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.

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