Solutions to Math 51 Midterm 1 July 6, 2016

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1 Solutions to Math 5 Midterm July 6, 26. (a) (6 points) Find an equation (of the form ax + by + cz = d) for the plane P in R 3 passing through the points (, 2, ), (2,, ), and (,, ). We first compute two non-collinear vectors parallel to the plane: 2 u = 2 = v = 2 = 2 A normal vector is found by taking the cross product of any two non-collinear vectors which are parallel to P ; taking the cross product of the above two vectors, we get the following normal vector: 2 n = u v = 3 The equation for the plane P then is which is equivalent to 2x y + 3z = 3. n x = (b) (5 points) Can you express this plane as the null space of a matrix? That is, is there a matrix A with N(A) = P? Why or why not? No. is not in the plane because Therefore the plane cannot be the null space of a matrix.

2 Math 5, Summer 26 Solutions to Midterm July 6, 26 Page 2 of 8 2. Suppose A is a 4 2 matrix whose column space C(A) admits the following description: 2 C(A) = {b R 4 b = and b = } (a) (5 points) Find with justification a basis for the the column space of A (hint: if b = express the condition for b C(A) as equations satisfied by the b i ). The conditions for b = b b2 following system of equations: 2 b b2 R 4, C(A) as described in the question above can be described as the b b2 2 b b2 = b + 2b 2 = () = b 2 + =. (2) The question is asking for us to find a basis for the set of all solutions (in vector form) to this pair of equations; this will give us a basis for the subspace C(A). The augmented matrix of this system is. By applying one row operation (subtracting twice the first column from the second column), we can put this augmented matrix into rref: rref( ) = 2 3. (3) This rref matrix has b and b 2 as pivot variables and and as free variables. (Note that since there are two free variables, there should be two vectors in the basis). Writing out the equations, we get b 3 = b 2 + =. Solving for the pivots in terms of frees, we get that b = 3 b 2 =. Writing the solutions in vector form (with the free variables arbitrary elements of R), the solutions can be described as C(A) = { 3b4 = { 3 = span( 3, R} +, R}, ). The vectors produced this way which span our subspace C(A) are linearly independent (this is a general fact, or by inspection in this case, since they are not collinear), and therefore are a basis 3 for C(A). So, a basis for C(A) is {, }.

3 Math 5, Summer 26 Solutions to Midterm July 6, 26 Page 3 of 8 (b) (5 points) Are there zero, one or many solutions to the equation Ax = 3? Explain. The equation Ax = b has solutions precisely when b C(A). The first thing to check is, is this 3 particular b = in C(A)? We need to check whether b satisfies the conditions of being in C(A) described in the problem statement; in particular note that b = = + 6 = 7, (4) so the conditions are not satisfied, and b / C(A). This, there are no solutions to Ax = (It would be sufficient also to note that for this b, b = 3 ). 3. (c) (5 points) Are there zero, one, or many solutions to the equation Ax =? Explain. First, note that b = is in C(A), which one can deduce either by directly verifying that 2 b = b =, or by noting that b is one of the basis vectors of C(A) computed in part (a). So there is at least one solution. When there are solutions to Ax = b, the question of whether there are exactly one solution or many solutions is a question about the null space of A: Recall that if x p is one solution to Ax = b, then the set of all solutions is the set {x p + w w N(A)}, which is a translation of N(A) by x p. So if N(A) = {} there is exactly one solution and if N(A) {} there are many solutions. Since we are not given the matrix A, we have to reason about N(A) indirectly, using the information we are given. In this case, we have determined from part (a) that dim C(A) = 2, so in particular since there are only two columns of A, these columns must be linearly independent. Therefore, by a result in the book, N(A) = {}, and there is only one solution. (Alternatively, since dim C(A) = 2, by the Rank-Nullity theorem dim N(A) = 2 dim C(A) =, so N(A) = {}).

4 Math 5, Summer 26 Solutions to Midterm July 6, 26 Page 4 of 8 3. (a) (4 points) Say what it means for a list of vectors {v,..., v k } to be linearly independent. A set of vectors {v,..., v k } is linearly independent if the only solution to the equation c v v k = is all c s equal to. An equivalent definition is as follows: for k >, a set of vectors is linearly independent if you can t write any one vector as a linear combination of the others. If k =, a vector is linearly independent if it is not the zero vector. (b) (6 points) Suppose {u, v, w} is a linearly independent set of vectors in R n. Is the collection {u + v, u 2v, w + u v} linearly independent or linearly dependent? Show your work. We want to solve Rearranging the terms we get : c (u + v) + c 2 (u 2v) + c 3 (w + u v) = (c + c 2 + c 3 )u + (c 2c 2 + c 3 )v + (c 3 )w = Because w, u, v are linearly independent it follows that the coefficients in the equation above must all be. This implies that that c + c 2 + c 3 = (5) c 2c 2 + c 3 =. (6) c 3 = (7) The last equation implies that c 3 = ; substituting this into the first two equations, and solving by a method of one s choice (for instance, using substitution) implies that c = c 2 =. Therefore since c = c 2 = c 3 = was the only solution, our initial vectors are linearly independent.

5 Math 5, Summer 26 Solutions to Midterm July 6, 26 Page 5 of 8 4. (a) (4 points) Say what it means for a function T : R n R m to be a linear transformation. The function T as above is a linear transformation if the following two conditions are satisfied: T (x + y) = T (x) + T (y) for any vectors x, y R n ; and T (cx) = ct (x) for any vector x R n and any scalar c R. (b) (5 points) Let v R n be any non-zero vector. Recall that on homework, we defined the projection onto the vector v, P roj v : R n R n, by the following formula: Proj v (x) = ( x v v v) v. It is a fact mentioned in class that P roj v : R n R n is a linear transformation. Now, let P R n be a plane through the origin, described as P = span(v, v 2 ) for a pair of perpendicular non-zero vectors v, v 2. We can define the projection onto the plane P via the following formula: Proj P : R n R n Proj P (x) := Proj v (x) + Proj v2 (x). Show directly from definitions that Proj P is a linear transformation. You may use the fact that Proj v is a linear transformation for any v without justification (though you should state how you are using it). We check the conditions in the previous part: For any vectors x, y R n, Proj P (x + y) = Proj v (x + y) + Proj v2 (x + y) (8) = (Proj v (x) + Proj v (y)) + (Proj v2 (x) + Proj v2 (y)) (9) = (Proj v (x) + Proj v2 (x)) + (Proj v (y) + Proj v2 (y)) () = Proj P (x) + Proj P (y). () where the first line uses the definition of Proj P, the second line used the fact that Proj v and Proj v2 were linear transformations, the third line redistributed terms, and the last line is again the definition of Proj P. For any vector x R n and any scalar c R, Proj P (cx) = Proj v (cx) + Proj v2 (cx) (2) = cproj v (x) + cproj v2 (x) (3) = c(proj v (x) + Proj v2 (x)) (4) = cproj P (x). (5) where once more the first line is the definition of Proj P, the second line uses the fact that Proj v and Proj v2 were linear transformations, the third line factored out the c, and and the last line is the definition of Proj P.

6 Math 5, Summer 26 Solutions to Midterm July 6, 26 Page 6 of 8 (c) (6 points) Now let n = 3 (so we are working in R 3 ), and let v = / 2 / 2 P = span(v, v 2 ). Compute the matrix associated to Proj P in this case., v 2 = / 2 / 2, and You may use the following fact if it is helpful: Both v and v 2 are unit-length vectors, meaning v = v 2 =. (Note that when m R n is unit length, the formulae for projection onto m simplifies to: Proj m (x) = (x m)m). For any linear transformation T : R n R n, the matrix associated to T is T (e ) T (e 2 ) T (e n ). Thus, the matrix for Proj P : R 3 R 3 is Proj P (e ) Proj P (e 2 ) Proj P (e 3 ), so we simply need to compute Using the simplified formula above, since v and v 2 are both unit-length vectors, we compute that Similarly, Proj P (e ) = Proj v (e ) + Proj v2 (e ) (6) / 2 / 2 / 2 / 2 = (e / 2 ) / 2 + (e / 2 ) / 2 (7) = (/ / 2 2) / 2 + (/ / 2 2) / 2 (8) = = e. (9) Finally, Proj P (e 2 ) = Proj v (e 2 ) + Proj v2 (e 2 ) (2) / 2 / 2 / 2 / 2 = (e 2 / 2 ) / 2 + (e 2 / 2 ) / 2 (2) = ( / / 2 2) / 2 + (/ / 2 2) / 2 (22) = = e 2. (23) Proj P (e 3 ) = Proj v (e 3 ) + Proj v2 (e 3 ) (24) / 2 / 2 / 2 / 2 = (e 3 / 2 ) / 2 + (e 3 / 2 ) / 2 (25) / 2 / 2 = () / 2 + () / 2 (26) = =, (27) So, the matrix associated to Proj P is e e 2 =.

7 Math 5, Summer 26 Solutions to Midterm July 6, 26 Page 7 of 8 5. (2 points) Find, showing all your work and using any method of your choice, all solutions (in vector form) to the system of equations: Describe your solutions as a point, line, or plane. 2 x y = z 3 We solve the system by writing the associated augmented matrix: The augmented matrix is : R 3 3R 5 R 3/6 2 2 R 2R 2 R R R2/5 R R We have two pivots variables x and x 2 ; solving for them in terms of the free variable x 3 (using the rref system), we see that x = 3x 3 + and x 2 = 2x 3 2. Thus, letting the free variable x 3 range over R, the set of solutions is 3x x 3 2, x 3 R = x , x 3 R x 3 This is the parametric representation of a line.

8 Math 5, Summer 26 Solutions to Midterm July 6, 26 Page 8 of 8 6. (2 points) (2 points each) Each of the statements below is either always true ( T ), or always false ( F ), or sometimes true and sometimes false, depending on the situation ( MAYBE ). For each part, decide which and circle the appropriate choice; you do not need to justify your answers. (a) Let A be a 3 4 matrix. If dim C(rref(A)) = 2, then dim C(A) = 2. T F MAYBE True. The dimension of C(A) is the number of pivots in rref(a), which is also the dimension of C(rref(A)). (b) If P R n is a plane going through the origin, and u, v, and w are three vectors in P, then P is spanned by u, v, and w. T F MAYBE Maybe. If at least two of them are linearly independent, then yes. If all of them are collinear, then no. (c) If the reduced row echelon form of the augmented matrix A b has two free variables, then there are many solutions to the associated system. T F MAYBE Maybe. If b is in the column space of A, then yes. Otherwise, there are no solutions. (d) If A is a 2 3 matrix, then N(A) {}. T F MAYBE True. There are at most two pivots for A, so there has to be at least one free variable, which implies the dimension of the null space is at least. Therefore there are infinitely many elements in N(A), so indeed N(A) {}. (e) If A is a 3 2 matrix, then N(A) {}. T F MAYBE Maybe. If rref(a) has two pivots (equivalently if the two columns of A are linearly independent), then N(A) = {}. If rref(a) has one or fewer pivor variables (equivalently if the two columns of A are linearly dependent), then indeed N(A) {}. (f) Suppose V is a subspace in R n of dimension d, and a collection of d vectors {v,..., v d } spans V. Then this collection is also linearly independent. True. This is an immediate application of Levandosky Proposition 2.3. T F MAYBE