Help with Autonne-Takagi factorization of a complex symmetric matrix.

Question

Let $A=A_1i+A_2$ with $A$ non singular. Now let

$$B =\begin{bmatrix} A_1 & A_2\\ A_2 & -A_1 \end{bmatrix}$$

With $A_1$, $A_2$ and $B$ symmetric. Is it true that:

1) $B$ is non singular

2) $B \begin{bmatrix} x \\ -y \end{bmatrix}=\lambda \begin{bmatrix} x \\ -y \end{bmatrix}$ if and only if $B \begin{bmatrix} x \\ -y \end{bmatrix}=-\lambda \begin{bmatrix} x \\ -y \end{bmatrix}$ so the eigenvalues of $B$ appear in $+-$ pairs.

3) Let $\begin{bmatrix} x_1 \\ -y_1 \end{bmatrix},\dots, \begin{bmatrix} x_n \\ -y_n \end{bmatrix}$ be the orthonormal eigenvectors of $B$ associated with its positive eigenvalues $\lambda_1,\dots,\lambda_n$. Let $X=\begin{bmatrix} x_1 & \dots & x_n \end{bmatrix}$, $Y=\begin{bmatrix} y_1 & \dots & y_n \end{bmatrix}$, $\Sigma=diag(\lambda_1,\dots,\lambda_n)$, $V =\begin{bmatrix} X & Y\\ -Y & X \end{bmatrix}$ and $\Lambda = \Sigma \oplus (-\Sigma)$. Then $V$ is real orthogonal and $B = V\Lambda V^T$. Let $U = X - iY$. Explain why $U$ is unitary and show that $U\Sigma U ^T=A$.

Answer 1

I am not going to do the homework for you. I have two remarks, though.

1. Part (1) is actually true for any real matrices $A_1$ and $A_2$. It has nothing to do with matrix symmetry. Note that $$ \pmatrix{I&-iI\\ 0&I} \pmatrix{A_1&A_2\\ A_2&-A_1} \pmatrix{I&0\\ iI&I} = \pmatrix{0&A_2+iA_1\\ A_2-iA_1&-A_1} $$ and $$ \det(A_2+iA_1)\,\det(A_2-iA_1)=\det(A_2+iA_1)\,\overline{\det(A_2+iA_1)}= |\det(A_2+iA_1)|^2. $$
2. Part (2) as it stands is false. Since $B$ is nonsingular, $\lambda\ne0$. Therefore $\lambda$ and $-\lambda$ are different eigenvalues and they cannot share the same eigenvector $\pmatrix{x\\ -y}$. The correct statement should be this: if $\lambda$ and $\pmatrix{x\\ -y}$ form an eigenpair of $B$, then $-\lambda$ and $\pmatrix{y\\ x}$ is also an eigenpair of $B$. Therefore, eigenvalues of $B$ occur in pairs of the form $\pm\lambda$.