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

Double Angle Trigonometric Formulas.

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.

PhD Comics
http://www.phdcomics.com/comics/archive.php?comicid=27

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!

Hydrobates

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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