A = u + V. u + (0) = u

Transcription

1 Recall: Last time we defined an affine subset of R n to be a subset of the form A = u + V = {u + v u R n,v V } where V is a subspace of R n We said that we would use the notation A = {u,v } to indicate such an affine subset Moreover, we proved that the subspace V is uniquely determined by A, but that the vector u is not In fact, we showed that A = u + V for any u A As I mentioned last time, the affine subsets of R n will be the objects we will study in this course Definition: Let A = {u,v } be an affine subset of R n The dimension of A is the dimension of the vector space V Remark: Inasmuch as the subspace V is uniquely determined by A, the definition makes sense It follows also from this definition that an affine subset of R n always has dimension n Observations: a) Since the only vector subspace of R n of dimension 0 is the subspace V = {0}, it follows that the affine subsets of R n of dimension 0 all have the form u + (0) = u ie all consist of a single point So, the 0-dimensional affine subsets of R n are individual points b) Since the only vector subspaces of R n of dimension 1 are subspaces of the form V = v, where v 0 (geometrically, the points on a line of R n through the origin) it follows that the only affine subsets of dimension 1 are subsets of the form u + v 1

2 As we have seen, these are precisely the points on a line through the point u all having the form u + λv, where λ R These lines can be described in two different ways: If u = (a 1,,a n ) and v = (b 1,,b n ) then a parametric description of the affine subset A = {(x 1,,x n ) = x x = u + λv} is given by x 1 = a 1 + λb 1 x n = a n + λb n for any λ R Ie since we know the a i and the b i already, it is enough to give a real number λ to find a point on this affine subset Notice that by describing the line parametrically in this way, when we give the value 0 to the parameter, the point that we get is u Ie from this parametric description of the line it is almost like we are thinking of u as the origin on this line (even though there is no origin on this line) Keeping this in mind, the remark I made earlier namely that we can use any point u of A to write A = u + v becomes even more significant If we write A in this way and then use that description to give a parametrization of the affine set, now the point u corresponds to the parameter 0 We can also write down an implicit description of this affine subset: using the notation above, we get that (x 1,,x n ) A if and only if (x 1 a 1,,x n a n ) = λ(b 1,,b n ) for some real number λ But this is the same thing as saying that the rows of the matrix x 1 a 1 x 2 a 2 x n a n b 1 b 2 b n 2

3 are linearly dependent, ie the matrix has rank 1 However, we also know that the vector (b 1,b 2,,b n ) is NOT the zero vector Thus, this matrix has rank exactly 1 Thus, since this matrix has rank exactly 1 we know, by Kronecker s Theorem, that this is equivalent to saying that all the 2 2 minors of this matrix that border (to border = orlare) a 1 1 non-zero minor, all have determinant zero So, for example, with the notation above: if b i 0 then we must have all the minors x j a j x i a i have determinant zero, for j i b j a j Example: Let s consider the 1-dimensional affine set in R 3 which is A = u + V where u = (2,0, 1) and V = ( 1,1,1) parametric description: (x,y,z) A there is an s R such that (x,y,z) = (2,0, 1) + s( 1,1,1) ie x = 2 + ( 1)s = 2 s y = 0 + (1)s = s z = 1 + (1)s = 1 + s implicit description: (x,y,z) A there is an s R such that (x,y,z) = (2,0, 1) + s( 1,1,1) ie (x,y,z) (2,0, 1) = (x 2,y,z + 1) = s( 1,1,1) ie rank x 2 y z + 1 = Since det[ 1] = 1 0, we have that the matrix has rank = 1 det x 2 y = 0 and det x 2 z + 1 =

4 (this last by Kronecker s Theorem) Ie (x 2) + y = 0 and (x 2) + (z + 1) = 0 ie x + y = 2 x + z = 1 This implicit description describes the points of our line as the points of intersection of two planes in R 3 In fact, it gives a description of this affine set as the set of solutions to a system of linear equations In fact, what we have seen in these examples is quite a general state of affairs I will express this fact in a general theorem Theorem: The affine subsets of R n are precisely those subsets of R n which arise as the solutions to a (not necessarily homogeneous) system of linear equations Ax = b ( ) where A is an m n real matrix, x = x 1 and b = b 1 Rm x n b m Proof: First I will show that sets of solutions for such a system of linear equations are affine subsets of R n Since the system of equations has n unknowns, the solutions can be viewed as points in R n You proved, last semester, that every solution to ( ) is of the form u + v where u is a particular solution to ( ), ie one solution to ( ), and v is a solution to the system of equations Ax = 0 ( ) 4

5 However, you also know that the set of solutions to a homogeneous system of linear equations in n unknowns is always a subspace of R n Thus the solutions to ( ) are all of the form and this is an affine subset of R n u + v where v V a subspace of R n In order to show the equivalence stated in the theorem it remains to prove that an affine subset A = u + V is the precise set of solutions to a system of linear equations The method to do this is as follows: we first construct a system of homogeneous linear equations whose solutions are exactly V ( I will do that a bit later) In fact, if dimv = d I will show how we can find n d homogeneous linear equations whose solutions are exactly V Assume, for the moment, that this has been done We then write that system of equations as Ax = 0 We set b = Au Then u is a particular solution to the system of equations and all solutions to Ax = b are given by A Ax = b It remains to prove that if V is a d-dimensional subspace of R n then V is exactly the set of solutions to a homogeneous system of n d equations in n unknowns Since the dimension of V is d, we can write V = v 1,,v d where the v i are linearly independent vectors If we write the coordinates of those vectors as the rows of a matrix, that matrix will have d rows and n columns and will have rank d Let s call that matrix v 1 v 2 M = v d 5

6 Now, the vector x = (x 1,,x n ) V if and only if the rank of N = x is also d M But, since M has rank d, there is a d d minor of M which has determinant different from 0 That minor of M is bordered ( = orlato) by exactly n d matrices of size (d + 1) (d + 1), which must all have determinant = 0 Those determinants give us precisely the n d homogeneous linear equations we were looking for Example: Let me illustrate the proof of the preceding theorem with an example I will just explain the last part of the proof with this example Let V = (1,0,1, 1),(0,1,1,0) The two vectors we have used to generate V are easily seen to be linearly independent and so dim V = 2 Also, V R 4 So, by what I have said above, V should be the precise set of solutions to a system of 2 (= 4-2) linear equations in 4 unknowns Following the reasoning above, we have: (x, y, z, w) V rank x y z w = Now notice that the 2 2 minor has determinant 0 and so, by Kronecker s 0 1 Theorem, this matrix has rank = 2 These give the equations: det x y z = 0 and det x y w = x y +z = 0 x +w = 0 which are two equations in 4 unknowns that we are looking for 6

7 We rewrite this system of equations as Ax = 0, ie x y = z w 0 0 and check that this is a system of equations which has V precisely as its set of solutions It is possible to use the ideas we have seen above to solve some nice geometric problems For example, how does one find equations (both parametric and implicit) for the line connecting two points in R n? That line is a 1-dimensional affine set and so we should have a way to find it How can we think about this set as an affine subset of R n? We have to think how to express our line in the form u + V, where V is a 1-dimensional subspace of R n Let s look in R 2 to get an idea of what we should do (Draw picture in R 2 of the line L through the two points P 1 and P 2 Use the fact that if P 1 and P 2 are in the affine set then P 1 P 2 is in the giacitura of the affine set Then do the following example in R 5 ) Example: In R 5 describe the line containing the point P 1 = (1,2,3,4,5) and the point P 2 = (1, 1,1, 1,1) This is an affine set L = u + V, where dimv = 1 We can write L = (1,2,3,4,5) + (0, 3, 2, 5, 4) We get a parametric description of L as follows: L = {(x 1,x 2,x 3,x 4,x 5 ) R 5 x 1 = 1 + 0s x 2 = 2 3s x 3 = 3 2s x 4 = 4 5s x 5 = 5 4s } if We get an implicit description of L observing that (x 1,x 2,x 3,x 4,x 5 ) L if and only rk x1 1 x 2 2 x 3 3 x 4 4 x 5 5 =

8 By Kronecker s Theorem, this last is true if and only if det x1 1 x 2 2 = 0,det 0 3 x2 2 x 3 3 = x2 2 x det 4 4 x2 2 x = 0,det 5 5 = These give the equations: 3x 1 = 3 2x 2 +3x 3 = 5 5x 2 +3x 4 = 2 4x 2 3x 5 = 7 8