vectors | Guzman's Mathematics Weblog

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 ${c}$ ,

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

${\mathbb{C}^{n}}$ 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:

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

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…

View original post 305 more words

Tag Archive

The Cauchy–Schwarz Inequality 1

Do you know these matrices?

Deciding linear independence

Categories

Archives

Recent Posts

Top Posts

Mathematical Comics

Mathematical websites

Blogroll

Tags

The Polymath Blog

PolyMath Wiki Recent Changes

Follow Guzman's Mathematics Weblog via Email

Meta