Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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.

share|cite|improve this answer
Could you explain me, Please, how you found the number of flops, analytically. Thanks in advance – user50017 Nov 23 '12 at 2:04
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. – copper.hat Nov 23 '12 at 18:33

Your Answer


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.