# How to find $A=UDU^H$ in this case

I am given a matrix $A$. I find out it is normal. And I compute $\det(A-\lambda)=0$ and find that not all $\lambda_i$ are different, i.e., the eigenvalues are not distinct. Thus, I am not sure if the eigenvectors are L.I.

Now, I want find a decomposition $$A=UDU^H$$ but how do I do it? And under what conditions can I do that?

Where

• $D$ is a diagonal matrix
• $^H$ is the hermitian conjucate
• $U$ is unitary matrix.
• What is $U^H$? What is $D$? – Brian Fitzpatrick Mar 17 '14 at 10:37
• @BrianFitzpatrick added a list at the end. – jacob Mar 17 '14 at 11:12
• @GitGud it should read "normal" instead of "not normal". edited that. – jacob Mar 17 '14 at 11:12

## 1 Answer

Since $A$ is normal, you can always find the decomposition $A=UDU^H$.

Assuming you know how to diagonalize a matrix, to find $U$ you simply take any diagonalizing matrix $P$ and apply Gram-Schmidt to the columns of $P$. The resulting vectors, disposed as the columns of $U$, make $U$unitary.