44.6© 2012 Pearson Education, Inc.Vector SpacesRANKSlide 4.6- 2© 2012 Pearson Education, Inc.THE ROW SPACE If A is an matrix, each row of A has n entries and thus can be identified with a vector in . The set of all linear combinations of the row vectors is called the row space of A and is denoted by Row A. Each row has n entries, so Row A is a subspace of . Since the rows of A are identified with the columns of AT, we could also write Col ATin place of Row A.mnnnSlide 4.6- 3© 2012 Pearson Education, Inc.THE ROW SPACE Theorem 13: If two matrices A and B are row equivalent, then their row spaces are the same. If B is in echelon form, the nonzero rows of B form a basis for the row space of A as well as for that of B. Proof: If B is obtained from A by row operations, the rows of B are linear combinations of the rows of A. It follows that any linear combination of the rows of B is automatically a linear combination of the rows of A.Slide 4.6- 4© 2012 Pearson Education, Inc.THE ROW SPACE Thus the row space of B is contained in the row space of A. Since row operations are reversible, the same argument shows that the row space of A is a subset of the row space of B. So the two row spaces are the same.Slide 4.6- 5© 2012 Pearson Education, Inc.THE ROW SPACE If B is in echelon form, its nonzero rows are linearly independent because no nonzero row is a linear combination of the nonzero rows below it. (Apply Theorem 4 to the nonzero rows of B in reverse order, with the first row last). Thus the nonzero rows of B form a basis of the (common) row space of B and A.Slide 4.6- 6© 2012 Pearson Education, Inc.THE ROW SPACE Example 1: Find bases for the row space, the column space, and the null space of the matrix Solution: To find bases for the row space and the column space, row reduce A to an echelon form:25 801713 51 5311 197 117135 3Slide 4.6- 7© 2012 Pearson Education, Inc.THE ROW SPACE By Theorem 13, the first three rows of B form a basis for the row space of A (as well as for the row space of B). ThusBasis for Row 13 5 1 50122700042000000AB:{(1,3, 5,1,5),(0,1, 2,2, 7),(0,0,0, 4,20)}A Slide 4.6- 8© 2012 Pearson Education, Inc.THE ROW SPACE For the column space, observe from B that the pivots are in columns 1, 2, and 4. Hence columns 1, 2, and 4 of A (not B) form a basis for Col A:Basis for Col Notice that any echelon form of A provides (in its nonzero rows) a basis for Row A and also identifies the pivot columns of A for Col A.250131:,,3117175A      Slide 4.6- 9© 2012 Pearson Education, Inc.THE ROW SPACE However, for Nul A, we need the reduced echelon form. Further row operations on B yield10 10 101 20 300 01 500 00 0ABCSlide 4.6- 10© 2012 Pearson Education, Inc.THE ROW SPACE The equation is equivalent to , that is, So , , , with x3and x5free variables.x0Ax0C13523545023050xxxxxxxx135xxx 23523xxx455xxSlide 4.6- 11© 2012 Pearson Education, Inc.THE ROW SPACE The calculations show thatBasis for Nul Observe that, unlike the basis for Col A, the bases for Row A and Nul A have no simple connection with the entries in A itself.1123:,100501A        Slide 4.6- 12© 2012 Pearson Education, Inc.THE RANK THEOREM Definition: The rank of A is the dimension of the column space of A. Since Row A is the same as Col AT, the dimension of the row space of A is the rank of AT. The dimension of the null space is sometimes called the nullity of A. Theorem 14: The dimensions of the column space and the row space of an matrix A are equal. This common dimension, the rank of A, also equals the number of pivot positions in A and satisfies the equationmnrank dim Nul AAnSlide 4.6- 13© 2012 Pearson Education, Inc.THE RANK THEOREM Proof: By Theorem 6, rank A is the number of pivot columns in A. Equivalently, rank A is the number of pivot positions in an echelon form B of A. Since B has a nonzero row for each pivot, and since these rows form a basis for the row space of A, the rank of A is also the dimension of the row space. The dimension of Nul A equals the number of free variables in the equation . Expressed another way, the dimension of Nul A is the number of columns of A that are not pivot columns.x0ASlide 4.6- 14© 2012 Pearson Education, Inc.THE RANK THEOREM (It is the number of these columns, not the columns themselves, that is related to Nul A). Obviously, This proves the theorem. number of number of number ofpivot columns nonpivot columns columns   Slide 4.6- 15© 2012 Pearson Education, Inc.THE RANK THEOREM Example 2:a. If A is a matrix with a two-dimensional null space, what is the rank of A?b. Could a matrix have a two-dimensional null space? Solution:a. Since A has 9 columns, , and hence rank .b. No. If a matrix, call it B, has a two-dimensional null space, it would have to have rank 7, by the Rank Theorem.7969(rank ) 2 9A7A69Slide 4.6- 16© 2012 Pearson Education, Inc.THE INVERTIBLE MATRIX THEOREM (CONTINUED) But the columns of B are vectors in , and so the dimension of Col B cannot exceed 6; that is, rank B cannot exceed 6. Theorem: Let A be an matrix. Then the following statements are each equivalent to the statement that A is an invertible matrix.m. The columns of A form a basis of .n. Colo. Dim Colp. rank6nnnnAAnAnSlide 4.6- 17© 2012 Pearson Education, Inc.RANK AND THE INVERTIBLE MATRIX THEOREMq. Nulr. Dim Nul Proof: Statement (m) is logically equivalent to statements (e) and (h) regarding linear independence and spanning. The other five statements are linked to the earlier ones of the theorem by the following chain of