Math 215 HW #4 Solutions


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1 Math 5 HW #4 Solutions. Problem.3.6. Choose three independent columns of U. Then make two other choice. Do the same for A. You have found bases for which spaces? U = and A = Solution: The most obvious way to choose three independent columns of U is to pick the three pivot columns, namely, 3 6, Another choice of three independent columns would be the first, third and fourth columns, which are independent because, if = c + c c 3 and 9 = 9 c + 4c + c 3 7c 9c 3 then it must be the case that c = c 3 = (from the second and third components), so the first component reduces to = c, so c = as well. We could also have chosen the second, third, and fourth columns. independent, suppose 3 4 3c + 4c + c 3 = c 6 + c 7 + c 3 9 = 6c + 7c 9c 3, To see that these are Then the third component implies that c 3 = and the first two component give the system of equations Interpreting via the augmented matrix 3c + 4c = 6c + 7c =. [ from row yields [ 3 4 and reducing by subtracting twice row.
2 Then c = and, since 3c + 4c =, this means that c = as well. So, as indicated, the second, third, and fourth columns of U are also independent. Since U is just the echelon form of A (in particular, you can get from A to U by subtracting twice row from row 4), the relationship between the columns of A is the same as the relationship between the columns of U. Hence, we can translate the three choices made above into three different choices of independent columns of A. Namely, the first, second, and fourth columns of A are independent, as are the first, third, and fourth columns of A, and the second, third, and fourth columns of A are also independent.. Problem.3.. Find two independent vectors on the plane x + y 3z t = in R 4. Then find three independent vectors. Why not four? This plane is the nullspace of what matrix? Solution: Notice that and are vectors in the plane x + y 3z t =. They re independent since 3 c + 3c = c + c = c c implies that c = = c. The vector 3 is also in this plane and we can see that this collection of three vectors is linearly independent as follows: suppose 3 = c + c + c 3 = c + 3c + c 3 c c c 3 Then it must be the case that c = c = c 3 =, so we see that the vectors 3,, and are linearly independent vectors in the plane x + y 3z t = in R 4. There cannot be four linearly independent vectors in this plane because any collection of four linearly independent vectors in R 4 must span all of R 4. Since there are clearly vectors in R 4 that are not in the plane (for example, ), this cannot be the case.
3 Any vector in the plane x + y 3z t = is a solution to the matrix equation 3 x 3 y 3 z = 3 t and, vice versa, any solution to this matrix equation definitely lies in the plane. Therefore, the plane x + y 3z t = is the nullspace of the matrix (Incidentally, this provides another check that there can t be four linearly independent vectors in the plane. Since this matrix clearly has rank, we know that the dimension of the nullspace is 4 = 3, so the plane x + y 3z t =, which is the same as the nullspace, is also threedimensional and so cannot contain four linearly independent vectors) 3. Problem.3.6. Suppose S is a fivedimensional subspace of R 6. True or false? (a) Every basis for S can be extended to a basis for R 6 by adding one more vector. Answer: True. Just pick any vector in R 6 that is linearly independent from the given basis (there must be lots of them, since R 6 is 6dimensional and S is 5dimensional). Then the set consisting of the given basis plus this new vector is, by construction, linearly independent and spans a 6dimensional space, so it must span all of R 6. Any set of linearly independent vectors that spans all of R 6 is a basis for R 6, so this is indeed a basis for R 6. (b) Every basis for R 6 can be reduced to a basis for S by removing one vector. Answer: False. Suppose S is the fivedimensional subspace described by the equation x + x + x 3 + x 4 + x 5 + x 6 =. (One way to see this is both a subspace and fivedimensional: by the same reasoning as in Problem, it is the nullspace of the 6 6 matrix with all entries equal to.) Consider the standard basis for R 6 :,,,,,. Then none of the elements of this basis lies in S, so deleting one of them cannot possibly give a basis for S. 4. Problem
4 (a) Find all functions that satisfy dy dx =. Solution: Integrating both sides of the equation, we see that y(x) = + C = C, where C is some constant. Thus, the functions that satisfy dy dx constant functions. (b) Choose a particular function that satisfies dy dx = 3. Solution: Notice that d = are exactly the dx (3x) = 3, so y p(x) = 3x is a particular function that solves dy dx = 3. (c) Find all functions that satisfy dy dx = 3. Solution: I claim that all functions that satisfy dy dx = 3 are of the form y(x) = y p (x) + C = 3x + C. One the one hand, this is completely obvious from our knowledge of Calculus. On the other hand, we can prove this just using parts (a) and (b) above. In particular, suppose y(x) is some solution to dy dx = 3. Then d dx (y(x) y p(x)) = dy dx dy p dx = 3 3 =, so y(x) y p (x) is a solution to the equation dy dx =. Hence, by part (a), we know that for some constant C, so we see that y(x) y p (x) = C y(x) = y p (x) + C = 3x + C. 5. Problem.4.. If Ax = b always has at least one solution, show that the only solution to A T y = is y =. Hint: What is the rank? Solution: If A is an m n matrix and Ax = b always has at least one solution for any choice of b R m, that means that any vector b R m must lie in the column space of A. In other words, col(a) = R m. This, then, means that the dimensions of these two spaces must be the same: dim col(a) = dim R m = m. Since the rank of A is the same as the dimension of the column space of A, we see that A has rank m. Now, the solutions to the equation A T y = are precisely those vectors y which lie in nul(a T ), otherwise known as the left nullspace of A. However, the dimension of the left nullspace of A is given by dim nul(a T ) = m rank(a) = m m =. Therefore, nul(a T ) is zerodimensional, so it only contains the zero vector, so the zero vector is the only solution to the equation A T y =. 4
5 6. Problem.4.8. Construct a matrix with (,, ) and (,, ) as a basis for its row space and its column space. Why can t this be a basis for the row space and nullspace? Solution: Consider the matrix A = 4 Notice, first of all, that A is symmetric, so a basis for its column space will also be a basis for its row space (and vice versa). Notice that the second column is twice the first column minus twice the third column, so we can ignore the second column for the purposes of finding a basis for the column space. Also, the vectors and are certainly linearly independent since they aren t multiples of each other. Therefore, these two columns do indeed give a basis for the column space of A (and, by symmetry, for the row space of A). Now, I claim that this cannot be a basis for the row space and the nullspace of any 3 3 matrix A. To see why, remember that the dimension of the row space of A is equal to the rank of A, whereas the dimension of the nullspace is equal to the number of columns minus the rank, namely 3 rank(a). If these two vectors are a basis for both the row space and the nullspace, then both the row space and the nullspace must be twodimensional, meaning that rank(a) = and 3 rank(a) =, meaning that the rank of A must be both and simultaneously. This is clearly impossible, so we see that the row space and the nullspace can t both be twodimensional, so no two vectors can be a basis for both the row space and the nullspace of any 3 3 matrix. 7. Problem.4.3. Describe the four subspace of R 4 associated with A = and I + A = Solution: The column space of A is just the space spanned by the vectors and, namely the xyplane. Notice that A is already in reduced echelon form, corresponding to the equations y = and z =. This leaves x as a free variable, so the solutions to the homogeneous equation are those vectors of the form x, 5
6 so the nullspace of A is the xaxis. The row space of A (i.e. the column space of A T ) is spanned by the vectors and, which is just the yzplane. The left nullspace of A (i.e. nul(a T )) consists of those vectors y such that A T y =. Since A T = is already in reduced echelon form, we can read off the equations x = and y =. Hence, z is a free variable, so the nullspace of A T consists of those vectors of the form z Thus, the left nullspace of A is the zaxis. (Notice that the column space and the left nullspace are perpendicular; similarly, the row space and the nullspace are perpendicular.) Now, turn to I + A. Notice that we don t have to do any elimination to see that I + A has three pivot columns (and three pivot rows), so the rank of I + A is equal to three. Thus, both the column space and the row space are threedimensional, so each is all of R 3. Moreover, dim nul(a) = 3 rank(a) = 3 3 =, so the nullspace is zerodimensional, meaning it consists only of the zero vector. Likewise, dim nul(a T ) = 3 rank(a) = 3 3 =, so the left nullspace is also zerodimensional and only contains the zero vector. 8. Problem.6.6. What 3 by 3 matrices represent the transformations that (a) project every vector onto the xyplane? Solution: To find the matrix for this transformation, we just need to apply the transformation to the standard basis for R 3 : the ith column of the matrix will be whatever vector results from applying the transformation to e i. Projecting the vector to the xyplane has no effect on it, so the first column of the matrix is 6
7 Likewise, projecting matrix is Finally, projecting is to the xyplane has no effect, so the second column of the to the xyplane yields Therefore, the matrix of this transformation is, so the third column of the matrix (b) reflect every vector through the xyplane? Solution: Again, we just apply the transformation to the standard basis. Reflecting through the xyplane has no effect, so the first column of the matrix is Likewise, reflecting through the xyplane has no effect, so the second column of the matrix is Finally, reflecting through the xyplane yields, so the third column of the matrix is Therefore, the matrix of this transformation is (c) rotate the xyplane through 9, leaving the zaxis alone? 7
8 Solution: Applying the transformation to the standard basis, we see that rotating through 9 (counterclockwise) yields, so the first column of the matrix is Rotating through 9 yields Finally, the transformation leaves the zaxis alone, so third column. Therefore, the matrix of this transformation is, so the second column of the matrix is is unchanged, and so is the (d) rotate the xyplane, then xz, then yz, through 9? Solution: First, apply this transformation to : the rotation in the xyplane yields, the rotation in the xzplane leaves this vector unaffected, and then the rotation in the yzplane yields Therefore, the first column of the matrix is Now, apply the transformation to Rotating this vector in the xzplane yields yzplane yields The rotation in the xyplane yields Finally, rotating this vector in the, so we see that the second column of the matrix is (notice that this means that the transformation actually has no effect on the yaxis). Finally, apply the transformation to The rotation in the xyplane has no effect on 8
9 this vector. Rotating in the xzplane takes this vector to, and then rotation in the yzplane has no further effect. Therefore, the third column of the matrix is Putting all of this together, the matrix of the transformation is (You can easily check that this is the same matrix you would get if the transformation had just been rotating the xzplane by 9 and leaving the yaxis alone, so the rotations in the xyplane and the yzplane were actually superfluous). (e) carry out the same three rotations, but each one through 8? Solution: Rotating through 8 in the xyplane yields Rotating this through 8 in the xzplane gets us back to, which is, in turn, unaffected by a 8 rotation in the yzplane. Thus, the first column of the matrix is Rotating in the xyplane yields, which is then unaffected by the 8 rotation in the xzplane. Rotating by 8 in the yzplane gets us back to, which is, therefore, the second column of the matrix. Finally, is unaffected by a 8 rotation in the xyplane. Rotating by 8 in the xzplane yields which is the third column of the matrix. Thus, the matrix of the transformation is ; rotating this by 8 in the yzplane takes us back to (Notice that this means we could have achieved the same effect by doing nothing, since this is also the matrix of the identity transformation). 9,
10 9. Problem.6.4. Prove that T is a linear transformation if T is linear (from R 3 to R 3 ). Proof. Suppose x, y R 3 and that a, b R. Then, in order to show that T is linear, we need to show that T (ax + by) = at (x) + bt (y). To that end, notice that T (ax + by) = T (T (ax + by)) = T (at (x) + bt (y)), since T is a linear transformations. In turn, again using the fact that T is a linear transformation, the above is equal to at (T (x)) + bt (T (y)) = at (x) + bt (y). Stringing these equalities together, we see that so T is indeed linear. T (ax + by) = at (x) + bt (y),. Problem.6.. Which of these transformations is not linear? The input is v = (v, v ). (a) T (v) = (v, v ). (b) T (v) = (v, v ). (c) T (v) = (, v ). (d) T (v) = (, ). Solution: I [ claim that (d) is [ the only one of these transformations that is not linear. To see v w this, let v = and w = be two vectors and let a, b R be scalars. v w To see that the transformation in (a) is linear, notice that ( [ [ ) ([ ) v w av + bw T (av + bw) = T a + b = T v w av + bw [ av + bw = av + bw [ [ v w = a + b Thus, the T from (a) is a linear transformation. v = at (v) + bt (w). Next, to see that the transformation from (b) is linear, we check that ( [ [ ) ([ ) v w av + bw T (av + bw) = T a + b = T v w av + bw [ av + bw = av + bw [ [ v w = a + b v = at (v) + bt (w). w w
11 Hence, the T from (b) is a linear transformation. Turning to (c), we see that T (av + bw) = T ( [ v a v so this is also a linear transformation. [ w + b w ) ([ ) av + bw = T av + bw [ = av + bw [ [ = a + b v = at (v) + bt (w), Finally, the transformation in (d) is not linear. To see this, notice that ([ [ ) ([ ) [ T + = T =, ( ) whereas T ([ ) ([ + T ) = [ [ + = [ w. ( ) Since the expressions in ( ) and ( ) are not equal, the transformation T cannot be linear.
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