# If a matrix A is both triangular and unitary, then it is diagonal

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!

• \pm gives $\pm$. Dec 23, 2018 at 6:26
• Your proof looks okay to me Dec 23, 2018 at 12:00

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. $$OE\Delta$$.

• Thanks for the alternate proof. would my conceptually correct answer be equivalent to normally correct here? Dec 23, 2018 at 21:12
• @H_1317: Yes, I think so; the only reason I said "conceptually correct" instead of simpley "correct" was that I was having a little trouble sorting out your equations $a_1 \ast a_n + 1 = 0$ etc. But I think I've got it now. Dec 23, 2018 at 21:14
• Ahh yes — those equations in the induction part were my biggest worry for a potential error or maybe just not convincing enough. Dec 23, 2018 at 21:16

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 $$n\times n$$ identity matrix by multiplying the $$i$$-th row by $$c\ne0$$; 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 $$i\ne j$$).

Note that $$E_i(c)$$ ($$c\ne0$$) and $$E_{ij}(d)$$ ($$i\ne j$$) 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.