I've been strugling with the following for a while. Will appreciate any direction.
If U is a unitary matrix and (U+iI) is hermitian, prove that U=-iI. (U is obviously square of order n).
Thanks,
SK
Think in terms of eigenvalues and eigenvectors. A unitary matrix is diagonalisable, so if all its eigenvalues are $\displaystyle -i$ then U must be equal to $\displaystyle -iI$. Also, each eigenvalue $\displaystyle \lambda$ has absolute value 1.
If $\displaystyle U + iI$ is hermitian then $\displaystyle U^*-iI = U+iI$. Now suppose that $\displaystyle \lambda$ is an eigenvalue of $\displaystyle U$, with eigenvector x. Then $\displaystyle Ux = \lambda x$ and $\displaystyle U^*x = U^{-1}x = \lambda^{-1}x = \overline{\lambda}x$. Now over to you: show from those facts that $\displaystyle \lambda = -i$.
Alternatively to [b]Opalg's[b] solution you can notice three things
1) The eigenvalues of $\displaystyle p(A)$ are $\displaystyle p(\lambda)$ for any polynomial, thus the eigenvalues of $\displaystyle U+iI$ are $\displaystyle \lambda+i$ where $\displaystyle \lambda\in\sigma\left(A\right)$
2) Since $\displaystyle U+iI$ is Hermitian it's eigenvalues must be real, thus $\displaystyle \lambda+i\in\mathbb{R}$ for each $\displaystyle \lambda\in\sigma\left(A\right)$. Thus, $\displaystyle \lambda=r-i$ for some $\displaystyle r\in\mathbb{R}$
3) But, since $\displaystyle U$ is unitary we have that for every $\displaystyle \lambda\in\sigma\left(A\right)$ that $\displaystyle 1=|\lambda|=|r-i|=r^2+1$ and thus $\displaystyle r=0$
Thus, all the eigenvalues of $\displaystyle U$ are $\displaystyle -i$.
Thanks for the idea. I think I understand the logic here. One thing that I'm missing is the jump from saying that if all the eigenvalues of U are -i then it is equal to -iI.
The only reasoning I can find is since U is unitary there exists an orthonormal base in which every single vector is an eigenvector, now if x is avector it can be represented as a linear combination of those base vectors and therfore one can show that U is kI (where k is the only eigenvalue).
BUT, I'm not sure that this is legitimate.
Thanks,
SK
Yes, that is a legitimate argument. Another way to see this result is to say that since there exists an orthonormal basis consisting of eigenvectors, the matrix of U with respect to that basis will be a diagonal matrix in which the diagonal elements are the corresponding eigenvalues. But if each of those eigenvalues is –i, then the matrix is –iI (where I is the identity matrix) and hence U = –iI (where I is the identity operator).
The easy part of the theorem I'm referring to says that if $\displaystyle \lambda$ is an eigenvalue of $\displaystyle A$ and $\displaystyle p(z)=a_0+\cdots+a_nz^n$ is a polynomial then $\displaystyle p(\lambda)$ is an eigenvalue of $\displaystyle p\left(A\right)=a_0I+a_1A+\cdots+a_nA^n$. This is easily verified since if $\displaystyle x_{\lambda}$ is the associated eigenvector then clearly $\displaystyle A^jx_{\lambda}=A^{j-1}\left(Ax_{\lambda}\right)=A^{j-1}\left(\lambda x_{\lambda}=\lambda A^{j-1}x_{\lambda}=\cdots=\lambda^n x_{\lambda}$ and thus
$\displaystyle \begin{aligned} p\left(A\right)x_{\lambda} &= \left(\sum_{j=0}^{n}a_j A^j\right)x_{\lambda}\\ &= \sum_{j=0}^{n}\left(a_jA^jx_{\lambda})\\ &=\sum_{j=0}^{n}\left(a_j \lambda^j x_{\lambda}\\ &= \left(\sum_{j=0}^{n}a_j\lambda^j\right)x_{\lambda} \\ &= p(\lambda)x_{\lambda}\end{aligned}$
And we used it by noticing that if $\displaystyle p(z)=z+i$ then $\displaystyle U+iI=p\left(U\right)$ so $\displaystyle \lambda$ being an eigenvalue of $\displaystyle U$ implies that $\displaystyle p\left(\lambda\right)=\lambda+i$ is an eigvenvalue of $\displaystyle U+iI$
Depending upon your knowledge this follows immediately since every unitary matrix is normal and thus we may appeal to the spectral theorem for normal matrices to conclude that $\displaystyle U=V\text{diag}(\lambda_1,\cdots,\lambda_n)V^{*}\qu ad (1)$ where $\displaystyle V$ is unitary. Thus, noticing that we have proven that $\displaystyle \lambda_1=\cdots=\lambda_n=-i$ we see that $\displaystyle (1)$ implies that $\displaystyle U=V\text{diag}(-i,\cdots,-i)V^{*}=V(-iI)V^{*}= -iVV^{*}=-iI$