Let $X = A + iB \in \Bbb C^{n \times n}$ be nonsingular, where $A$ and $B$ are real $n \times n$ matrices. Show that $X^{-1}$ can be expressed in terms of the inverse of the real matrix of order $2n$ $$ Y = \begin{pmatrix} A & -B \\ B & A \\ \end{pmatrix}. $$ Compare the economics of real versus complex matrix inversion.

THANKS in advance


1 Answer 1


In the following, $A,B,x,y,w,z$ are all real quantities of the appropriate dimension.

It is clear that $(A+iB)(x+iy) = w+iz$ $\iff$ $(Ax-By) + i (Ay+Bx) = w + iz$ $\iff$ $\begin{bmatrix}Ax-By \\Ay+bx \end{bmatrix} = \begin{bmatrix} w \\z \end{bmatrix}$ $\iff$ $\begin{bmatrix}A& -B \\ B & A \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}w\\ z \end{bmatrix}$

The 'economics' depend on the matrix and the means of inversion. LU factorization is a reasonable benchmark, and takes $\frac{4}{3}n^3$ operations (in the appropriate field) to invert. Consequently, it would take $\frac{4}{3}n^3$ 'complex flops' vs. $\frac{4}{3}(2n)^3= 8\frac{4}{3}n^3$ real flops. Since a 'complex flop' costs around 2-6 real flops, it is clear that with this rough measure, you are better off with complex inversion.

  • $\begingroup$ Could you explain me, Please, how you found the number of flops, analytically. Thanks in advance $\endgroup$
    – user50017
    Nov 23, 2012 at 2:04
  • $\begingroup$ First, I had a mistake in the above, my flop count arithmetic was off by 2x. LU factorization with partial pivoting costs around $\frac{1}{3}n^3$ flops to compute the factorization, and then $n n^2$ flops to compute the inverse (by solving with right hand sides $e_i$). Look at Golub & Van Loan, Matrix Computations" Algorithm 4.4-2 and homework problem P4.4-5. $\endgroup$
    – copper.hat
    Nov 23, 2012 at 18:33

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