By Jean Gallier
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Extra resources for Applications of Scientific Computation [Lecture notes]
N. Indeed, it can be shown directly (by induction) that an upper (or lower) triangular matrix is invertible iff all its diagonal entries are nonzero. If A is lower-triangular, we solve the system from top-down by forward-substitution. Thus, what we need is a method for transforming a matrix to an equivalent one in uppertriangular form. This can be done by elimination. Let us illustrate this method on the following example: 2x + y + z = 5 4x − 6y = −2 −2x + 7y + 2z = 9. We can eliminate the variable x from the second and the third equation as follows: Subtract twice the first equation from the second and add the first equation to the third.
INTRODUCTION TO VECTORS AND MATRICES for some scalars λ1 , . . , λn , µ ∈ R. We must have µ = 0, because otherwise v = (−λ1 /µ)u1 + · · · + (−λn /µ)un contradicting the fact that v is not a linear combination of the ui s. But then, λ1 u1 + · · · + λn un = 0, and since u1 , . . , un are linearly independent, we must have λ1 = · · · = λn . Therefore u1 , . . , un , v are linearly independent. Now, u1 , . . 6. Therefore, u1 , . . , un span Rn . 6. 8. Let u1 , . . , up and v1 , . . , vq be any vectors in Rn .
Also, we can check easily that (λx) · y = λ(x · y) = x · (λy). It follows that if x = 0 and y = 0, then x·y = x y (x · y). 32 CHAPTER 1. 6: A unit vector in R2 Therefore, we just have to figure out what is the inner product of two unit vectors in the plane. However, a unit vector x in the plane corresponds to a point on the unit circle. Thus, its coordinates are of the form (cos α, sin α), where α is the angle betwen the x-axis and the line supported by x. 7. It follows that x · y = (cos α, sin α) · (cos β, sin β) = cos α cos β + sin α sin β = cos(β − α).