Ytukyg

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 $B = VLambda V^T$. Let $U = X - iY$.
Explain why $U$ is unitary and show that $USigma U ^T=A$.

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, $lambdane0$. 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 $pmlambda$.