Csdrhrt

I have a proof I was hoping someone could review. Thanks in advance!

Let A be a matrix that is both triangular and unitary.
Then the columns of A are of length 1 and are orthogonal to each other.
Assume A is upper triangular, in this case.
Then set a$_n$ = column $n$ in A.

Clearly, since A is upper triangular and the columns are of length 1, then $ a$_1$ = $$
begin{bmatrix}
pm 1 \
0 \
0 \
vdots \
0 \
end{bmatrix}
$$
$

Now column $a_2$ has a similar composition with $a_1$. That is, on the diagonal of the entire matrix A, its second component here, $y_2$, has the value $pm$ 1 and zeros in all other components. This is due to first only two options exists for non-zero entries in the columns since A is upper triangular. Second since $a_1$*$a_2$ = 0 by orthogonality we have:

$$
1*y_1 + 0*y_2 + 0 + 0 + ... + 0 = 0
$$

Where $a_2$ is made up of components $y_1$, $y_2$ , ... $y_n$. So clearly $y_1$ must be zero, and for i > 2 we know $y_i$ = 0, by the upper triangular property

$ a$_2$ = $$
begin{bmatrix}
0 \
pm 1 \
0 \
vdots \
0 \
end{bmatrix}
$$
$

The remainder of the proof uses induction after examining these first two cases as the base case.

Let the following claim hold for $a_n$, we wish to prove that $a_(n+1)$ is a zero vector except in its n+1 component (where $pm$ 1 resides). Then given the $a_(n+1)$ column we note that by the orthogonality with all the previous n columns we can show that 0 must be the value of the first n components in $a_(n+1)$.

for any ith component in the column for i < n+1 we have:

$a_1$ * $a_n+1$ = 0 => the 1st component must be zero (similarly to above)

$a_2$ * $a_n+1$ = 0 => the 2nd component must be zero (similarly to above)
and so on ... until we conclude that each component up to the n+1 component must be equal to zero by orthogonality.

Hence only the $n+1$ component in the column $a_(n+1)$ can be equal to $pm$ 1 since all other inner products with the previous n columns indicate the remaining components must be zero (and any components below are zero due to A being upper triangular).

Hence for the upper triangular case, the matrix A must be diagonal.

I believe the lower triangular case to be a similar proof, but have not delved into it yet.

Thanks!

I think our OP H_1317's proof is conceptually correct.

Here is a more somewhat more abstract proof:

Suppose $A$ is upper triangular; then I claim that $A^{-1}$ is also upper triangular; for we may write

$A = D + T, tag 1$

where $D$ is diagonal and $T$ is strictly upper triangular; that is, the diagonal entries of $T$ are all zero; we observe that, since $A$ is triangular, $det(A)$ is the product of the diagonal entries of $A$; since $A$ is unitary, it is non-singular and thus $det(A) ne 0$, so none of the diagonal entries of $A$ vanish, and the same applies to $D$; therefore $D$ is invertible and we may write

$A = D(I + D^{-1}T); tag 2$

we next observe that $D^{-1}T$ is itself strictly upper triangular, hence nilpotent; in fact we have

$(D^{-1}T)^n = 0, tag 3$

where $n = text{size}(A)$; the nilpotence of $D^{-1}T$ allows us to write an explicit inverse for $I + D^{-1}T$; indeed, we have the well-known formula

$(I + D^{-1}T) displaystyle sum_0^{n - 1} (-D^{-1}T)^k = I + (-1)^n (D^{-1}T)^n = I; tag 4$

thus,

$(I + D^{-1}T)^{-1} = displaystyle sum_0^{n - 1} (-D^{-1}T)^k; tag 5$

since every matrix $(-D^{-1}T)^k$ occurring in this sum is upper triangular, we see that $(I + D^{-1}T)^{-1}$ is upper triangular as well; from (2),

$A^{-1} = (I + D^{-1}T)^{-1}D^{-1}, tag 6$

which shows that $A^{-1}$ is upper triangular.

Having established my claim, we now invoke the unitarity of $A$:

$A^dagger A = AA^dagger = I, tag 7$

i.e.,

$A^dagger = A^{-1}; tag 8$

we have by definition

$A^dagger = (A^ast)^T = ((D + T)^ast)^T = (D^ast + T^ast)^T = (D^ast)^T + (T^ast)^T, tag 9$

from which we see that $A^dagger$ is lower triangular when $A$, and hence $A^{-1}$, is upper triangular;
then only way (8) can hold is with $T = 0$; therefore we see that

$A = D tag{10}$

is a diagonal matrix.

Of course, if $A$ is lower triangular the same result binds, the proof almost identical to that given above. $OEDelta$.

Here is a proof that an invertible lower triangular matrix has a lower triangular inverse only using elementary row operations. From this it clearly follows that a unitary triangular matrix must be diagonal.

Suppose a lower triangular matrix $A$ is invertible, say
$$
A=begin{bmatrix}
a_{11} & 0 & 0 & dots & 0 \
a_{21} & a_{22} & 0 & dots & 0 \
vdots & vdots & vdots & ddots & vdots \
a_{n1} & a_{n2} & a_{n3} & dots & a_{nn}
end{bmatrix}
$$
Then $a_{ii}ne0$, for $i=1,2,dots,n$.

Denote by $E_i(c)$ the matrix obtained from the $ntimes n$ identity matrix by multiplying the $i$-th row by $cne0$; similarly, denote by $E_{ij}(d)$ the matrix obtained from the identity by adding to the $i$-th row the $j$-th row multiplied by $d$ (here it is assumed that $ine j$).

Note that $E_i(c)$ ($cne0$) and $E_{ij}(d)$ ($ine j$) are invertible lower triangular matrices and their inverses are, respectively, $E_i(c^{-1})$ and $E_{ij}(-d)$, again lower triangular.

Then it is easily seen that multiplying a matrix on the left by $E_i(c)$ or $E_{ij}(d)$ is the same as performing the corresponding elementary row operation on $A$. You can also verify that the product
$$
E_1(a_{11})E_{21}(a_{21})dotsm E_{n1}(a_{n1}) E_{2}(a_{22})dotsm E_{n}(a_{nn})I_n
$$
gives back the matrix $A$, just by performing successively the elementary row operations: starting from the right, the first row operations places $a_{nn}$ at position $(n,n)$; the following one is $E_{n,n-1}(a_{n,n-1})$ will place $a_{n-1,n}$ at position $(n-1,n)$. These two entries will be unaffected by the subsequent row operations and the same is for all the other matrix multiplications/row operations. You can see that the matrix is “filled in” starting from the lower left corner, going then left and up until the diagonal is reached.

Therefore $A$ is the product of lower triangular matrices each of them having a lower triangular inverse; hence also $A^{-1}$ is lower triangular as well.