8

I was reading about the Cauchy–Schwarz inequality from Courant, Hilbert - Methods Of Mathematical Physics Vol 1 and I can not understand what they mean when they said the line that has been highlighted with red in the picture given below

I can not understand why a and b has to be proportional and why is this so crucial for the roots to be imaginary and why we want the roots to be imaginary in the first place.

enter image description here

noir1993
  • 701
  • Try having a look at the very first pages of Steele, The Cauchy-Schwarz Master Class, here. You might find it useful. – Giuseppe Negro Jan 28 '13 at 18:19
  • I am familiar with the proof give in Steele's book. But I am trying to understand this particular proof (i.e., the one form Hilbert and Courant.) – noir1993 Jan 28 '13 at 18:30

3 Answers3

9

Hint: The quadratic equation $ax^2+bx+c=0$ has real roots if and only if $b^2-4ac\ge 0$.

Added: The quadratic polynomial $\sum_1^n (a_ix+b_i)^2$ is a sum of squares. Thus this polynomial is always $\ge 0$.

Recall that a quadratic $ax^2+bx+c$, where $a\gt 0$, is always $\ge 0$ if and only if the discriminant $b^2-4ac$ is $\le 0$. Compute the discriminant of the messy quadratic. The inequality $b^2-4ac\le 0$ turns out to be precisely the C-S Inequality (well, we have to divide by $4$).

As to when we have equality, the quadratic has a real root $k$ if and only if $a_ik+b_i=0$ for all $i$. This is the case iff $b_i=-ka_i$, meaning that the $a_i$ and $b_i$ are proportional.

By the way, things are I think marginally prettier if we look at the polynomial $\sum_1^n (a_ix-b_i)^2$.

André Nicolas
  • 507,029
  • Yes, I know that! I can't figure out why are they avoiding the real roots and from the language that is being employed by the authors, does it not mean that "The roots can not be real unless a and b are proportional" And just before that they have stated a and b must be proportional. Sorry, but I am confused. – noir1993 Jan 28 '13 at 18:22
  • 1
    @AchiralSarkar If you're referring to the statement "where the equality holds if and only if the $a_i$ and the $b_i$ are proportional", the author means that the inequality is strict iff they are not proportional, and the $\leq$ becomes $=$ iff they are proportional – ebsddd Jan 28 '13 at 18:28
  • 1
    The "messy" quadratic in your post is a sum of squares, so can only be $0$ if $a_ix+b_i=0$ for all $i$. That gives a single solution $x$, and $x=b_i/a_i$ for all $i$, so we get a solution iff all the $b_i/a_i$ are equal, i.e. there is a constant $k$ such that $b_i=ka_i$ for all $i$. In all other cases, the discriminant $b^2-4ac\lt 0$. If you plug in what $a$, $b$, and $c$ are into the inequality $b^2-4ac\lt 0$, you will get the C-S Inequality. – André Nicolas Jan 28 '13 at 18:32
  • @valtron : No, I was talking about the roots of the quadratic equation.....for the unknown x can never be real and distinct, but must be imaginary,unless a and b are proportional. – noir1993 Jan 28 '13 at 18:34
  • 2
    As I wrote above, the polynomial $\sum (b_ix+a_i)^2$ is a sum of squares. If $x$ is real, $b_ix+a_i$ is real, so $(b_ix+a_i)^2\ge 0$. Thus the only way the sum can be $0$ is if all the terms are $0$. – André Nicolas Jan 28 '13 at 18:38
  • @Andre Nicolas: I understand it now. If you want you could elaborate your first comment and add it to the answer. – noir1993 Jan 28 '13 at 18:47
  • 1
    @AndréNicolas why is a quadratic always $\geq 0 $ if and only if the discriminant $b^2 - 4ac \leq 0$? This part is what I don't understand. Could you please explain? – Kamil Jul 20 '15 at 20:24
  • 4
    @Kamil: I was being imprecise. Our quadratic has lead coefficient which is a sum of squares, so $a\gt 0$. Thus the curve $y=ax^2+bx+c$ is an upward facing parabola. It is always $\ge 0$ if the equation $ax^2+bx+c=0$ has no real roots, or just one real root. By the quadratic formula, there are no real roots if $b^2-4ac$ is negative, and one real root (a "double" root) if $b^2-4ac=0$. – André Nicolas Jul 20 '15 at 20:43
6

The expression $\displaystyle\sum_{i=1}^n (a_i x + b_i)^2$ cannot be negative since it's a sum of squares of real numbers, and cannot be zero unless every term is zero, in which case, since $a_i x=b_i$, the vectors $\vec{a}$ and $\vec{b}$ are proportional, and $x$ is the constant of proportionality.

Therefore the quadratic equation can have a real solution for $x$ only if the two vectors are proportional, in which case it has only one solution. If a quadratic equation with real coefficients has no real solutions or only one, then its discriminant is non-positive. In this case the discriminant is $$ \left(2\sum_{i=1}^n a_i b_i\right)^2 - 4\left(\sum_{i=1}^n a_i \right)\left(\sum_{i=1}^n b_i\right). $$ That expression must therefore be $\le0$, and $=0$ only if the two vectors are proportional. From that, the Cauchy-Scharz inequality follows.

Michael Hardy
  • 1
2

We have the equation $$ \sum_{i=1}^n(a_ix+b_i)^2 =x^2\sum_{i=1}^na_i^2+2x\sum_{i=1}^na_ib_i+\sum_{i=1}^nb_i^2\tag{1} $$ If there were two distinct real roots of the right hand side of $(1)$, then the minimum of the quadratic would be at their average and would be less than $0$; this because the positive coefficient of $x^2$ yields strict convexity. However, if there were an $x$ so that the right hand side were less than $0$, the left hand side would yield a sum of squares that was negative.

$\Rightarrow$ the roots cannot be real and distinct

The only possibility for a real root would be in the case where the left hand side of $(1)$ were $0$. for that to occur, each term in the sum would need to be $0$. That is, for all $0\le i\le n$, $$ b_i=-a_ix\tag{2} $$ $\Rightarrow$ $a$ and $b$ must be proportional.

If the roots of the right hand side of $(1)$ are equal or non-real, we have, from the quadratic formula, that $$ \left(2\sum_{i=1}^na_ib_i\right)^2-\ 4\left(\sum_{i=1}^na_i^2\right)\left(\sum_{i=1}^nb_i^2\right)\le0\tag{3} $$ which upon rearrangement is Cauchy-Schwarz.

$\Rightarrow$ we want the roots of the right hand side of $(1)$ to be equal or non-real.

robjohn
  • 345,667