Alg1231W2T4 - Column space of a matrix

Wednesday, 25 July 2018

8:46 PM

Machine generated alternative text: For the matrix consider the set An element of S has the form A 3 -4 1 1 -1 2 -8 10 -6 s Ax for some y , which can be written as the linear combination of vectors So indeed S deserves name of the column space O of A, as it consists of all linear combinations ot columns Recall: the Maple notation for a vector in R is< a,b,c >. O of A.

 

For the matrix the set is the column space of A. The vector v has [select all that apply] a unique solution infinitely many solutions Cl no solution. A 2 2 2 -3 -1 3 2 3 5 S b e R : b = Ax for some y belongs to this set whenever the augmented matrix 2 2 2 -3 -1 3 By row reducing, we find that saying v e S is equivalent to and z satisfying a linear condition of the form ax + by + cz In this case the column space of A is a plane tnrougn tne origin Note: • the Maple syntax for the equation ax + by + cz = 0 is • make sure to explicitly specify multiplication using an asterisk * 0 _ Specifically, this condition is equivalent to

 

Machine generated alternative text: For the matrix the set is the column space of A. The vector v has [select all that apply] a unique solution infinitely many solutions Cl no solutions 2 8 4 4 16 8 3 12 6 S b e R : b = Ax for some xeR3} y belongs to this set whenever the augmented matrix 2 8 4 {Z-æx=o, 4 16 8 12 y There are ftvo linear conditions that must be satisfied; namely: Note: enter the conditions above in the form of a set In this case the column space of A is a line (e.g. O with parametric vector form x = Ab where Recall: the Maple notation for a vector in IR3 is< a,b,c

 

Machine generated alternative text: For the matrix consider the set s : b = Ax 3 Let's show that S is indeed a subspace of R 1 -4 1 So a + b e S tor some 4 -2 2 -6 10 -4 -1 4 -1 for some i) S contains the zero-vector. Since Recall: the Maple syntax for a vector in R ii) S is closed under addition If a, b e S Ax where is< a,b,c,d then: • a = Ax tor some . b = Ay tor some O x e IR and 4 O y €zlR Thus a + b = Ax + Ay = Az where z iii) S is closed under scalar multiplication Suppose b e S. and so, if * e R , then = M Ax) = Az where z o e IR4 Then b = Ax Rx xeR4} 4 O x e IR, O e IR and so Abe S. 3 Thus; using the Subspace Theorem; we can conclude that S is a vector subspace of R

 

Machine generated alternative text: For the matrix consider the set A 4 -3 4 2 -1 4 The vector The vector The vector 4 -3 4 2 -1 4 6 -4 8 e S since e S since e S since 4 -3 4 2 -1 4 6 -4 8 s Ax where Ay where Az where Ax for some xeR2} Recall: the Maple syntax for a vector in R is< a,b

 

 

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