in my course. I think my professor meant — given $A$ is a unitary upper triangular matrix and $A^{-1}$ and $A^*$ is also upper triangular, then $A$ must be a diagonal matrix with diagonal entries having absolute value one. $\endgroup$
    – user733666
    Commented Apr 29, 2021 at 17:06

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