Need help in understanding the notation used in the following Lemma.

Question

The author is proving this inequality through Cauchy-Schwarz inequality. I don't understand what the $a_{i,j}$ mean on both sides of the inequality. I would be really grateful if someone could expand one of the terms just to give me a way into understanding the proof.

Answer 1

$a_{i,j}$ is just a double-indexed sequence of numbers. For example, I can denote the elements in the list $4, 100, -3, 20$ as $a_{1}, a_{2}, a_{3}, a_{4}$. I chose the subscripts of $a$ to denote the position of the number. $a_{1}$ represents the first number in the list, for example.

Suppose you have a matrix $A = \begin{bmatrix} 3 & 1 & 16 \\ 2 & 5 & 9 \\ 21 & -35 & 0 \end{bmatrix}$.

What is a really nice and clear way to refer to the position of the elements in the matrix? Well, $3$ is in the first row and first column, so if I wanted to, I could denote it $a_{1,1}$. $16$ is in the first row and third column, so I could denote it $a_{1,3}$. $-35$ is in the third row and second column, so I could denote it as $a_{3,2}$. Now if I am talking to you, instead of referring to $-35$ I could just say "Take a look at $a_{3,2}$.

Answer 2

For example,

$$\sum_{j=1}^3\prod_{i=1}^3a_{i,j}=\sum_{j=1}^3a_{1,j}a_{2,j}a_{3,j}=a_{1,1}a_{2,1}a_{3,1}+a_{1,2}a_{2,2}a_{3,2}+a_{1,3}a_{2,3}a_{3,3}$$

Answer 3

$$ \big(\sum_{j=1}^{n}\prod_{i=1}^{n}a_{i,j}\big)^2=\big(\sum_{j=1}^{n}a_{1,j}a_{2,j}\cdots a_{n,j}\big)^2= \\ \\ =\big(a_{1,1}a_{2,1}\cdots a_{n,1}+a_{1,2}a_{2,2}\cdots a_{n,2}+\cdots+a_{1,n}a_{2,n}\cdots a_{n,n}\big)^2 $$ while $$ \prod_{i=1}^{n}\big(\sum_{j=1}^{n}a_{i,j}^{2}\big)= \prod_{i=1}^{n}\big(a_{i,1}^{2}+a_{i,2}^{2}+\cdots+a_{i,n}^{2}\big) = \\ \\ =\big(a_{1,1}^{2}+a_{1,2}^{2}+\cdots+a_{1,n}^{2}\big)\big(a_{2,1}^{2}+a_{2,2}^{2}+\cdots+a_{2,n}^{2}\big)\cdots\big(a_{n,1}^{2}+a_{n,2}^{2}+\cdots+a_{n,n}^{2}\big) $$