By Genov G.K.

**Read or Download A basis of identities of the algebra of third-order matrices over a finite field PDF**

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**Extra info for A basis of identities of the algebra of third-order matrices over a finite field**

**Example text**

The sum of two vectors x and y of the same order, written as x + y, is defined to be the vector x1 + y1 x2 + y2 x + y := . . xm + ym Multiplication of a vector x by a scalar λ is defined by means of the relation λx1 λx2 λx := . , . . λxm which can also be written is xλ. 1. 1 — Vector addition and scalar multiplication in R3 . formed by x, y and the origin. The product λx is obtained by multiplying the magnitude of x by λ and retaining the same direction if λ > 0 or the opposite direction if λ < 0.

Solution Let A := diag(a1 , a2 , . . , an ) and B := diag(b1 , b2 , . . , bn ). Then, AB = diag(a1 b1 , . . , an bn ) = diag(b1 a1 , . . , bn an ) = BA. A diagonal matrix is the simplest generalization of a scalar, and essentially all properties of scalars also hold for diagonal matrices. 21 (Triangular matrix) (a) Consider the lower triangular matrices 1 0 0 A = 1 1 0 and 0 0 1 1 B= 0 0 0 1 −2 0 0 . 1 Show that AB and BA are lower triangular, but that AB = BA. (b) Show that the product of two lower triangular matrices is always lower triangular.

For real matrices A, B and Solution (a) Clearly, A = O implies A A = O. Conversely, assume A A = O. Then, for all j, the j-th diagonal element of A A is zero, that is, i a2ij = 0. This implies that aij = 0 for all i and j, and hence that A = O. 8(b). (b) Clearly, AB = O implies A AB = O. Conversely, if A AB = O, then (AB) (AB) = B A AB = O and hence AB = O, by (a). (c) This follows by replacing B by B − C in (b). (d) Consider a = (1 + i, 1 − i) . Then a a = (1 + i)2 + (1 − i)2 = 0, even though a = 0.