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Basis and Dimension (II.B) Basis and dimension How would you explain that a plane has two dimensions? Well, you can go in two independent directions, and no more. To make this idea precise, we formulate the DEFINITION 1. A (finite) subset f~v1,...,~vmg of a vector space V (over a field F) is a basis of V if (a) they are linearly independent (over F), and (b) they span V (over F).1 The “over F” means that the linear combinations are understood to be of the form ∑ ai~vi, with ai 2 F. As usual, our “default” will be F = R. EXAMPLE 2. A basis of Rm is given by the standard basis vectors m feˆigi=1. Of course, one could say the same for any field: the standard basis vectors are also basis for Cm, viewed as a vector space over C. But what if we view Cm as a vector space over R? Then you need the 2m vectors eˆ1,..., eˆm and ieˆ1,..., ieˆm! (Why?) EXAMPLE 3. Let P2(x, y) denote the vector space of quadratic polynomials in two variables x and y, with real coefficients. This has 2 6 2 2 the “monomial basis” fri (x, y)gi=1 = f1, x, y, xy, x , y g. Similarly, 3 10 2 2 2 2 3 3 fri (x, y)gi=1 = f1, x, y, xy, x , y , x y, y x, x , y g gives a basis for the space P3(x, y) of cubic polynomials. We’ll give a geometric application of this example below. 1I am not saying that you can’t have infinite bases; indeed, non-finitely-generated vector spaces (like P¥(x) := all polynomials) may have such a basis. I’m just not defining or discussing them at this stage. 1 2 (II.B) BASIS AND DIMENSION EXAMPLE 4. Let A be an m × n matrix, and V be the vector sub- space of Rn consisting of solutions to A~x = 0. How do you compute a basis of V? By making use of the row-reduction algorithm, of course! For instance, we might have 0 0 0 1 −1 −1 1 0 1 2 0 3 0 1 B C B − C B 2 4 2 4 2 C B 0 0 1 1 0 C A = B C 7−! rre f (A) = B C . @ 2 4 3 3 3 A @ 0 0 0 0 1 A 3 6 6 3 6 0 0 0 0 0 The free variables are x2 and x4 , and an arbitrary solution is given by plugging arbitrary values in for these and solving for the pivot variables x1, x3, x5 . To get a basis , you plug in (x2 = 1, x4 = 0) and (x2 = 0, x4 = 1) to obtain 0 1 0 1 −2 −3 B C B C B 1 C B 0 C B C B C B 0 C , B 1 C . B C B C B C B C @ 0 A @ 1 A 0 0 More generally, the ith basis element is given by setting the ith free variable to 1 and the rest to zero. EXAMPLE 5. Suppose A is an invertible m × m matrix. I claim m m that its columns f~cigi=1 are a basis for R . To establish this, check that: • f~cig are linearly independent : g1~c1 + ... + gm~cm = 0 =) A~g = 0 =) ~g = 0 (since A is invertible). m m • f~cig span R : show every ~y 2 R can be written ~y = g1~c1 + ... + −1 gm~cm (= A~g). This is easy: just put ~g = A ~y . We say that V is finitely generated if some finite subset f~v1,...,~vmg spans V . Here is the precise reason why dimension makes sense: PROPOSITION 6. If V is finitely generated then (i) it has a (finite) basis, and (ii) any 2 bases of V have the same number of elements. (II.B) BASIS AND DIMENSION 3 DEFINITION 7. We define the dimension dimF(V) of V over F to be the number of elements in (ii).2 We get (i) by going through the generating subset f~v1,...,~vmg and throwing out any ~vi that is a linear combination of the earlier vectors. What remains is a basis. In order to show (ii) we first prove the following LEMMA 8. If ~v1,...,~vm span V then any independent set ⊆ V has no more than m vectors. PROOF. Consider any set ~u1,...,~un , n > m (of vectors in V ). Since ~v1,...,~vm span V , each ~uj is a linear combination of the ~vi : m ~uj = ∑ Aij~vi , for A m × n. i=1 Since m < n, there is a nontrivial solution to A~x = ~0, some nonzero ~x 2 Rn. That is, n ∑ Aijxj = 0 for each i, j=1 and so n n m m n ! m ~ ∑ xj~uj = ∑ xj ∑ Aij~vi = ∑ ∑ xj Aij ~vi = ∑ 0~vi = 0. j=1 j=1 i=1 i=1 j=1 i=1 n We conclude that f~ujgj=1 is not an independent set! Now let f~v1,...,~vmg and fw~ 1,..., w~ ng be two bases for V . The f~vig span V and the fw~ jg are independent, so the Lemma =) n ≤ m . The same argument with f~vig and fw~ jg swapped =) n ≥ m, and so we’ve proved the Proposition. We can rephrase the Lemma as follows: if dim(V) = m then (a) less than m vectors cannot span V , while (b) more than m vectors are linearly dependent. A Geometric Application. Given 5 points A, B, C, D, E “in gen- eral position” in the plane, there is a unique conic passing through 2The F is typically omitted when there is no ambiguity. That said, in the first n n example we had dimC(C ) = n while dimR(C ) = 2n. 4 (II.B) BASIS AND DIMENSION them. Here’s why: the equation of a conic Q has the form 6 2 2 2 0 = fQ(x, y) = ∑ airi (x, y) = a1 + a2x + a3y + a4xy + a5x + a6y , i=1 and the statement that it passes through A is 0 = fQ(xA, yA) = ∑ airi(xA, yA). Continuing this for B thru E we get 5 equations in 6 unknowns ai , a 3 system which “generically” has a “line” of solutions: if (a1,..., a6) solves the system then so does (c · a1,..., c · a6). But this just gives multiples of the same fQ , which won’t change the shape of the curve, and so there is really only one nontrivial solution: through five general points in the plane there exists a unique conic Q . Similarly, given 8 points A, B, C, D, E, P1, P2, P3 what can we say about the “vector space” of cubics passing through them (i.e. of cubic polynomials in (x,y) vanishing at all 8 points)? First of all, the vector space of all cubics 10 3 3 3 ∑ airi (x, y) = a1 + a2x + a3y + ... + a9x + a10y i=1 should be thought of in terms of the coefficient vector~a . The 8 con- straints give a linear system = 10 3( ) 9 0 ∑i=1 airi xA, yA > . = . 8 equations > = 10 3( ) ; 0 ∑i=1 airi xP3 , yP3 whose space of solutions “in general” (i.e., under the assumption of maximal rank[=8]) is two-dimensional – there are 8 leading 1 s in rre f of the 8 × 10 [r] matrix, and therefore 2 parameters to choose freely. So start with A, B, C, D, E in “general position”; we’d like to construct points on the (unique) conic through them, using a straight- edge. Begin by drawing the lines AB , DE , CD , and BC ; label 3more precisely, “generically” means here that the matrix [r] (with rows indexed by A thru E , columns by i = 1 thru 6 ) is of maximal rank (= 5 ). (II.B) BASIS AND DIMENSION 5 AB \ DE by P1.(Q is drawn in a background color because we don’t know what it looks like yet.) B Q C A D E BC AB CD DE P1 Now draw (almost) any line ` through A – the choice is free ex- cept that ` shouldn’t go through B, C, D, or E . All the rest of the construction depends on this choice; in particular the final point q depends on ` , and so by varying ` we can vary q and in this way (we hope) construct enough points on Q to get an idea of what it looks like. Label ` \ CD =: P2 ; then draw P1P2 and label P1P2 \ BC =: P3 ; finally set q := EP3 \ `. We must prove q 2 Q . Adding dotted lines to our diagram for the lines depending on `, Q B C A D q E BC AB CD EP l 3 DE P P P 1 2 3 6 (II.B) BASIS AND DIMENSION Now consider the following three cubic equations (where e.g. fAB(x, y) = 0 is the linear equation of the line AB , and so on): f1(x, y) := fQ(x, y) · fP1P2 (x, y) = 0, f2(x, y) := fAB(x, y) · fCD(x, y) · fEP3 (x, y) = 0, f3(x, y) := f`(x, y) · fBC(x, y) · fDE(x, y) = 0. The graphical solutions of these equations are (respectively) Q [ P1P2 , AB [ CD [ EP3 , ` [ BC [ DE. All 3 of these unions contain the 8 points A, B, C, D, E, P1, P2, P3; that is, the cubic polynomials f1 , f2 , and f3 each vanish at all 8 points. Above we explained that the (vector) space of such cubic polynomials was 10 − 8 = 2 dimensional; it follows that 3 “vectors” f1 , f2 , f3 cannot be linearly independent and there is a nontrivial relation a f1(x, y) + b f2(x, y) + g f3(x, y) = 0 .
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