# Complex matrix decomposition

If I have a block matrix of complex matrices

$$\begin{bmatrix} P &Q\\ Q^T & P \end{bmatrix}$$ while Q being skew symmetric, the decomposition is $$\begin{bmatrix} I & -iI\\ . & I \end{bmatrix} \begin{bmatrix} P-iQ & .\\ -Q & I \end{bmatrix} \begin{bmatrix} I & iI\\ . & P+iQ \end{bmatrix}$$ These are results from (R.A Wooding 1956). Can anyone please explain what kind of complex decomposition is this or please provide any reference explaing this above decomposition. I have looked into Block LU decomposition but this does not explain how the above stated decomposition is working.

Appreciate your suggestions !

• Hi. What's the dot supposed to mean? – mathreadler Jul 19 '16 at 7:18
• I think thats just dont care symbol – Vendetta Jul 19 '16 at 7:34
• Ah, ok I assumed it was 0. I should add that to the hint. – mathreadler Jul 19 '16 at 7:37

Hint: the matrix valued function of a scalar $a$ : $$m(a) = \left[\begin{array}{cc}1&0\\a&1\end{array}\right]$$ has the property that $$m(a)\cdot m(b) = m(a+b)$$ and this property carries over to block form with $\bf I$ corresponding to $1$ and $\bf 0$ corresponding to $0$: $$m({\bf A}) = \left[\begin{array}{cc}\bf I&\bf 0\\{\bf A}&{\bf I}\end{array}\right]$$ so that: $$m({\bf A})m({\bf B}) = m({\bf A+B})$$
• Maybe the $i$ are used to make it cancel somehow thanks to $i^2=-1$, I don't know. Why don't you try evaluate the multiplication. – mathreadler Jul 19 '16 at 8:30
• Its 'The Multivariate distribution of complex normal variables by R.A Wooding (1956)'. What I think is that Schur complement might be formulated as $$\begin{pmatrix} I & BD^{-1}\\ 0& I \end{pmatrix} \begin{pmatrix} A-BD^{-1}C & 0\\ -B & I \end{pmatrix} \begin{pmatrix} I & D^{-1}C\\ 0 & A+BD^{-1}C \end{pmatrix}$$ Again like you said twice multiplication by $i$ should be cause of -Q but still that -B in second matrix should not exist it should be zero. There is not reference explaining this too, perhaps this is some trivial case. – Vendetta Jul 19 '16 at 10:59