18.06 Problem Set 4 Solution

Transcription

1 8.6 Poblem Set 4 Solution Total: points Section 3.5. Poblem 2: (Recommended) Find the lagest possible numbe of independent vectos among ) ) ) v = v 4 = v 5 = v 6 = v 2 = v 3 =. Solution (4 points): Since v 4 = v 2 v, v 5 = v 3 v, and v 6 = v 3 v 2, thee ae at most thee independent vectos among these: futhemoe, applying ow eduction to the matix [v v 2 v 3 ] gives thee pivots, showing that v, v 2, and v 3 ae independent. Section 3.5. Poblem 2: Find a basis fo the plane x 2y + 3z = in R 3. Then find a basis fo the intesection of that plane with the xy plane. Then find a basis fo all vectos pependicula to the plane. Solution (4 points): This plane is the nullspace of the matix The special solutions 2 3 A = 2 3 v = v 2 = give a basis fo the nullspace, and thus fo the plane. The intesection of this plane with the xy plane is a line: since the fist vecto lies in the xy plane, it must lie on the line and thus gives a basis fo it. Finally, the vecto v 3 = 2 3 is obviously pependicula to both vectos: since the space of vectos pependicula to a plane in R 3 is one-dimensional, it gives a basis. Section 3.5. Poblem 37: If AS = SA fo the shift matix S, show that A must have this special fom: a b c a b c a b c If d e f = d e f, then A = a b g h i g h i ai The subspace of matices that commute with the shift S has dimension.

2 pset4-s-soln: page 2 Solution (4 points): Multiplying out both sides gives a b d e f d e = g h i g h Equating them gives d = g = h =, e = i = a, f = b, i.e. the matix with the fom above. Since thee ae thee fee vaiables, the subspace of these matices has dimension 3. Section 3.5. Poblem 4: Wite a 3 by 3 identity matix as a combination of the othe five pemutation matices. Then show that those five matices ae linealy indpendent. (Assume a combination gives c P + +c 5 P 5 =, and check enties to pove c i is zeo.) The five pemutation matices ae a basis fo the subspace of 3 by 3 matices with ow and column sums all equal. Solution (2 points): The othe five pemutation matices ae P 2 =, P 3 =, P 32 =, P 32 P 2 =, P 2 P 32 = Since P 2 + P 3 + P 32 is the all s matix and P 32 P 2 + P 2 P 32 is the matix with s on the diagonal and s elsewhee, I = P 2 + P 3 + P 32 P 32 P 2 P 2 P 32. Fo the second pat, the combination above gives c 3 c + c 4 c 2 + c 5 c + c 5 c 2 c 3 + c 4 = c 2 + c 4 c 3 + c 5 c Setting each element equal to fist gives c = c 2 = c 3 = along the diagonal, then c 4 = c 5 = on the off-diagonal enties. Section 3.5. Poblem 44: (An aside in the text, followed by) dimension of outputs + dimension of nullspace = dimension of inputs. Fo an m by n matix of ank, what ae those 3 dimensions? Outputs = column space. This question wil be answeed in Section 3.6, can you do it now? Solution (2 points): You should think about the aside in the text, as well as poblem 43: the actual question asked, hee, howeve is quite simple. The dimension of inputs fo an m by n matix is n (the matix takes n-vectos to m-vectos), while the dimension of the nullspace is n and the dimension of outputs = dimension of column space is. Since n + = n, we have the given elation. Section 3.6. Poblem : A is an m by n matix of ank. Suppose thee ae ight sides b fo which Ax = b has no solution. (a) What ae all the inequalities (< o ) that must be tue between m, n, and? (b) How do you know that A T y = has solutions othe than y =? Solution (4 points): (a) The ank of a matix is always less than o equal to the numbe of ows and columns, so m and n. Moeove, by the second statement, the column space is smalle than the space of possible output matices, i.e. < m. (b) These solutions make up the left nullspace, which has dimension m > (that is, thee ae nonzeo vectos in it). Section 3.6. Poblem 24: A T y = d is solvable when d is in which of the fou subspaces? The solution is unique when the contains only the zeo vecto.

3 pset4-s-soln: page 3 Solution (4 points): It is solvable when d is in the ow space, which consists of all vectos A T y, and is unique when the left nullspace contains only the zeo vecto (as any two solutions diffe by an element in the left nullspace). Section 3.6. Poblem 28: Find the anks of the 8 by 8 checkeboad matix B and the chess matix C: n b q k b n p p p p p p p p B= and C= p p p p p p p p n b q k b n The numbes, n, b, k, q, p ae all diffeent. Find bases fo the owspace and left nullspace of B and C. Challenge poblem: Find a basis fo the nullspace of C. Solution (4 points): In both cases, elimination kills all but the top two ows, so, if p =, both matices have ank 2 as well as owspace bases given by the top two ows (o couse, if p =, C has ank with owspace geneated by the top ow). B is symmetic, so its left nullspace is the same as the nullspace, and the special solutions ae: = v, v 2 =, v 3 =, v 4 =, v 5 =, v 6 = Finally, the nullspace of C T is given by = w, w 2 =, w 3 =, w 4 =, w 5 =, w 6 = if p =, and, w 2 =, w 3 =, w 4 =, w 5 =, w 6 =, w 7 = w =

4 pset4-s-soln: page 4 if p =. Solution (2 points): (Challenge subpat) Thee ae thee obvious special solutions of C: u =, u 2 =, u 3 = If p =, the othe solutions ae similaly staightfowad: b k q n u 4 =, u 5 =, u 6 =, u 7 = Othewise, simultaneously solving c + c 2 n + b = and (c + c 2 + )p = (and similaly fo q and k instead of b), we get n b n q n k b q k u 4 =, u 5 =, u 6 = Section 3.6. Poblem 3: If A = uv T is a 2 by 2 matix of ank, edaw Figue 3.5 to show clealy the Fou Fundamental Subspaces. If B poduces those same fou subspaces, what is the exact elation of B to A? Solution (2 points): One daws the same diagam as in the book, but now each space has dimension, the column space is the set of multiples of u, the ow space is the set of multiples of v T, the nullspace is the plane pependicula to v, and the left nullspace is the plane pependicula to u. If B = u v T poduces the same fou subspaces, u is a multiple of u and v is a multiple of v, i.e. B is a multiple of A. Section 3.6. Poblem 3: M is the space of 3 by 3 matices. Multiply each matix X in M by A =. Notice: A =.

5 pset4-s-soln: page 5 (a) Which matices X lead to AX =? (b) Which matices have the fom AX fo some matix X? (a) finds the nullspace of that opeation AX and (b) finds the column space. What ae the dimensions of those two subspaces of M? Why do the dimensions add to (n ) + = 9? Solution (2 points): (a) A clealy has ank 2, with nullspace having the basis [] T. AX = pecisely when the columns of X ae in the nullspace of A, i.e. when they ae multiples of the all s vecto. a b c X = a b c a b c (b) On the othe hand, the columns of any matix of the fom AX ae linea combinations of the columns of A. That is, they ae vectos whose components all sum to, so a matix has the fom AX if and only if all of its columns individually sum to. a b c AX = B if and only if B = d e f a d b e c f The dimension of the nullspace is 3, while the dimension of the column space is 6. They add up to 9, which is the dimension of the space of inputs of this matix, when teated as a linea map on matices.

6 MIT OpenCouseWae Linea Algeba Sping 2 Fo infomation about citing these mateials o ou Tems of Use, visit: