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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!

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

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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. $OE\Delta$.

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    $\begingroup$ Thanks for the alternate proof. would my conceptually correct answer be equivalent to normally correct here? $\endgroup$
    – H_1317
    Dec 23, 2018 at 21:12
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    $\begingroup$ @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. $\endgroup$ Dec 23, 2018 at 21:14
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    $\begingroup$ Ahh yes — those equations in the induction part were my biggest worry for a potential error or maybe just not convincing enough. $\endgroup$
    – H_1317
    Dec 23, 2018 at 21:16
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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.

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