Bases for column and null spacesBas es for null spacesTo find a basis for Nul(A):• find the parametric form of the solutions to Ax = 0,• express solutions x as a linear combination of vectors with the free variables ascoefficients;• these vectors form a basis of Nul(A).Example 1. F i nd a basis for Nul(A) withA =3 6 6 3 96 12 13 0 3.Solution.Bas es for column spacesRecall that the columns ofA are i ndependentAx = 0 has only the trivial s o lution (namely, x = 0),A has no free variables.A basis for Col(A) is given by the pivot columns of A.Armin [email protected] 2. F i nd a basis for Col(A) withA =1 2 0 42 4 −1 33 6 2 224 8 0 16.Solution.Warning: For the basis of Col(A), you have to take the colu mns of A, not the columnsof an echelon form.Dimension of Col(A) and Nul(A)Theorem 3. Let A be an m × n matrix. Then:• dim Col(A) is the number of pivots of A• dim Nul(A) is the number of free variables of A• dim Col(A) + dim Nul(A) =Proof. Example 4. Suppose A =1 2 3 42 4 7 8. Find the dimension of Col(A) and N u l(A).Solution.Armin [email protected] four fundamental subspacesRow space and le ft null spaceDefinition 5.•The row space of A is the column space of AT.• The left null space of A is the null space of AT.Definition 6. The rank of a A is the number of its pivots.Theorem 7. (Fundamental Theorem of Linear Algebra, Part I)Let A be an m × n matrix of rank r.• dim Col(A) = r• dim Col(AT) = r• dim Nul(A) = n − r (number of free variables of A)• dim Nul(AT) = m − rExample 8. Obs er ve that the colum n space and the row space have the same dimension!Easy to see for a matrix in echelon f orm2 1 3 00 0 1 20 0 0 7,but not obvious for a ra n dom matr ix.In particular, ann ×n matrix has independent columns if and only if it ha s independentrows.Armin [email protected] 9. (Existence o f inverses)Let A be an n × n matrix. Then th e f ollowing statements are equivalen t:(a) A is invertible.(b) A is row eq uivalent to In.(c) A has rank n. (that is, A has n pivots)(d) The columns of A span Rn.In o ther words, for everyb, the system Ax = b has at least one solution.(e) The columns of A are inde pendent.In o ther words, the systemAx = 0 has only the solution x = 0.(f) For every b, the system Ax = b has a u nique soluti on.Practice problemsExample 10. Suppose A is a 5 × 5 matrix, and that v is a vector in R5which i s nota linear combination of the columns of A.What can you say about the nu mber of s olutions toAx = 0?Solution.True or false?• An n × n matrix A is invertible if and only if Nul(A) = {0}.• An n × n matrix A is invertible if and only if the rows of A span Rn.• An n × n matrix A is invertible if and only if the rows of A are independent.Armin
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