# $U\in{Mat(\mathbb{C})} , $$U is unitary and U+iI is self adjoint, prove: U = -iI I'm preparing for a test and this question was a real pain to do. I get a bit confused by all the terms, so I'd appreciate if you guys peek at my attempt to solve this, and see if its correct. What I did: First of all, U+iI is self adjoint, therefore its normal and unitary diagonalizable, which means there exists a unitary matrix P and a diagonal matrix D so:$$U+iI = PDP^{-1}$$Now because U+iI is diagonalizable it has n distinct eigenvectors, and for every such vector v_i there is an eigenvalue \lambda_k so:$$(U+iI)v_i = {\lambda_k}v_iUv_i + iIv_i = {\lambda_k}v_iUv_i = (\lambda_k-i)v_i$$Therefore$(\lambda_k-i)$is an eigenvalue of$U$, but$U$is unitary which means that$(\lambda_k-i) (\overline{\lambda_k-i}) = 1 $and finally$\lambda_k = 0$. This occurs for all$v_i$eigenvectors, therefore$U+iI$has one eigenvalue and its zero. Therefore$D$has only zeroes on its diagonal, and$U+iI = 0$therefore$U = -iI$. I'm not really sure if the argument I used about the eigenvectors\values is correct, if its not, would love to hear a few tips on how to proceed, thanks! ## 1 Answer This solution is correct as far as I can tell. For a simpler solution:$U + iI$is self-adjoint, so it is equal to its complex-conjugate transpose. This means that all elements on the diagonal of$U + iI$must be real. Thus the imaginary part of all the diagonal elements of$U$must be$-1$. Since$U$is unitary, each row and column must have a vector length of 1, but since$-i$has magnitude 1, all other terms have to be zero (adding a real part to a diagonal element or any nonzero element off the diagonal will give that row and column a length greater than one). Therefore$U = -i I\$.

• Well that makes sense as well, thank's a lot :)
– Xsy
Jun 18, 2014 at 16:48