explain with examples and simple terms please The n × n n×n Identity Matrix Definition 2.62: For each n ≥ 2 n≥2, the n × n n×n identity matrix, denoted I n I n ​ , is the matrix having ones on its main diagonal and zeros elsewhere, and is defined for all n ≥ 2 n≥2. Example 2.63: I 2 = [ 1 0 0 1 ] , I 3 = [ 1 0 0 0 1 0 0 0 1 ] I 2 ​ =[ 1 0 ​ 0 1 ​ ],I 3 ​ = ⎣ ⎢ ⎡ ​ 1 0 0 ​ 0 1 0 ​ 0 0 1 ​ ⎦ ⎥ ⎤ ​ Definition 2.64: Let n ≥ 2 n≥2. For each j j, 1 ≤ j ≤ n 1≤j≤n, we denote by e j e j ​ the j th j th column of I n I n ​ . Example 2.65: When n = 3 n=3, e 1 = [ 1 0 0 ] , e 2 = [ 0 1 0 ] , e 3 = [ 0 0 1 ] e 1 ​ = ⎣ ⎢ ⎡ ​ 1 0 0 ​ ⎦ ⎥ ⎤ ​ ,e 2 ​ = ⎣ ⎢ ⎡ ​ 0 1 0 ​ ⎦ ⎥ ⎤ ​ ,e 3 ​ = ⎣ ⎢ ⎡ ​ 0 0 1 ​ ⎦ ⎥ ⎤ ​ . Theorem 2.66: Let A A be an m × n m×n matrix Then A I n = A AI n ​ =A and I m A = A . I m ​ A=A. Proof. The ( i , j ) (i,j)-entry of A I n AI n ​ is the product of the i t h i th row of A = [ a i j ] A=[a ij ​ ], namely [ a i 1 a i 2 ⋯ a i j ⋯ a i n ] [ a i1 ​ ​ a i2 ​ ​ ⋯ ​ a ij ​ ​ ⋯ ​ a in ​ ​ ] with the j t h j th column of I n I n ​ , namely e j e j ​ . Since e j e j ​ has a one in row j j and zeros elsewhere, [ a i 1 a i 2 ⋯ a i j ⋯ a i n ] e j = a i j [ a i1 ​ ​ a i2 ​ ​ ⋯ ​ a ij ​ ​ ⋯ ​ a in ​ ​ ]e j ​ =a ij ​ Since this is true for all i ≤ m i≤m and all j ≤ n j≤n, A I n = A AI n ​ =A. The proof of I m A = A I m ​ A=A is analogous—work it out! ♣ Instead of A I n AI n ​ and I m A I m ​ A we often write A I AI and I A IA, respectively, since the size of the identity matrix is clear from the context: the sizes of A A and I I must be compatible for matrix multiplication. Thus A I = A and I A = A AI=A and IA=A which is why I I is called an identity matrix – it is an identity for matrix multiplication.