By Hefferon J.

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Thus this is a trivial vector space, and has dimension zero. 8. 28 The first chapter defines a plane — a ‘2-flat’ — to be a set of the form {p + t1 v1 + t2 v2 t1 , t2 ∈ R} (also there is a discussion of why this is equivalent to the description often taken in Calculus as the set of points (x, y, z) subject to some linear condition ax + by + cz = d). When the plane passes throught the origin we can take the particular vector p to be 0. Thus, in the language we have developed in this chapter, a plane through the origin is the span of a set of two vectors.

Un ∈ U, c1 · h(u1 ) + · · · + cn · h(un ) = h(c1 · u1 ) + · · · + h(cn · un ) = h(c1 · u1 + · · · + cn · un ) which is itself in h(U ) as c1 · u1 + · · · + cn · un is in U . Thus this set is a subspace. 48 Linear Algebra, by Hefferon (d) The natural generalization is that the inverse image of a subspace of is a subspace. Suppose that X is a subspace of W . Note that 0W ∈ X so the set {v ∈ V h(v) ∈ X} is not empty. To show that this set is closed under combinations, let v1 , . . , vn be elements of V such that h(v1 ) = x1 , .

To see that the space is the direct sum of these two, just note that each member of P2 has the unique decomposition m + nx + px2 = (m + px2 ) + (nx). 24 Each of these is R3 . (a) These are broken into lines for legibility. 26 It is W2 . 8. 28 Two distinct direct sum decompositions of R4 are easy to find. Two such are W1 = [{e1 , e2 }] and W2 = [{e3 , e4 }], and also U1 = [{e1 }] and U2 = [{e2 , e3 , e4 }]. ) In contrast, any partition of R1 ’s single-vector basis will give one basis with no elements and another with a single element.