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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

 Zhang, Fuzhen. Matrix theory: basic results and techniques. Springer Science & Business Media, 2011.     Euler Formula

The Euler formula, sometimes also called the Euler identity, states $\displaystyle e^{ix}=cosx+isinx,$

where ${i}$ is the imaginary unit. Note that Euler’s polyhedral formula is sometimes also called the Euler formula, as is the Euler curvature formula.

The special case of the formula with ${x=\pi}$ gives the beautiful identity $\displaystyle e^{i\pi}=-1,$

an equation connecting the fundamental numbers ${i,\pi,e,1,}$ and ${0}$, the fundamental operations ${+}$, ${\times}$, and exponentiation, the most important relation ${=}$, and nothing else. Gauss is reported to have commented that if this formula was not immediately obvious, the reader would never be a first-class mathematician. 