can anybody help me with this question?
Show that if a matrix A is both triangular and unitary, then it is diagonal.
thanks
i'll assume that the matrix $\displaystyle A=[a_{ij}], \ 1 \leq i,j \leq n,$ is upper triangular. the lower triangular case is the same. proof is by induction over $\displaystyle n$: for $\displaystyle n=2$ it's easy. now if $\displaystyle [b_{ij}]=AA^*=I_n,$ where $\displaystyle A^*$
is the complex conjugate of $\displaystyle A,$ then we'll have $\displaystyle b_{in}=0, \ \forall \ i \leq n-1,$ but $\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ and $\displaystyle a_{nn} \neq 0.$ therefore $\displaystyle b_{in}=0, \ \forall \ i \leq n-1.$ now apply the induction hypothesis for the $\displaystyle (n-1) \times (n-1)$
matrix $\displaystyle C=[a_{ij}], \ 1 \leq i,j \leq n-1$ to finish the proof.
then we'll have $\displaystyle b_{in}=0, \ \forall \ i \leq n-1,$ but $\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ and $\displaystyle a_{nn} \neq 0.$ therefore $\displaystyle b_{in}=0, \ \forall \ i \leq n-1.$
I don't really understand how
$\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ and $\displaystyle a_{nn} \neq 0.$
From what I'm reading here is that for a say a 2X2 matrix, the element$\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ and $\displaystyle a_{nn} \neq 0.$
$\displaystyle b_{i2}$ is the $\displaystyle a_{i2}$ element multiplied by the
$\displaystyle b_{22}$ element.
Is that correct?
i'm going to try that on paper.
SOrry, I mean the conjugate of the $\displaystyle a_{22}$ element.
not $\displaystyle b_{aa}$ element.
Don't worry,I will do some paper work here.
I think I'm starting to understand some of your proof using a 2X2 Identity matrix.
thanks for your response.