Lectures in Abstract Algebra: II. Linear Algebra by Nathan Jacobson

By Nathan Jacobson

The current quantity is the second one within the author's sequence of 3 facing summary algebra. For an knowing of this quantity a definite familiarity with the elemental suggestions handled in quantity I: teams, jewelry, fields, homomorphisms, is presup­ posed. even if, we've attempted to make this account of linear algebra self reliant of an in depth wisdom of our first quantity. References to express effects are given sometimes yet a number of the primary innovations wanted were handled back. briefly, it truly is was hoping that this quantity may be learn with whole knowing by way of any scholar who's mathematically sufficiently mature and who has a familiarity with the normal notions of contemporary algebra. Our viewpoint within the current quantity is largely the summary conceptual one. besides the fact that, now and then we now have deviated slightly from this. sometimes formal calculational equipment yield sharper effects. in addition, the result of linear algebra are usually not an result in themselves yet are crucial instruments to be used in different branches of arithmetic and its purposes. it's for this reason helpful to have to hand tools that are confident and which might be utilized in numerical difficulties. those tools occasionally necessitate a a little lengthier dialogue yet we have now felt that their presentation is justified at the grounds indicated. A stu­ dent good versed in summary algebra will absolutely become aware of brief cuts. a few of these were indicated in footnotes. we have now integrated lots of workouts within the text.

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Ai are all different, . is in L(, n). (Hint: The determinant of this matrix is a so-called Vandermonde determinant. ) i>i 2. Prove that if Ll. = is a field and (a) e L(, n), then the transposed matrix (a)' e L(, n). 3. Prove the following converse of Ex. , 2), then Ll. is a field. 25 FINITE DIMENSIONAL VECTOR SPACES 4. ] -: -5 1 7 9. Factor spaces. Any subspace to of ~ is, of course, a subgroup of the additive group~. Since ~ is commutative, we can define the factor group lR = ~/t0.

J. Hence dim [XI, X2, ... , xrl = dim [YI, Y2, .. ·,Yrl· In a similar fashion we define the column rank of (a). Here we introduce a right vector space m' of r dimensions with basis (el', e2', "', e/). Then we define the column rank of (a) to be the rank of the set (Xl" X2', ... , x n ') where x/ = "1;e/ aji. The x/ are called column vectors of (a). We shall prove in the next chapter that the two ranks of a matrix are always equal. In the special case Ll. = cf> a field (commutative) this equality can be established by showing that these ranks coincide with still another rank which can be defined in terms of determinants.

Zi = L Zj e @5i j r'i + + ... + + ... + n (e;l + ... + e;i-l + e;i+l + ... + e;r). Hence Zi = 0 and Yi = Y/. The converse of this result holds also; for if (9) fails for some i, then there is a vector Zi ~ 0 in this intersection. Thus Zi = L Zh and we have two distinct j r"i representations of this element as a sum of elements out of the spaces elk. We have therefore proved Theorem 10. A necessary and sufficient condition that the spaces e;l, e;2, ... , e;r be independent is that every vector in @5 = e;l e;2 e;r have a unique representation in the form ~Yi, Yi in @5i.

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