EXAM Exam # Math 6 Summer II Morning Class Nov 5 ANSWERS
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Problem Consider the matrix 6 pts A = 6 4 9 5 7 6 5 5 5 4 The RREF of A is the matrix R = A Find a basis for the nullspace of A Solve the homogeneous system Ax = The system Rx = has the same solutions Call the variables x x x 6 Looking at R we see that x x and x 4 are leading variables and the rest are free variables Set x = α x 5 = β and x 6 = γ Working up from the bottom of the matrix we get x 4 x 5 x 6 = = x 4 = x 5 + x 6 = β + γ x + x + x 5 = = x = x x 5 = α β x x x 5 + x 6 = = x = x + x 5 + x 6 = α + β + γ This gives the following parameterization of the nullspace x x x x 4 x 5 x 6 = α + β + γ α β α β + γ β γ = α + β + γ Thus a basis for the nullspace of A is
B Find a basis for the rowspace of A A basis for the rowspace of A is given by the nonzero rows of the RREF Hence a basis for the rowspace of A is [ ] [ ] [ ] C Find a basis for the columnspace of A The leading entries in R are in columns and 4 A basis of the columnspace of A is formed by the corresponding columns of A Thus a basis for the columnspace of A is 6 5 pts Problem Let A be a 5 6 matrix and let B be a 4 4 matrix A What is the largest possible value of the rank of A? The rank of A is defined as the value of two equal quantities: the dimension of the rowspace of A and the dimension of the columnspace of A Thus the rank must be less than or equal to the number of rows and less than or equal to the number of columns In the present case the largest possible value for the rank is 5 B If the nullspace of A has dimension what is the rank of A? The rank of A plus the nullity of A is equal to the number of columns in A The nullity is the dimension of the nullspace Thus we must have rank + = 6 so the rank of A is C If the rowspace of B has dimension what is the dimension of the nullspace of B? The dimension of the rowspace is the same as the rank of B The rank plus the nullity is equal to the number of columns Thus we have + nullity = 4 so the nullity (which is the dimension of the nullspace) is
4 pts Problem Consider the following three vectors in R 4 v = v = v = In each part determine if the given vector is in span(v v v ) If so express it as a linear combination of v v and v A B w = 7 w = 5 6 First put the vectors v v v w and w into a matrix A Thus 7 A = 5 6 The RREF of A is R = In R we have col 4 (R) = col (R) col (R) + col (R) The columns of A have the same relation Expressing this in terms of the original vectors we see that w = v v + v
so w span(v v v ) In R the last column is not in the span of the first three columns (any linear combination of the first three columns would have a zero in the bottom row) Thus the same holds for the columns of A We conclude that w / span(v v v ) Alternatively one can view the one in the fourth row of the last column of R as showing that the system Ax = w is inconsistent so w is not a linear combination of the columns of A and so it s not a linear combination of v v v 4 pts Problem 4 Determine if the following vectors in R 4 are linearly independent Justify your answer 4 v = 4 7 v = 5 9 v = 4 Put these vector in a matrix A so 4 A = 4 5 7 9 4 The RREF of A is R = The columns of R are independent so the same must be true of the columns of A Thus v v and v are independent Another way of looking at this computation is to regard R as saying there are no free variables in the system Ax = so this system has only the trivial solution Thus the columns of A satisfy only the trivial linear relation and so are independent 4
6 pts Problem 5 Pare the following set of vectors down to a basis of R Express the vectors that are not in the basis as linear combinations of the basis vectors 4 v = v = v = v 4 = v 5 = Put the vectors in as the columns of a matrix A so 4 A = The RREF of A is R = The leading entries of R are in columns and 4 Thus the corresponding columns of A form a basis for the column space of A which is the same as S = span(v v v v 4 v 5 ) Thus v v and v 4 form a basis of S Since S is a -dimensional subspace of R S = R Thus v v and v 4 are a basis of R We need to express v and v 5 as linear combinations of our basis vectors We can read off these linear relationships from the columns of R Thus we have col (R) = col (R) + col (R) so we must have v = v + v Similarly col 5 (R) = col (R) + col (R) + col 4 (R) so we have v 5 = v + v + v 4 4 pts Problem 6 The two matrices A = B = are row equivalent so they have the same rowspace Do they have the same columnspace? Explain As many people pointed out row equivalent matrices need not have the same column space But in a particular example they may or may not have the 5
same column space you have to check your particular matrices For example the matrices [ ] [ ] are row equivalent and have the same columnspace (namely R ) A quick reference to a calculator shows that B is the RREF of A This gives the vectors a = a = () as a basis of the columnspace of A and the vectors e = e = () as a basis for the columnspace of B Certainly these are different lists of vectors but what we want to compare are the spans of these sets of vectors It is certainly possible to have span(v v ) = span(w w ) even though v v and w w are different lists of vectors (think of different bases for R ) Thus merely noting that we have different lists of vectors in () and () is not enough to show that the columnspaces are different; we have to consider the particular vectors involved If the columnspaces of the matrices where equal we would have span(a a ) = span(e e ) If this were true we would have (in particular) a span(e e ) If this is true we would be able to write a as a linear combination of e and e ie we would have a = αe + βe for some scalars α and β Then we would have = α + β = But this is impossible because the right-hand side will always have a zero in the bottom row Thus a is not in the columnspace of B so the columnspaces of the two matrices are not equal α β 6