can anybody help me with this question?

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

thanks

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- Apr 13th 2009, 12:30 AMBucephalusUnitary and triangular = diagonal matrix
can anybody help me with this question?

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

thanks - Apr 13th 2009, 02:15 AMNonCommAlg
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. - Apr 13th 2009, 03:05 AMBucephalus
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 $\displaystyle a_{nn} \neq 0.$ - Apr 13th 2009, 03:10 AMBucephalusQuote:

$\displaystyle b_{in}=a_{in} \overline{a_{nn}}$ 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. - Apr 13th 2009, 03:18 AMBucephalus
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.