Shilov's Linear Algebra: canonical form of symmetric Hermitian quadratic form

Question

9.16. a. Given a symmetric Hermitian quadratic form $A(x, x)$ in an $n$-dimensional complex space $\mathbf C_n$, there exists a basis in $\mathbf C_n$ in which $A(x, x)$ can be written in the canonical form $$A(x, x) = \sum_{k = 1}^n \lambda_k\,\eta_k\,\bar\eta_k = \sum_{k = 1}^n \lambda_k\,|\eta_k|^2$$ with real coefficients $\lambda_1, \lambda_2, \cdots, \lambda_n$.

In the proof of this proposition, the book says, on page 251, that the following transformation $$\xi_1 = \xi_1' + \xi_2',\\ \xi_2 = \xi_1' + i\,\xi_2'$$ carries the sum $a_{12}\xi_1 \bar{\xi_2} + \bar a_{12} \bar{\xi_1} \xi_2 \;(a_{12} \neq 0)$ into the expression $$(a_{12} + \bar a_{12}) \xi_1' \bar{\xi_1'} - i(a_{12} - \bar a_{12}) \xi_2' \bar{\xi_2'} + \cdots.$$ Note the ellipsis at the end of the above expression. The complete expression is $$(a_{12} + \bar a_{12}) \xi_1' \bar{\xi_1'} - i(a_{12} - \bar a_{12}) \xi_2' \bar{\xi_2'} + (a_{12} + i\bar a_{12}) \bar{\xi_1'} \xi_2' + (\bar a_{12} - i a_{12}) \xi_1' \bar{\xi_2'}.$$

But, in my understanding, we should find a transformation that could carry the sum into a canonical form like $\lambda_1 \xi'_1 \bar{\xi_1'} +\lambda_2 \xi_2' \bar{\xi_2'}$, where $\lambda_1, \lambda_2$ are real.

And, if I'm right, how can we construct such a transformation? Thank you!

Answer 1

The construction is already done on p.250, the page preceding p.251. Given $A(x,x)$, there are only three possibilities:

1. Some $a_{ii}\ne0$.
2. $a_{11}=a_{22}=\cdots=a_{mm}=0$ and $a_{ij}=0$ for all $i\ne j$.
3. $a_{11}=a_{22}=\cdots=a_{mm}=0$ and $a_{ij}\ne0$ for some $i\ne j$.

Shilov has dealt with case 1 on p.250, and nothing needs to be done in case 2 because $A(x,x)=0$ in that case.

The purpose of the transformation on p.251 isn't to express $A(x,x)$ as a sum of squares, but to reduce case 3 to case 1. By relabelling the indices if necessary, we may assume that $a_{12}\ne0$. It follows that at least one of $a_{12}+\bar{a}_{12}$ or $i(a_{12}+\bar{a}_{12})$ is nonzero. Therefore, by the change of variables mentioned in your question, we may write $$ A(x,x)=\underbrace{(a_{12}+\bar{a}_{12})}_{a_{11}'}\,\xi_1'\bar\xi_1' \, \underbrace{- i(a_{12}-\bar{a}_{12})}_{a_{22}'}\,\xi_2'\bar\xi_2' + \cdots $$ and now at least one of $a_{11}'$ or $a_{22}'$ is nonzero among the new coefficients $a_{ij}'$.

Answer 2

By the following coordinate transformation $$\begin{align} \xi_1 &= {1 \over a_{12}} (\xi_1' + \xi_2'),\\ \xi_2 &= \xi_1' - \xi_2', \end{align}$$ $a_{12} \xi_1 \bar{\xi_2} + \bar a_{12} \bar{\xi_1} \xi_2 \;(a_{12} \neq 0)$ can be transformed into a canonical form directly, as $$\begin{align} a_{12} \xi_1 \bar{\xi_2} + \bar a_{12} \bar{\xi_1} \xi_2 &= a_{12} {1 \over a_{12}} (\xi_1' + \xi_2') \overline{(\xi_1' - \xi_2')} + \bar a_{12} \overline{{1 \over a_{12}} (\xi_1' + \xi_2')} (\xi_1' - \xi_2') \\ &= a_{12} {1 \over a_{12}} (\xi_1' + \xi_2') (\bar{\xi_1'} - \bar{\xi_2'}) + \bar a_{12} {1 \over \bar a_{12}} (\bar{\xi_1'} + \bar{\xi_2'}) (\xi_1' - \xi_2') \\ &= (\xi_1' + \xi_2') (\bar{\xi_1'} - \bar{\xi_2'}) + (\bar{\xi_1'} + \bar{\xi_2'}) (\xi_1' - \xi_2') \\ &= 2 \xi_1' \bar{\xi_1'} - 2 \xi_2' \bar{\xi_2'} \end{align}$$

The transformation is guaranteed to be nonsingular. But, since it depends on a specific coefficient ($a_{12}$ here), I'm not sure if it lends itself well to numerical computation.