Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is well known that a symmetric matrix over field $\Bbb F$ is congruent to a diagonal matrix, i.e., there exists some A s.t. $A^TUA=D$ with $U$ symmetric and $D$ diagonal. If $\Bbb F=\Bbb C$ then we can make $D=I$.

Recently I learned that if $U$ is unitary that we can do one step further by requiring $A$ to be unitary too. A similar result holds for unitary skew matrices. But I fail to figure out a proof myself.

Can anyone provide a proof of this or at least help me to locate some references? Many thanks!

share|improve this question
2  
At least over the complex numbers, where I assume you are working, the result is best understood in terms of inner products and orthonormal bases. (By the way, you probably mean to use ${\bar A}^{T}UA$ in the complex case and use Hermitian matrices in place of symmetric ones). A complex square matrix $A$ is called normal if $A$ commutes with ${\bar A}^{T}.$ This includes unitary and Hermitian and skew-Hermitian. The relevant result is that a normal matrix has an orthonormal basis of eigenvectors. The change of basis matrix from the standard basis is then unitary –  Geoff Robinson Jul 20 '12 at 12:02
4  
any normal matrix can be unitarily diagonalized. –  chaohuang Jul 20 '12 at 13:35

1 Answer 1

Let $U$ be complex symmetric unitary, i.e., $U^T = U$ and $U^* U = I$. From Takagi factorization we know that $U = V D V^T$, i.e.,

$$D = V^{-1} U V^{-T},$$

for some unitary $V$ and real diagonal $D$. Here, we do not have $D = I$, but we do have the unitarity of $V$.

Using the unitarity of $U$ and $V$, we see that

\begin{align*} D^* D &= (V^{-T})^* U^* V^{-*} V^{-1} U V^{-T} = \overline{V^{-1}} U^* (V V^*)^{-1} U V^{-T} = \overline{V^{-1}} U^* U V^{-T} \\ &= \overline{V^{-1}} V^{-T} = \overline{(V^* V)^{-1}} = {\rm I}. \end{align*}

In other words, $D$ is real diagonal unitary, which means that it has diagonal elements $-1$ and $1$. Define a complex diagonal matrix $X = \operatorname{diag}(x_1, x_2, \dots, x_n)$:

$$x_k = \begin{cases} 1, & D_{kk} = 1, \\ {\rm i}, & D_{kk} = -1. \end{cases}$$

Obviously, $D = X X^T$. Defining $A = VX$ we get

$$U = V D V^T = (VX) (VX)^T = A {\rm I} A^T,$$

which is the required form. Notice that $X$ is unitary, so $A$ is unitary as well.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.