Unitary and triangular = diagonal matrix

**Bucephalus** · April 13th 2009, 12:30 AM

can anybody help me with this question?

Show that if a matrix A is both triangular and unitary, then it is diagonal.

thanks

**NonCommAlg** · April 13th 2009, 02:15 AM

Originally Posted by Bucephalus

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 $A=[a_{ij}], \ 1 \leq i,j \leq n,$ is upper triangular. the lower triangular case is the same. proof is by induction over $n$ : for $n=2$ it's easy. now if $[b_{ij}]=AA^*=I_n,$ where $A^*$

is the complex conjugate of $A,$ then we'll have $b_{in}=0, \ \forall \ i \leq n-1,$ but $b_{in}=a_{in} \overline{a_{nn}}$ and $a_{nn} \neq 0.$ therefore $b_{in}=0, \ \forall \ i \leq n-1.$ now apply the induction hypothesis for the $(n-1) \times (n-1)$

matrix $C=[a_{ij}], \ 1 \leq i,j \leq n-1$ to finish the proof.

**Bucephalus** · April 13th 2009, 03:05 AM

then we'll have $b_{in}=0, \ \forall \ i \leq n-1,$ but $b_{in}=a_{in} \overline{a_{nn}}$ and $a_{nn} \neq 0.$ therefore $b_{in}=0, \ \forall \ i \leq n-1.$

I don't really understand how

$b_{in}=a_{in} \overline{a_{nn}}$ and $a_{nn} \neq 0.$

**Bucephalus** · April 13th 2009, 03:10 AM

$b_{in}=a_{in} \overline{a_{nn}}$ and $a_{nn} \neq 0.$

From what I'm reading here is that for a say a 2X2 matrix, the element
$b_{i2}$ is the $a_{i2}$ element multiplied by the
$b_{22}$ element.

Is that correct?
i'm going to try that on paper.

**Bucephalus** · April 13th 2009, 03:18 AM

SOrry, I mean the conjugate of the $a_{22}$ element.
not $b_{aa}$ 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.

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