By Charles A. Weibel

A portrait of the topic of homological algebra because it exists at the present time

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**Extra resources for An Introduction to Homological Algebra**

**Example text**

1. Give details of the geometrical interpretations of 1 R and 2 R. Solution. Of course R is essentially the same as R. Represent segment drawn from an arbitrary point u + a as the x -axis. u Thus elements of a in R by a line on the x -axis to the point R are represented by line segments along the x -axis . a l o € R , represent this vector by a line segment in the xy 2 -plane with arbitrary initial point (u^ u^j and end point (u, + a^ u^ + o a 2 ). Thus line segments in the plane represent vectors in R .

1 ) + ... + a^y' + aQ(x)y = 0. Solution. 2: Vector Spaces and Subspaces 0 53 0 and have determinant 0 1 So there is no rule of addition here. 0. (b) X+ and X~ are solutions of This is not a vector space. B , then X1 + X2 If is not a solution. = B + B = 2B t 0. ) = cB . eX, AX = A(X1 + X^j = AX^ + AX2 will not be a solution if c # 1 since Also there is no zero vector. This is a vector space. If y, equation, then so is For 1. In this case we have neither a rule of addition nor a rule of scalar multiplication.

3, 4. ' 3,' 4. un for Solution. Let -4 be an n x n matrix. The expansion of det(^4) by row 1 is n det(4) = V a-. 4.. . Now A. is an (n - 1) - square determinant, so % =1 we need u * operations to compute it. Hence to form all the a1 . A, • 71—1 calls for 11 n (u _. + 1) n numbers to get + 271-1. Obviously operations. det(i4). ^ = 0; Hence 4w3 + 7 = 63. Finally, we still have to add up these u = n (u , + l) + 7 i - l = using the formula, we obtain (which we can also see directly). 3 6 1. For the matrices identity 3 A = and 4 -3 det(i4)det(B).