Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:50 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://orion.math.iastate.edu/hentzel/class.307.05 Text: Linear Algebra and its Applications, Third Edition David C. Lay Friday, February 18 2.9 Main Idea: We label the vectors using an index system. Key Words: Dimension, basis. Goal: Learn the basic results about linearly independent spanning sets. Previous Assignment Assignment Page 175 Problems 5,6,31,32,33,34,35,36 Page 175 Problem 5 | 2 | | -4 | | 8 | Let V1 = | 3 |, V2 = | -5 |, and W = | 2 | |-5 | | 8 | |-9 | Determine if W is in the subspace of R^3 generated by V1 and V2. | 2 -4 8 | | 3 -5 2 | |-5 8 -9 | | 1 -2 4 | | 3 -5 2 | |-5 8 -9 | | 1 -2 4 | | 0 1 -10| | 0 -2 11| | 1 -2 4 | | 0 1 -10| No solution therefore W is not in | 0 0 -9| the span of V1 and V2. -------------------------------------------- Page 175 Problem 6 | 1 | | 4 | | 5 | | -4 | Let V1 = | -2 |, V2 = | -7 |, V3 = | -8 |, and U = | 10 | | 4 | | 9 | | 6 | | -7 | | 3 | | 7 | | 5 | | -5 | Determine if U is in the subspace of R^4 generated by {V1,V2,V3}. | 1 4 5 -4 | | -2 -7 -8 10 | | 4 9 6 -7 | | 3 7 5 -5 | | 1 4 5 -4 | | 0 1 2 2 | | 0 -7 -14 9 | | 0 -5 -10 7 | | 1 4 5 -4 | | 0 1 2 2 | | 0 0 0 23 | No solution. Therefore U is not in the span | 0 0 0 17 | of {V1, V2, V3}. ------------------------------------------------ Page 175 Problem 31 Suppose F is a 5x5 matrix whose column space is not 5 equal to R . What cay you say about the Null Space of F. The null space is not trivial. ------------------------------------------------ Page 175 Problem 32 If R is a 6x6 matrix and Null Space of R is not the zero subspace, what can you say about the Column space of R. 6 The column space of R cannot be all of R . ------------------------------------------------- Page 175 Problem 33 4 If Q is a 4x4 matrix and Column Space of Q = R . What can you say about solutions of equations of the form 4 QX = B for B in R . The solution always exists and is unique. ------------------------------------------------- Page 175 Problem 34 If P is a 5x5 matrix and Null Space of P is the zero subspace. What can you say about solutions of 5 equations of the form PX = B for B in R . The solution always exists and is unique. ----------------------------------- Page 175 Problem 35 What can you say about Null Space of B when B is a 5x4 matrix with linearly independent columns. The Null Space of B is the zero space. ------------------------------------------- Page 175 Problem 36 What cay you say about the shape of an mxn matrix A m when the columns of A form a basis for R . The matrix must be square. Equivalently: n = m. ------------------------------------------------------ New Material ------------------------------------------------------------------ There are many many bases for a particular vector space. There is nothing unique about bases except for one thing. Two bases of the same space will always have the same number of vectors. We call this number the dimension of the space. So our first task is to show that it is really true that two bases of the same vector space always have the same number of elements. We actually show something stronger. An independent set is always smaller than a spanning set. Theorem. Consider vectors V1, V2, ..., Vr and W1, W2, ..., Ws n in a subspace of R . If the V's are linearly independent and the W are a spanning set, then r <= s. Proof: [V1 V2 ... Vr ] = [W1 W2 ... Ws] [X1 X2 ... Xr] nxr nxs sxr Independent Spanning set Coefficients <----has this shape---> _______________________ | | If r > s this matrix [X1 X2 ... Xr ] = | | sxr |_____________________| sxr Then there is a vector X =/= 0 such that [X1 X2 ... Xr] X = 0. But then [V1 V2 ... Vr] X = 0 as well and this contradicts the fact that the Vi's are linearly independent. Thus we have to agree that r <= s. That means, the number of vectors in any spanning set must be at least as many as the number of vectors in any linearly independent set. n Theorem: All bases of a subspace V of R have the same number ****** of vectors. Proof: Given a basis V1, V2, ..., Vp and W1, W2, ..., Wq, then p <= q and q <= p. Thus p = q. Definition: The number of elements in any basis of a vector subspace V is called the dimension of V. n Theorem: Consider a subspace V of R with dim(V) = m. (a) We can find at most m linearly independent vectors in V. (b) We need at least m vectors to span V. (c) If m vectors in V are linearly independent, then they form a basis of V. (d) If m vectors span V, then they form a basis of V. Find a basis of the kernel of the matrix. A = | 1 2 0 3 0 | | 2 4 1 9 5 | a b c x y z w u | 1 2 0 3 0 | | 0 0 1 3 5 | | x | |-2 | | -3| | 0 | | y | | 1 | | 0| | 0 | | z | = a| 0 | + b | -3| + c |-5 | | w | | 0 | | 1| | 0 | | u | | 0 | | 0| | 1 | <----These are a basis--> of the Null Space Find a basis of the Range of a matrix. A = | 1 2 0 3 0 | | 2 4 1 9 5 | RCF(A) = | 1 2 0 3 0 | | 0 0 1 3 5 | x x Use these columns of A for a basis of the Range of A. The Range of A is also the Column Space of A. A basis of Range(A) is | 1 | | 0 | | 2 |, | 1 | Notice that the elements of the null space provide the dependence relations for the columns which do not contain stair step ones. -------------------------------------------------------------- When you do Row Canonical form, you are finding a basis of the row space. When you multiply by an invertible matrix P, the row space is preserved. Proof: If P is any matrix, then the rows of PA are linear combinations of the rows of A. Thus RS(PA) c RS(A). When P is invertible, then we get the reverse inclusion since -1 RS(A) = RS(P P A) c RS(PA) c RS(A). Thus RS(PA) = RS(A) when P is invertible. In particular, RS(A) = RS(RCF(A)) and the non zero rows of RCF(A) are linearly independent due to the stair step ones. Thus the dimension of the RS(A) is the number of non zero rows in the RCF(A). Theorem: The Dimension of the Row Space of a Matrix is always the same as the Dimension of the column space. Given a matrix A, choose columns Ci1 Ci2 ... Cir to be a basis of the column space. | x11 ... x1n | | | A = [ Ci1 Ci2 ... Cir ] | | | | | xr1 ... xrn | rxn So A has a spanning set of the rows with r vectors. Thus Dimension of the Column Space is >= the Dimension of the Row Space. We can reverse the argument and prove the reverse inequality. Thus the dimension of the Row Space and Column Space are equal.