By Wu X.

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**Example text**

5). 60) for a typical basis vector vh we obtain that (vhx)e = (Ebhh (x)v h )0 = E bhk(x)tkj Vj > j,k and (vhB)x = (E thiuil x = E thi aij (x)uj . 60) for a G-homomorphism 0 reduces to the matrix equation TA(x) = B(x)T (x E G). 60). 62) are essentially equivalent, and we denote the correspondence between 0 and T briefly by 0 4--)T, assuming that definite bases have been chosen for V and U. 63) GROUP REPRESENTATIONS We recall that if 0: V -+ U is a linear map, then the kernel of O, that is ker O = {voE VjvoO =0}, is a subspace of V, and the image of 0, that is im 0 = {u 1 E UIu1= v10 for some v 1 }, is a subspace of U.

4 . . X 2 e lf X O If, Xe 1If Xe 2If ... XCelf where X = (x if ) is an arbitrary e x e matrix over K. 70) where ✓ 6 is the complete matrix algebra over K of degree e. If T1 = X 1 E+ I f and T2= X 2 I f are two elements of W(D), then, by the properties of the tensor product, k1T1+k2T2 = (k 1X1 + k2X2)O If, and T1T2 = (X1X2)O If. Hence, as an algebra, (D) has the same structure as Ale ; the factor If in the tensor product merely `blows up' the mat rices, but does not affect the abstract laws of composition.

60) for all v E Vandx E G. Suppose now that U and V afford the matrix representations A (x) and B(x) relative to the bases U= [u 1 , u 2 , ... , u m ] and V = [ Vi, V2, • • . > Vn] respectively. Thus m u i x = E a ij (x )uj (i = 1, 2, . . , m ) j=1 and vhx = E bhk (x)vk (h = 1, 2, . . , n). 12), p. 5). 60) for a typical basis vector vh we obtain that (vhx)e = (Ebhh (x)v h )0 = E bhk(x)tkj Vj > j,k and (vhB)x = (E thiuil x = E thi aij (x)uj . 60) for a G-homomorphism 0 reduces to the matrix equation TA(x) = B(x)T (x E G).