Math 250B Midterm II Review Session Spring 2019 SOLUTIONS

Transcription

1 Math 250B Midterm II Review Session Spring 2019 SOLUTIONS [ Problem #1: Find a spanning set for nullspace SOLUTION: The row-reduced form of this matrix is Setting the variables corresponding to the columns as x y z u and v respectively we see that x = 2y + 2u 3v Hence the nullspace is given by {( 2y + 2u 3v y z u v) : y z u v R} To obtain a spanning set we bust the variables out to get {( ) ( ) ( ) ( )} Problem #2: Decide whether the set {2 x x + x 2 3 x} is linearly dependent or linearly independent Does this set span P 2 (R)? SOLUTION: These vectors exist in P 2 (R) which is 3-dimensional Since 3 vectors are given that is the correct number of vectors They might be linearly independent To check this we arrange the coefficients of each polynomial into the columns of a matrix: We can quickly check that this matrix has a non-zero determinant Thus the vectors form of a basis for P 2 (R) which means that they set is both linearly independent and spans P 2 (R) Problem #3: Show the following is a subspace (or argue why it is not): (a): V = M 23 (R) and S = {A M 23 (R) : the rows of A sum to 1 } SOLUTION: NOT A SUBSPACE The zero matrix does not belong to S so Axiom (A5) fails (b): V = M 23 (R) and S = {A M 23 (R) : the rows of A sum to 0 } a b (a + b) SOLUTION: The general form for the matrices in S is We c d (c + d) claim that S is a subspace which is verified by showing that S is closed under addition and scalar multiplication

2 Closure under Addition: Let a b (a + b) A = c d (c + d) and B = [ x y (x + y) z w (z + w) be in S Then note that A + B = [ a + x b + y (a + b + x + y) c + z d + w (c + d + z + w) Since the rows of A + B each add up to zero we conclude that A + B S as needed Closure under Scalar Multiplication: Let A be as above and let k R be a scalar Then ka kb k(a + b) k A = kc kd k(c + d) Again observe that the rows of k A sum to zero so we conclude that k A S as needed (c): Find a spanning set for S in part (b) SOLUTION: Busting out the four variables from the general form of the matrix in part (b) we get the spanning set immediately: { } Problem #4: Let A = Find bases for nullspace(a) rowspace(a) and colspace(a) SOLUTION: The row-echelon form of A is For the nullspace of A we set z = t and y = r as free variables Using REF(A) we then find that x = t Thus vectors in the nullspace of A have the form (t r t) Busting out the variables the basis is {(1 0 1) (0 1 0)}

3 A basis for rowspace(a) is given by {(1 0 1)} and a basis for colspace(a) is given 1 by 3 1 Problem #5: Suppose T : P 2 (R) M 2 (R) is a linear transformation such that T (1) = T (x + 1) = T (x 2 + x + 1) = Compute T (ax 2 + bx + c) SOLUTION: Note that T (x) = T (x + 1) T (1) = [ and Therefore [ = a [ T (x 2 ) = T (x 2 + x + 1) T (x + 1) = T (ax 2 + bx + c) = at (x 2 ) + bt (x) + ct (1) = [ a 2b c 2a 2b + 2c + b + c = a + 2b b Problem #6: Let T : P 2 (R) P 2 (R) be defined by T (p(x)) = xp (x) + 2p (x) Show that T is a linear transformation SOLUTION: We must show that T respects addition and scalar multiplication: T respects addition: We have T (p 1 (x) + p 2 (x)) = x(p 1 + p 2 ) (x) + 2(p 1 + p 2 ) (x) = as needed (xp 1(x) + 2p 1(x)) + (xp 2(x) + 2p 2(x)) = T (p 1 (x)) + T (p 2 (x))

4 T respects scalar multiplication: We have T (k p(x)) = x(kp) (x)+2(kp) (x) = xkp (x)+2kp (x) = k(xp (x)+2p (x)) = kt (p(x)) as needed Problem #7: Find a basis for V := span{1 2x+3x 3 2 5x 3x 2 +6x 3 x+3x 2 2 x+4x 2 7x 3 5 8x+x 2 +2x 3 } SOLUTION: The five given polynomials already span V by the very construction of V itself The only thing we need to do is remove any linear dependencies To find them we arrange the coefficients of the polynomials on a column-by-column basis in a matrix augmented by zero: Upon row-reduction this becomes The 3rd and 5th columns are un-pivoted which means they introduce linear dependencies Therefore we should eliminate the 3rd and 5th polynomial from our set of polynomials in order to get a basis: {1 2x + 3x 3 2 5x 3x 2 + 6x 3 2 x + 4x 2 7x 3 } Problem #8: (a): Let V := M 2 (R) Consider the vectors { [ S := [ }

5 Are the vectors linearly independent spanning set for V both or neither? SOLUTION: LINEARLY INDEPENDENT ONLY Since V is 4-dimensional and we are only given 3 vectors we know that S cannot possibly span V The only question is whether it is linearly independent or not Placing the vectors into the columns of an augmented matrix and row-reducing we obtain We see that no free variable arises here so the set is linearly independent (b): Repeat part (a) for V := P 3 (R) and the set S := {1 + x + 3x 2 + 3x 3 x + x 2 2 x + x x + 6x 2 + x 3 } SOLUTION: NEITHER Let us check linear independence first We are seeking a non-trivial solution to a(1 + x + 3x 2 + 3x 3 ) + b(x + x 2 ) + c(2 x + x 3 ) + d( 3 + 6x + 6x 2 + x 3 ) = 0 That is (a + 2c 3d) + (a + b c + 6d)x + (3a + b + 6d)x 2 + (3a + c + d)x 3 = 0 So we must have a + 2c 3d = 0 a + b c + 6d = 0 3a + b + 6d = 0 3a + c + d = 0 Putting these equations into an augmented matrix yields Putting this into row-echelon form yields

6 If we set d = t then the third equation gives c = 2d = 2t Then the second equation gives b = 3c 9d = 6t 9t = 3t Then the first equation gives a = 3d 2c = 3t 4t = t So we have (a b c d) = ( t 3t 2t t) Setting t = 1 gives a = 1 b = 3 c = 2 and d = 1 That is the vectors are linear dependent and we have derived the linear dependency: (1 + x + 3x 2 + 3x 3 ) 3(x + x 2 ) + 2(2 x + x 3 ) + ( 3 + 6x + 6x 2 + x 3 ) = 0 Since the vectors are linearly dependent they cannot span a 4-dimensional space (even though they are 4 vectors) Thus they also cannot span V which is 4-dimensional