# Thread: Unitary matrices

1. ## Unitary matrices

1. Find Unitary matrice $U$ and Diagonal matrice $D$ That $U^*AU=D$
For $A=\begin{bmatrix}1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

2.Let $A\in M_n(C)$ be a matrice with $\lambda_1,\lambda_2...\lambda_n$ eigenvalues so that
$|\lambda_1|...|\lambda_n|=1$.
Is $A$ Unitary?

So I've got no idea how to approach these problems.

2. Originally Posted by cursedbg
1. Find Unitary matrice $U$ and Diagonal matrice $D$ That $U^*AU=D$ For $A=\begin{bmatrix}1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$
The matrix $A$ is nornal, that is $AA^*=A^*A$ . This mean that eigenvectors associated to distinct eigenvalues are orthogonal. The eigenvalues of $A$ in this case are simple, so find the corresponding eigenvectors $u_1,u_2,u_3$ and divide between the norm to obtain $e_1,e_2,e_3$. Then,

$U=[e_1,\;e_2,\;e_3]$

Fernando Revilla

3. Originally Posted by cursedbg
1. Find Unitary matrice $U$ and Diagonal matrice $D$ That $U^*AU=D$ For $A=\begin{bmatrix}1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

The matrix $A$ is nornal, that is $AA^*=A^*A$ . This mean that eigenvectors associated to distinct eigenvalues are orthogonal. The eigenvalues of $A$ in this case are simple, so find the corresponding eigenvectors $u_1,u_2,u_3$ and divide between the norm to obtain $e_1,e_2,e_3$. Then,

$U=[e_1,\;e_2,\;e_3]$

Fernando Revilla

Edited: Sorry I published by mistake twice the same post.

4. Originally Posted by cursedbg
2.Let $A\in M_n(C)$ be a matrice with $\lambda_1,\lambda_2...\lambda_n$ eigenvalues so that
$|\lambda_1|...|\lambda_n|=1$. Is $A$ Unitary?

No, it isn't. Choose as a counterexample:

$A=\begin{bmatrix}{1}&{1}\\{0}&{1}\end{bmatrix}$

Fernando Revilla