I am trying to prove that a matrix that is both "unitary and upper triangular" must be a diagonal matrix.

I am thinking that the fact that columns of all unitary matrices form an orthonormal basis of F^n will ensure that all columns of this matrix are mutually orthogonal.

Also, the fact that the matrix is upper triangular will then give me the required result....but I am trying to derive a formal proof. Please suggest if my approach is OK.

Thanks in advance.


1 Answer 1


Your approach is OK. As you said, the columns of $A$, denote them by $C^A_1,...,C^A_n$, form an orthonormal basis of $F^n$ w.r.t the standard inner product. Denote $C^A_j=(a_{1,j},...,a_{n,j})$.
Since $A$ is upper-triangular, $a_{k,1}=0$ for all $k>1$. Since $A$ is unitary, $a_{1,1}\neq0$. Now compute the inner-product $\langle C^A_1,C^A_j\rangle$ for all $j>1$: you get: $$0=\langle C^A_1,C^A_j\rangle=a_{1,1}\cdot a_{1,j}+0+...+0$$ Since $a_{1,1}\neq0$, we have $a_{1,j}=0$ for all $j>1$. (this shows that the first row is $(a_{1,1}\hspace{2pt}0\hspace{2pt}...\hspace{2pt}0)$)
Now use induction on columns.

  • $\begingroup$ Thanks a lot for giving an outline of the proof. $\endgroup$
    – Tav
    Commented Mar 1, 2012 at 17:15
  • $\begingroup$ Can you elaborate on why the statement "Since $a_{1,1} \neq 0$, we have $a_{1,j} = 0$ for all $j>1$." I can't see why $a_{1,2} = 0$ from the inner-product. $\endgroup$
    – holala
    Commented Aug 26, 2021 at 20:54
  • $\begingroup$ Because Since A is unitary, the product of any two different columns is, so <C_1,C_2>=0, expanding this you will end up with a_1,1 * a_1,j=0, since a_1,1 is not 0, since A is triangular (the first column must be almost full of 0's) and unitary(every column is l.i, so none is 0), you'll end up with a_1,j=0 $\endgroup$ Commented Oct 8, 2022 at 16:18

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