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M1GLA: Geometry and Linear Algebra Lecture Notes
... Definition (Scalar product). The scalar product (or dot product) of two vectors x, y ∈ R2 is (x · y) = x1 y1 + x2 y2 E.g. If x = (1, −1), y = (1, 2) then (x · y) = 1 + (−2) = −1. Easy properties: For any vectors x, y, z ∈ R2 • x · (y + z) = x · y + x · z (distribution over +) • ||x||2 = (x · x) = x2 ...
... Definition (Scalar product). The scalar product (or dot product) of two vectors x, y ∈ R2 is (x · y) = x1 y1 + x2 y2 E.g. If x = (1, −1), y = (1, 2) then (x · y) = 1 + (−2) = −1. Easy properties: For any vectors x, y, z ∈ R2 • x · (y + z) = x · y + x · z (distribution over +) • ||x||2 = (x · x) = x2 ...
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Chapter 15. The Kernel of a Three-by
... To prove (i), assume that det(A) 6= 0. Then we know from the previous section that A−1 exists. If u is any vector in ker A then by definition we have Au = 0. We apply A−1 to both sides of this and get u = A−1 Au = A−1 0 = 0. This shows that u = 0. Hence, the only vector in ker A is 0. This means tha ...
... To prove (i), assume that det(A) 6= 0. Then we know from the previous section that A−1 exists. If u is any vector in ker A then by definition we have Au = 0. We apply A−1 to both sides of this and get u = A−1 Au = A−1 0 = 0. This shows that u = 0. Hence, the only vector in ker A is 0. This means tha ...
![arXiv:math/0604168v1 [math.CO] 7 Apr 2006](http://s1.studyres.com/store/data/017890502_1-2c1abc75bd42752544cdf6d7b46b6ed7-300x300.png)
