Recently I studied Cauchy Schwarz inequality within inner product space.I found a proof in Hoffmann Kunze linear algebra book which constructs a vector $\gamma=\beta-\frac{<\beta,\alpha>}{||\alpha||^2} \alpha$ all of a sudden and then uses the fact that $||\gamma||^2 \geq 0$.But I cannot see why they have constructed like this,what is the intuition behind such construction and how do I interpret the proof so that I can do it intuitively or visually so that I can work it out on my own and need not memorise the process of the proof.

I have seen the link of mathstackexchange Intuition behind how the Cauchy-Schwarz inequality's proof was obtained but still I am not satisfied.Can someone provide me with a different view?

Note that I have no problem in visualization of the result/statement of CS inequality but I am looking for visualization behind the process of the proof.

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    $\begingroup$ An inner product basically allows you to use the tools familiar from geometry in $\mathbb{R}^n$ in a more general context. Going with this fact then the second term in the definition of $\gamma$ is how you define the projection of $\beta$ onto $\alpha$.The reason for looking at this is that now the vectors $\beta $, the above projection, and their difference form a "right triangle". Now Cauchy-Schwarz is equivalent to the triangle inequality (expand both sides in the triangle inequality), but by Pythagoras Theorem we can explicitly compute the different norms involved. Hope this helps. $\endgroup$ – Jose27 Jan 16 at 17:10
  • $\begingroup$ This is the thread you should be looking at: math.stackexchange.com/questions/436559/… I think the answer I wrote is quite visual: math.stackexchange.com/a/436672/40119 $\endgroup$ – littleO Jan 17 at 1:43

The Wikipedia article inner product space gives the properties that an inner product has to satisfy. If you already know and understand these properties then you can skip to the summary at the end of this answer.

The article doesn't mention why we have inner products. Suppose we have a finite dimensional real vector space such as $\,\mathbb{R}^n\,$. In this form, each vector is a tuple $\,(a_1,a_2,\dots,a_n)\,$ real numbers. However, in general, although a vector space $\,V\,$ is assumed to have a finite basis, how to find the coefficients $\,a_k\,$ of any vector? The solution is to use the dual space $\,V^*\,$ with its corresponding dual basis. The Wikipedia article states

The pairing of a functional $\,\varphi\,$ in the dual space $\,V^∗\,$ and an element $\,x\,$ of $\,V\,$ is sometimes denoted by a bracket: $\,\varphi(x) =[x,\varphi]\,$ or $\,\varphi(x)=\langle\varphi,x\rangle.\,$ This pairing defines a nondegenerate bilinear mapping $\,\langle,\cdot,\cdot\rangle: V^* \times V \to F\,$ called the natural pairing.

Later in the article it states:

If $\,V\,$ is finite-dimensional, then $\,V\,$ is isomorphic to $\,V^*\,$. But there is in general no natural isomorphism between these two spaces. Any bilinear form $\,\langle\cdot,\cdot\rangle\,$ on $\,V\,$ gives a mapping of $\,V\,$ into its dual space via $\,v\mapsto\langle v,\cdot\rangle\,$ where the right hand siade is defined as the functional on $\,V\,$ taking each $\,w\in V\,$ to $\,\langle v,w\rangle\,$.

Given a nondegenerate inner product we choose an rthonormal basis $\,\{e_i\}\,$ and we can now write any vector $\,v\in V\,$ as $\,v = a_1\,e_1+a_2\,e_2+\cdots+a_n\,e_n\,$. where the coefficients $\,a_i\,$ are the equivalents to the elements of $n$-tuples in $\,\mathbb{R}\,$.

With this background, consider any vector $\,v\in V\,$. Without loss of generality, choose any unit vector $\,w\,$. We can write $$v = \langle v,w\rangle\, w + t \tag{1} $$ where $\,t\,$ consists of the sum of the remaining components of $\,v\,$. We find that $\;t\!\perp\!w\;$ because from equation $(1)$ we get that

$$\langle v,w\rangle = \langle v,w\rangle \langle w,w \rangle+ \langle t,w\rangle\tag{2}$$

and since $\,\langle w,w\rangle=1\,$ by assumption on $\,w\,$, implies that $\,\langle t,w\rangle=0\,$. Also from equation $(1)$ we get that $\,\langle v,t\rangle = \langle v,w\rangle \langle w,t\rangle + \langle t,t\rangle\,$. But we showed that $\,\langle t,w\rangle=0\,$ which implies that $\,\langle v,t\rangle = \langle t,t\rangle\,$ and this implies $\,\langle v-t,w\rangle=0\,$ or $\;(v-t)\!\perp\!w\;$.

This proves that $\,0,v\,$ and $\,t\,$ form the vertices of a right triangle with the right angle at $\,t\,$. The Pythagorean theorem in this case states that

$$ ||v-t||^2 + ||0-t||^2 = ||0-v||^2. \tag{3}$$

By equation $(1)$ we get $\,v-t = \langle v,w\rangle\, w\,$ which by assumption on $\,w\,$ implies $\,||v-t||^2 = \langle v,w\rangle^2\,$ and from equation $(3)$ we get that $\, \langle v,w\rangle^2\le ||v||^2\,$. Without the assumption that $\,\langle w,w\rangle=1\,$ we get the general inequality

$$ \langle v,w\rangle^2 \le ||v||^2\,||w||^2. \tag{4}$$

Summary: The visual interpretation of the Cauchy Schwarz inequality and its standard proof is that the vector $\,v\,$ is decomposed into a vector $\,t\,$ perpendicular to $\,w\,$ and the vector $\,v-t\,$. These three vectors form a right triangle. The Pythagorean theorem implies the Cauchy Schwarz inequality since one consequence of the theorem is that in any right triangle the length of any leg is bounded by the length of the hypoteneuse.

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  • $\begingroup$ I understood the summary but I do not know dual spaces,anyways I found what I wanted to know. $\endgroup$ – Kishalay Sarkar Jan 17 at 3:27

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