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.

Let $A,B \in \mathbb{R}^{n,n}$.

Now $C = \begin{pmatrix} A & iB \\ -iB & A \end{pmatrix}$ and $D = \begin{pmatrix} A & B \\ -B & A \end{pmatrix}$.

Show that $\det(C) \in \mathbb{R}$ and $\det(D) \ge 0$.

I tried to transform $C$ so I could use $\det \begin{pmatrix} E & F \\ 0 & G \end{pmatrix} = \det(E)\det(G)$, but I didn't manage to. I have no clue how to show $\det(D) \ge 0$.

I'd rather have hints than fully-fledged solutions. Thanks.

share|improve this question
1  
If you write "x det y" in TeX, you see $x det y$, but if you write "x\det y", you see $x\det y$. Two differences, at least: "$\det$" is not italicized, and proper spacing before and after $\det$ appears. –  Michael Hardy Dec 13 '12 at 23:34
1  
I appreciate all the help on part 2, but your solutions were too advanced for me. But I managed to solve it on my own. First I scaled the last $n$ rows by $i$ and the last $n$ columns by $i$. Then I used column/row-operations and the multilinearity of the derminant. I ended up with a matrix where I could use the above identity, which resulted in $$ \det (-B+iA) \det (-B-iA) = \det (-B+iA) \overline{\det (-B+iA)} = \left| \det (-B+iA) \right|^2$$ And this is obviously positive. –  user49797 Dec 14 '12 at 11:52

3 Answers 3

up vote 3 down vote accepted

Here's a hint for (1): think about scaling the first $n$ rows by $-i$.

Still thinking about (2). I haven't worked this out, but I think considering the signed permutations way of calculating the determinant, and pairing $\sigma$ with $\sigma^{-1}$ might be helpful.

edit: Yeah, okay, just got back, have something kinda similar to levap. A matrix of the form $D$ is exactly a complex linear one, under the identification $\mathbb{R}^{2n} \rightarrow \mathbb{C}^n$, where $\mathbb{R}^{2n}$ has ordered basis $x_1, \ldots, x_n, y_1, \ldots, y_n$ and $\mathbb{C}^n$ $z_i = x_i + i y_i$.

With this, one knows as an operator on $\mathbb{C}^n$ one can write it in Jordan canonical form. If one has a Jordan block: $$\begin{pmatrix} \lambda & - & \ldots \\ & \lambda & \ldots \end{pmatrix}$$with basis $w_i$, setting $w_i = x'_i + iy'_i$, in the ordered basis $x_1', y_1', x_2', y_2' \ldots$, it's easy to check that $D$ has determinant $|\lambda|^r$, where $r$ is the size of the Jordan block.

share|improve this answer
    
I managed (1) now, thanks to your hint. I got no time left for (2) now, I'll have to try that later. Thank you. –  user49797 Dec 13 '12 at 23:10
    
no worries, glad to help. I'll post (a hint) if I figure it out –  uncookedfalcon Dec 13 '12 at 23:11
  1. Consider $\begin{pmatrix}I\\&zI\end{pmatrix}C\begin{pmatrix}I\\&\bar{z}I\end{pmatrix}$ for some complex numbers $z$ of modulus 1.
  2. As invertible matrices are dense in the matrix space and determinant is a continuous function in matrix entries, we may assume that $A$ is invertible. Using the block determinant formula, we get $\det D = \det A \det(A+BA^{-1}B) = (\det A)^2\det(I+A^{-1}BA^{-1}B)$. Now for any real matrix $X$, argue why $\det(I+X^2)$ must be nonnegative.
share|improve this answer

I'll answer the question about $\det(D)$. Define the matrix $$ J_0 = \left( \begin{array}{cc} 0 & -I \\ I & 0 \end{array} \right) \in \mathrm{GL}_{2n}(\mathbb{R})$$ and note that $J_0^2 = -I$ and that the matrix $J_0$ commutes with your matrix $D$. Now consider all matrices (and maps) as matrices over the complex numbers (on $\mathbb{C}^{2n}$). The matrix $J_0$ is diagonalizable over the complex number with eigenvalues $\pm i$. Denote the eigenspaces by $$ V^{\pm} = \mathrm{ker}(J \pm iI) \subset \mathbb{C}^{2n}. $$ Each eigenspace is of complex dimension $n$ and the antilinear conjugation map $S : \mathbb{C}^{2n} \rightarrow \mathbb{C}^{2n}$ given by $$ S(z_1, \ldots, z_{2n}) = (\bar{z}_1, \ldots, \bar{z}_{2n})$$ exchanges between the eigenspaces $V^{+}$ and $V^{-}$. Since $J_0$ and $D$ commute, the spaces $V^{\pm}$ are $n$ dimensional complex invariant spaces of $D$. Considering $D$ as a linear map over $\mathbb{C}^{2n}$, we have $$ \det(D) = \det(D_+) \det(D_{-}) = d_{+} d_{-} $$ where the maps $D_{\pm} : V^{\pm} \rightarrow V^{\pm}$ are just the restrictions of $D$ to the invariant subspaces $V^{\pm}$. Let us show that $\bar{d}_{+} = d_{-}$ and so $\det(D) = d_{+} \bar{d}_{+} = |d_{+}|^2 \geq 0$. Because $D$ is a real matrix, it commutes as a linear map with the complex conjugation $S$. This means we have the diagram $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} V^{+} & \ra{D_{+}} & V^{+} \\ \da{S} & & \da{S} \\ V^{-} & \ra{D_{-}} & V^{-} \\ \end{array} $$ where $S$ is an antilinear isomorphism. If $S$ would be a linear isomorphism, it would imply that $\det(D_{+}) = \det(D_{-})$. Since $S$ is antilinear, it implies that the determinants must be conjugate.


This might look highly unmotivated if you are unfamiliar with the notion of a complex linear structure on a vector space. Under the identification of $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ given by $$ (z_1, ..., z_n) = (x_1 + iy_1, ..., x_n + iy_n) \to (x_1, ..., x_n, y_1, ..., y_n) $$ the complex matrices $A + iB \in \mathrm{GL}_n(\mathbb{C})$ are identified with the matrices of the form $D$ in $\mathrm{GL}_{2n}(\mathbb{R})$. This matrices are complex linear and so commute with "multiplication by $i$", which corresponds under the identification to multiplication by $J_0$.

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.