A short linear algebra book (answers) - download pdf or read online

By Hefferon J.

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16. (b) Here is the point located in the lattice. 42β2 . The picture on the right shows where that appears inside of the crystal lattice, taking as the origin the lower left corner of the hexagon in the lower left. So this point is in the next column of hexagons over, and either one hexagon up or two hexagons up, depending on how you count them. 99), but it doesn’t seem to have to do much with the physical structure that we are studying. 3 In terms of the basis the locations of the corner atoms are (0, 0, 0), (1, 0, 0), .

No sub-basis of E2 forms a basis for the subspace U of R2 that is the line y = x. (b) It is a basis (for its span) because the intersection of linearly independent sets is linearly independent (the intersection is a subset of each of the linearly independent sets). It is not, however, a basis for the intersection of the spaces. For instance, these are bases for R2 : 1 0 2 0 , and B2 = , B1 = 0 1 0 2 2 and R2 ∩ R2 = R2 , but B1 ∩ B2 is empty. All we can say is that the intersection of the bases is a basis for a subset of the intersection of the spaces.

Remark. There is a small point about this argument that is worth mention. In a set, repeats collapse, that is, strictly speaking, this is a one-element set: {v, v}, because the things listed as in it are the same thing. Observe, however, the use of the subscript n in the above argument. In moving from the domain set {v1 , . . , vn } to the image set {f (v1 ), . . ) Answers to Exercises 43 To show that if f : V → W is an isomorphism and if U is a subspace of the domain V then the set of image vectors f (U ) = {w ∈ W w = f (u) for some u ∈ U } is a subspace of W , we need only show that it is closed under linear combinations of two of its members (it is nonempty because it contains the image of the zero vector).

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A short linear algebra book (answers) by Hefferon J.


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