Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.
In case you don't remember what a unitary matrix is, please refer to wikipedia.
Possibly relevant theorem from the textbook:
Theorem 6.18. Let be a linear operator on a finite-dimensional inner product space . Then the following statements are equivalent:
(a) .
(b) for all .
(c) If is an orthonormal basis for , then is an orthonormal basis for .
(d) There exists an orthonormal basis for such that is an orthonormal basis for .
(e) for all .
Every time I start into this, it turns into something horribly messy. I feel like there's a trick I'm missing. Any thoughts? Or perhaps a full solution?
Thanks!
So let's make sure I understand the proof correctly...
Choose an upper-triangular, unitary matrix. Partition so that
,
with upper-triangular unitary matrices. Then we have
.
So .
First of all, is that correct so far?
If yes, then I have one remaining question: Why can we assume and are unitary? Obviously they are upper-triangular, but how do I prove that if I arbitrarily partition , we have unitary?
Thanks!
let be an unitary and upper-triangular matrix. since is upper triangular, we have thus now let be the column with in the -th
entry and 0 everywhere else. note that is the -th column of let then and thus for all so all the entries in the first
row of except are 0. next, let then and thus for all so all the entries in the second row of except are 0.
continuing this way we see that for any the only non-zero entry in the -th row of is and hence is diagonal.
Alternative:
Let the matrix be .
If you know that 'the inverse of an upper triangular matrix is upper triangular', the proof is simpler.
Inverse of unitary is . But will be lower triangular. Since 'the inverse of an upper triangular matrix is upper triangular', must be upper triangular too. Thus is both upper triangular and lower triangular, thereby diagonal. So is