In the previous post, two proofs were given for the Cauchy-Schwarz inequality. We will now consider another proof.

Definition 1 Let ${M}$ be an ${n\times n}$ matrix written as a ${2\times2}$ block matrix

$\displaystyle M=\left[\begin{array}{cc} A & B\\ C & D \end{array}\right],$

where ${A}$ is a ${p\times p}$ matrix, ${D}$ is a ${q\times q}$ matrix, ${B}$ is a ${p\times q}$, and ${C}$ is a ${q\times p}$, so ${n=p+q}$. Assuming ${A}$ is nonsingular, then

$\displaystyle M/A=D-CA^{-1}B$

is called the Schur complement of ${A}$ in ${M}$; or the Schur complement of ${M}$ relative to ${A}$.

The Schur complement probably goes back to Carl Friedrich Gauss (1777-1855) (for Gaussian elimination). To solve the linear system

$\displaystyle \left[\begin{array}{cc} A & B\\ C & D \end{array}\right]\left[\begin{array}{c} x\\ y \end{array}\right]=\left[\begin{array}{c} c\\ d \end{array}\right],$

that is

$\displaystyle \begin{array}{rcl} Ax+By & = & c\\ Cx+Dy & = & d, \end{array}$

by mimicking Gaussian elimination, that is, if ${A}$ is square and nonsingular, then by eliminating ${x}$, by multiplying the first equation by ${CA^{-1}}$ and subtracting the second equation, we get

$\displaystyle (D-CA^{-1}B)y=d-CA^{-1}c.$

Note, the matrix ${D-CA^{-1}B}$ is the Schur complement of ${A}$ in ${M}$, and if it is square and nonsingular, then we can obtain the solution to our system.

The Schur complement comes up in Issai Schur’s (1875-1941) seminal lemma published in 1917, in which the Schur determinate formula was introduced. By considering elementary operations of partitioned matrices, let

$\displaystyle M=\left[\begin{array}{cc} A & B\\ C & D \end{array}\right],$

where ${A}$ is square and nonsingular. We can change ${M}$ so that the lower-left and upper-right submatrices become ${0}$. More precisely, we can make the lower-left and upper-right submatrices ${0}$ by subtracting the first row multiplied by ${CA^{-1}}$ from the second row, and by subtracting the first column multiplied by ${A^{-1}B}$ from the second column. In symbols,

$\displaystyle \left[\begin{array}{cc} A & B\\ C & D \end{array}\right]\rightarrow\left[\begin{array}{cc} A & B\\ 0 & D-CA^{-1}B \end{array}\right]\rightarrow\left[\begin{array}{cc} A & 0\\ 0 & D-CA^{-1}B \end{array}\right],$

and in equation form,

$\displaystyle \left[\begin{array}{cc} I & 0\\ -CA^{-1} & I \end{array}\right]\left[\begin{array}{cc} A & B\\ C & D \end{array}\right]\left[\begin{array}{cc} I & -A^{-1}B\\ 0 & I \end{array}\right]=\left[\begin{array}{cc} A & 0\\ 0 & D-CA^{-1}B \end{array}\right].$

Note that we have obtain the following factorization of ${M}$:

$\displaystyle \left[\begin{array}{cc} A & B\\ C & D \end{array}\right]=\left[\begin{array}{cc} I & 0\\ CA^{-1} & I \end{array}\right]\left[\begin{array}{cc} A & 0\\ 0 & D-CA^{-1}B \end{array}\right]\left[\begin{array}{cc} I & A^{-1}B\\ 0 & I \end{array}\right].$

By taking the determinants

$\displaystyle \left|\begin{array}{cc} A & B\\ C & D \end{array}\right|=\left|\begin{array}{cc} I & 0\\ CA^{-1} & I \end{array}\right|\left|\begin{array}{cc} A & 0\\ 0 & D-CA^{-1}B \end{array}\right|\left|\begin{array}{cc} I & A^{-1}B\\ 0 & I \end{array}\right|.$

we obtain the Schur’s determinant formula for ${2\times2}$ block matrices,

$\displaystyle \det M=\det A\;\det(M/A).$

Mathematician Emilie Virginia Haynsworth (1916-1985) introduced a name and a notation for the Schur complement of a square nonsingular (or invertible) submatrix in a partitioned (two-way block) matrix. The term Schur complement first appeared in Emily’s 1968 paper On the Schur Complement in Basel Mathematical Notes, then in Linear Algebra and its Applications Vol. 1 (1968), AMS Proceedings (1969), and in Linear Algebra and its Applications Vol. 3 (1970).

We will now present a ${2\times2}$ block matrix proof, focusing on ${m\times n}$ complex matrices.

Proof: Let ${A,B\in\mathbb{C}^{m\times n}}$. Then

$\displaystyle M=\left[\begin{array}{cc} I & A^{*}\\ B & I \end{array}\right]\left[\begin{array}{cc} I & B^{*}\\ A & I \end{array}\right]=\left[\begin{array}{cc} I+A^{*}A & A^{*}+B^{*}\\ A+B & I+BB^{*} \end{array}\right]\geq0.$

By taking the Schur complement of ${I+A^{*}A}$, we arrive at

$\displaystyle S=I+BB^{*}-(A+B)(I+A^{*}A)^{-1}(A^{*}+B{}^{*})\geq0$

and hence

$\displaystyle (I+A^{*}A)(I+BB^{*})\geq(A+B)(A+B)^{*}.$

which ensures, when ${A}$ and ${B}$ are square, that

$\displaystyle \left|\det(A+B)\right|^{2}\leq\det(I+A^{*}A)\;\det(I+BB^{*}).$

Equality occurs if and only if rank ${M=}$ rank ${A}$; that is, by the Guttman rank additivity formula, rank ${M=}$ rank ${A+}$rank ${S}$ if and only if ${S=0}$. When ${A}$ is nonsingular, ${M}$ is nonsingular if and only if ${S}$ is nonsingular.$\Box$

References

[1] Zhang, Fuzhen. Matrix theory: basic results and techniques. Springer Science & Business Media, 2011.