Solutions to Section 2.9 Homework Problems Problems 1 5, 7, 9, 10 15, (odd), and 38. S. F. Ellermeyer June 21, 2002

Transcription

1 Solutions to Section 9 Homework Problems Problems 9 (odd) and 8 S F Ellermeyer June The pictured set contains the vector u but not the vector u so this set is not a subspace of The pictured set contains the vectors u and v but not the vector u v so this set is not a subspace of The picture set contains some short vectors in the second quadrant but not all scalar multiples of these vectors so this set is not a subspace of The pictured set contains the vector u u so this set is not a subspace of but not the vector To determine if w is in the subspace generated by v and v we must determine if w can be written as a linear combination of v and v The equivalence shows that w is not in the subspace generated by v and v (because the equation v v x w is inconsistent) a There are three vectors in the set v v v b There are infinitely many vectors in ColA c The equivalence shows that p is in ColA In particular we see that p 9 v v (and there are infinitely many other ways to write p as a linear

2 combination of v v and v ) 9 NulA consists of all solutions of the equation Ax Note that 6 Ap so p is not in NulA Au 6 6 so u is in NulA NulA is a subspace of and ColA is a subspace of NulA is a subspace of and ColA is a subspace of Using the equivalence 9 9 we see that a non zero vector in NulA is x (This was obtained by setting both free variables x and x equal to ) Obtaining a non zero vector in ColA is easy: We just take a linear combination of the columns of A such as a a a a Using the equivalence

3 we see that a non zero vector in NulA is x A non zero vector in ColA is a a a This is a set of two linearly independent vectors in and hence is a basis for Since 6 these vectors form a basis for 9 Since has dimension every basis for must contain exactly three vectors Thus the given vectors do not form a basis for (since there are only two vectors in the given set) Since columns and are the pivot columns of A a basis for ColA is a a where Also since a A and a 8 we see that the general solution of Ax is x x x and since the vectors that appear on the right hand side of the above 8

4 equation are linearly independent a basis for NulA is v v where v and v Since columns and are the pivot columns of A a basis for ColA is a a a where 8 a a and a 6 Also since 9 A we see that the general solution of Ax is x x x and since the vectors that appear on the right hand side of the above equation are linearly independent a basis for NulA is v v where v and v The subspace of spanned by the given set of vectors is the column space of the matrix 9 9

5 Since A A 8 we see that the pivot columns of A are columns and Thus a basis for the subspace spanned by the given vectors is a a a where a a a If A is a x matrix with three pivot columns then the vectors in these three columns are a basis for ColA Thus ColA is three dimensional and is a subset of This means that ColA Since the equation Ax has two free variables NulA is two dimensional However note that NulA is a subset of Thus NulA is a two dimensional subspace of In this case we should not say that NulA but rather that NulA a subspace of that is isomorphic to We picture NulA as a plane in passing through the origin in (although it is impossible for us to actually visualize a five dimensional space) 9 The coordinate vector of x relative to the basis B is the solution of the equation Ax b which shows that x B Note that the subspace in this problem is a two dimensional subspace of (a plane passing through the origin in ) Also

6 so x B If the null space of a x matrix is four dimensional then the column space of this matrix is three-dimensional (This is by the Rank Theorem) If A is a x6 matrix and the solution space of Ax has a basis consisting of two vectors then NulA is two dimensional which means that ColA is four dimensional (by the Rank Theorem) Thus RankA a False (or really just stated in a way that does not make sense) A subspace of n is a subset H of n such that i The zero vector is in H ii If u and v are any two vectors in H then the vector u v is in H iii If u is any vector in H and c is any scalar then the vector cu is in H b True c False It is a subspace of m d False The pivot columns of A form a basis for ColA e True 8 a False See the answer to problem a where a correct definition of the term subspace is given b False The column space of A is the space spanned by the columns of A c True d True e True 6