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Deriving double angle trigonometric formulas using transformations. A really nice elementary proof.

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.

Definition 1 A vector space ${V}$ over the number field ${\mathbb{C}}$ or ${\mathbb{R}}$ is called an inner product space if it is equipped with an inner product ${\left\langle \cdot,\cdot\right\rangle }$ satisfying for all ${u,v,w\in V}$ and scalar ${c}$,

1. ${\left\langle u,u\right\rangle \geq0}$, ${\left\langle u,u\right\rangle =0}$ if and only if ${u=0}$,
2. ${\left\langle u+v,w\right\rangle =\left\langle u,v\right\rangle +\left\langle v,w\right\rangle }$,
3. ${\left\langle cu,v\right\rangle =c\left\langle u,v\right\rangle }$, and
4. ${\left\langle u,v\right\rangle =\overline{\left\langle v,u\right\rangle }}$.

${\mathbb{C}^{n}}$ is an inner product space over ${\mathbb{C}}$ with the inner product

$\displaystyle \left\langle x,y\right\rangle =y^{*}x=\overline{y_{1}}x_{1}+\cdots+\overline{y_{n}}x_{n}.$

An inner product space over ${\mathbb{R}}$ is usually called a Euclidean space.

The following properties of an inner product can be deduced from the four axioms in Definition 1:

1. ${\left\langle x,cy\right\rangle =\bar{c}\left\langle x,y\right\rangle }$,
2. ${\left\langle x,y+z\right\rangle =\left\langle x,y\right\rangle +\left\langle x,z\right\rangle }$,
3. ${\left\langle ax+by,cw+dz\right\rangle =a\bar{c}\left\langle x,w\right\rangle +b\bar{c}\left\langle y,w\right\rangle +a\bar{d}\left\langle x,z\right\rangle +b\bar{d}\left\langle y,z\right\rangle }$,
4. ${\left\langle x,y\right\rangle =0}$ for all ${y\in V}$ if and only if ${x=0}$, and
5. ${\left\langle x,\left\langle x,y\right\rangle y\right\rangle =\left|\left\langle x,y\right\rangle \right|^{2}}$.

An important property shared by all inner products is the Cauchy-Schwarz inequality and, for an inner product space, one of the most useful inequalities in mathematics.

Theorem 1 (Cauchy-Schwarz Inequality) Let ${V}$ be an inner product space. Then for all vectors ${x}$ and ${y}$ in ${V}$ over the field ${\mathbb{C}}$ or ${\mathbb{R}}$,

$\displaystyle \left|\left\langle x,y\right\rangle \right|^{2}\leq\left\langle x,x\right\rangle \left\langle y,y\right\rangle .$

Equality holds if and only if ${x}$ and ${y}$ are linearly dependent.

The proof of this can be done in a number of different ways. The most common proof is to consider the quadratic function in ${t}$

$\displaystyle \left\langle x+ty,x+ty\right\rangle \ge0$

and derive the inequality from the non-positive discriminant. We will first present this proof.

Proof: Let ${x,y\in V}$ be given. If ${y=0}$, the assertion is trivial, so we may assume that ${y\neq0}$. Let ${t\in\mathbb{R}}$ and consider

$\displaystyle \begin{array}{rcl} p(t) & \equiv & \left\langle x+ty,x+ty\right\rangle \\ & = & \left\langle x,x\right\rangle +t\left\langle y,x\right\rangle +t\left\langle x,y\right\rangle +t^{2}\left\langle y,y\right\rangle \\ & = & \left\langle x,x\right\rangle +2t\, Re\left\langle x,y\right\rangle +t^{2}\left\langle y,y\right\rangle , \end{array}$

which is a real quadratic polynomial with real coefficients. Because of axiom (1.), we know that ${p(t)\geq0}$ for all real ${t}$, and hence ${p(t)}$ can have no real simple roots. The discriminant of ${p(t)}$ must therefore be non-positive

$\displaystyle (2\, Re\left\langle x,y\right\rangle )^{2}-4\left\langle y,y\right\rangle \left\langle x,x\right\rangle \leq0$

and hence

$\displaystyle (Re\left\langle x,y\right\rangle )^{2}\leq\left\langle x,x\right\rangle \left\langle y,y\right\rangle . \ \ \ \ \ (1)$

Since this inequality must hold for any pair of vectors, it must hold if ${y}$ is replaced by ${\left\langle x,y\right\rangle y}$, so we also have the inequality

$\displaystyle (Re\left\langle x,\left\langle x,y\right\rangle y\right\rangle )^{2}\leq\left\langle x,x\right\rangle \left\langle y,y\right\rangle \left|\left\langle x,y\right\rangle \right|^{2}$

But ${Re\left\langle x,\left\langle x,y\right\rangle y\right\rangle =Re\overline{\left\langle x,y\right\rangle }\left\langle x,y\right\rangle =Re\left|\left\langle x,y\right\rangle \right|^{2}=\left|\left\langle x,y\right\rangle \right|^{2}}$, so

$\displaystyle \left|\left\langle x,y\right\rangle \right|^{4}\leq\left\langle x,x\right\rangle \left\langle y,y\right\rangle \left|\left\langle x,y\right\rangle \right|.^{2} \ \ \ \ \ (2)$

If ${\left\langle x,y\right\rangle =0}$, then the statement of the theorem is trivial; if not, then we may divide equation (2) by the quantity ${\left|\left\langle x,y\right\rangle \right|^{2}}$ to obtain the desired inequality

$\displaystyle \left|\left\langle x,y\right\rangle \right|^{2}\leq\left\langle x,x\right\rangle \left\langle y,y\right\rangle .$

Because of axiom (1.), ${p(t)}$ can have a real (double) root only if ${x+ty=0}$ for some ${t}$. Thus, equality can occur in the discriminant condition in equation (1) if and only if ${x}$ and ${y}$ are linearly dependent. $\Box$

We will now present a matrix proof, focusing on the complex vector space, which is perhaps the simplest proof of the Cauchy-Schwarz inequality.

Proof: For any vectors ${x,y\in\mathbb{C}^{n}}$ we noticed that,

$\displaystyle \left[x,y\right]^{*}\left[x,y\right]=\left[\begin{array}{cc} x^{*}x & x^{*}y\\ y^{*}x & y^{*}y \end{array}\right]\geq0.$

By taking the determinant for the ${2\times2}$ matrix,

$\displaystyle \begin{array}{rcl} \left|\begin{array}{cc} x^{*}x & x^{*}y\\ y^{*}x & y^{*}y \end{array}\right| & = & (x^{*}x)(y^{*}x)-(x^{*}y)(y^{*}x)\\ & = & \left\langle x,x\right\rangle \left\langle x,y\right\rangle -\left\langle y,x\right\rangle \left\langle x,y\right\rangle \\ & = & \left\langle x,x\right\rangle \left\langle x,y\right\rangle -\overline{\left\langle x,y\right\rangle }\left\langle x,y\right\rangle \\ & = & \left\langle x,x\right\rangle \left\langle x,y\right\rangle -\left|\left\langle x,y\right\rangle \right|^{2} \end{array}$

the inequality follows at once,

$\displaystyle \left|\left\langle x,y\right\rangle \right|^{2}\leq\left\langle x,x\right\rangle \left\langle y,y\right\rangle .$

Equality occurs if and only if the ${n\times2}$ matrix ${\left[x,y\right]}$ has rank 1; that is, ${x}$ and ${y}$ are linearly dependent. $\Box$

References

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

Do you know these matrices described by Alan Rendall? If so, please point out a source where he may find more information about them. I am interested in knowing too!

I have come across a class of matrices with some interesting properties. I feel that they must be known but I have not been able to find anything written about them. This is probably just because I do not know the right place to look. I will describe these matrices here and I hope that somebody will be able to point out a source where I can find more information about them. Consider an $latex n\times n$ matrix $latex A$ with elements $latex a_{ij}$ having the following properties. The elements with $latex i=j$ (call them $latex b_i$) are negative. The elements with $latex j=i+1\ {\rm mod}\ n$ (call them $latex c_i$) are positive. All other elements are zero. The determinant of a matrix of this type is $latex \prod_i b_i+(-1)^{n+1}\prod_i c_i$. Notice that the two terms in this sum always have opposite signs. A property of these matrices which I…

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