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 Subject: LINEAR ALGEBRA questions Category: Science > Math Asked by: alison28-ga List Price: \$50.00 Posted: 26 Apr 2005 16:29 PDT Expires: 26 May 2005 16:29 PDT Question ID: 514646
 ```1) Find the angle between the diagonal of a cube and the diagonal of one of its sides. Please do this using the ideas of vector spaces. 2) Let following bases be given for a vector space V: B = {(1,3,4),(2,-5,2),(-4,2,-6) B' = {(1,2,-2),(4,1,-4),(-2,5,8)} a. Find the transition matrix P from B to B'. b. Find the transition matrix Q from B' to B. c. Verify that the matrices P and Q are inverses of one another. d. Find [x]B given [x]B' = (-1,0,2). Please respond before midnight (i put a nice amt of \$). Thank you very much.``` Clarification of Question by alison28-ga on 26 Apr 2005 17:25 PDT ```Just wanted to remind you that this is *not* for anything that's graded. ...im just trying to prepare for a final in this class, and I'm having trouble w/ these ones. Also, if you can't do one of them, just give me what you can do. Thank you!```
 ```Hi, alison28-ga: (1) The angle A between two linearly independent vectors in R^3 is easily determined by the formula for the dot product in terms of the vectors lengths and the cosine of the included angle: = cos(A) * ||u|| * ||v|| Taking the cube to have unit-length edges parallel to the coordinate axes, we can position the cube in the "positive" octant (first orthant), so that the coordinates of the eight vertex are all the possible triples than can be formed from entries in {0,1}. That is, the vector from the origin to the opposite corner (1,1,1) gives as a diagonal of the cube: u = [1 1 1] Here I'm making a notational distinction between the row-vector [1 1 1] and the two points in the three dimensional Euclidean space (0,0,0) and (1,1,1) which define the origin and terminus of that vector. Likewise one of the three "face" diagonals that meets the "long" diagonal at the origin is: v = [1 1 0] Solving the relation above and using dot product and the lengths ||u|| and ||v|| found by the usual "distance" formula: cos(A) = --------------- ||u||*||v|| = 2 / (SQRT(3)*SQRT(2)) = SQRT(2/3) We then deduce that A = arccos( SQRT(2/3) ) = 0.84106867... radians or 48.1896851... degrees. (2) The computation of the "transition matrix" P which converts coordinates with respect to the ordered basis: B = { (1,3,4), (2,-5,2), (-4,2,-6) } into coordinates for the same point with respect to the ordered basis: B'= { (1,2,-2), (4,1,-4), (-2,5,8) } can mostly easily be conceived of in two stages. The first and easy stage is the conversion from B-coordinates into "standard" coordinates, that is coordinates with respect to the standard basis: E = { e_1 = (1,0,0). e_2 = (0,1,0), e_3 = (0,0,1) } which we will show is accomplished by multiplying by a matrix M. The second stage is the conversion from standard coordinates to B'-coordinates, also accomplished by matrix multiplication, and the appropriate product of the two matrices then gives the transition matrix P which is the first objective in this problem. A brief reflection of what it means for a point x to have B-coordinates of (a,b,c) tells us that the point x is a linear combination of the standard basis vectors as follows: x = a*(1,3,4) + b*(2,-5,2) + c*(-4,2,-6) = (1a+2b-4c)*e_1 + (3a-5b+2c)*e_2 + (4a+2b-6c)*e_3 In particular the B-coordinates (1,0,0) correspond to standard coordinates (1,3,4), etc. Although it is essentially an arbitrary decision, the coordinates with respect to an ordered basis are often represented by a column vector. With this convention the transition matrix M that converts from B-coordinates to standard coordinates turns out to have three columns, each of which has entries transposed from the corresponding basis vector's entries: / 1 2 -4 \ M = | 3 -5 2 | \ 4 2 -6 / With this convention of column-vectors for coordinates, one then has the conversion from B-coordinates to standard coordinates represented as follows: / 1a+2b-4c \ / a \ coord. wrt E | 3a-5b+2c | = M * | b | \ 4a+2b-6c / \ c / The point here is that M can be written down by inspection. Similarly the matrix that converts B'-coordinates to standard coordinates could also be written down by inspection: / 1 4 -2 \ G = | 2 1 5 | \-2 -4 8 / However we need to convert in the reverse direction, from standard coordinates to B'-coordinates in order to accomplish the overall purpose for matrix P, and thus we need to find the multiplicative inverse of G so that we may set: P = G^-1 M One rather convenient way to do this is by constructing a "super-augmented" matrix: / 1 4 -2 | 1 2 -4 \ [ G | M ] = | 2 1 5 | 3 -5 2 | \-2 -4 8 | 4 2 -6 / then applying the "Swiss Army knife" of numerical linear algebra, Gaussian elimination/reduction to reduced row echelon form, we find after the appropriate sequence of elementary row operations that the reduced row echelon form of this is [ I | P ]. I shall leave this for the moment as a calculation for the reader to try his or her hand at. The calculation of Q is similar to that of P, except that we would apply Gaussian elimination instead to the extended matrix [ M | G ] and obtain from it [ I | Q ]. One then expects and can easily verify that P and Q are multiplicative inverses, as multiplying by Q is naturally undoing what multiplying by P does. Finally we are asked to determine what B-coordinates [x]_B correspond to the point x whose B'-coordinates are given by [x]_B' = (-1,0,2). For this it suffices to multiply Q times the column (-1,0,2)'. [Here the prime ' denotes transpose, giving a column in place of "row" (-1,0,2).] If we had not already at this point found Q (as well as P), then a shortcut calculation which finds only [x]_B can be arranged by reducing the augmented matrix: [ M | (-5,8,18)' ] whose right-hand side (-5,8,18) is G * (-1,0,2)'. Thus to solve: / 1 2 -4 | -5 \ | 3 -5 2 | 8 | \ 4 2 -6 | 18 / I will return shortly with the various step by step calculations, but the answer to part (d) is that the B-coordinates of x work out to: [x]_B = ( 14 11/13, 11 8/13, 10 10/13 ) Possibly these fractional values indicate that the problem was not copied correctly (or possibly someone simply picked fairly challenging numbers for hand calculation!), but in any case I thought it most valuable to sketch out the approach for all parts of this second question before diving into the details of the computations. regards, mathtalk-ga``` Request for Answer Clarification by alison28-ga on 26 Apr 2005 23:28 PDT ```One thing.. on angle, I got arccos(Squareroot2/3) = 35.2 instead of what you got. Can you confirm 35.2 degrees?``` Clarification of Answer by mathtalk-ga on 27 Apr 2005 06:36 PDT ```Hi, alison28-ga: Okay, you are totally right about the inverse cosine calculation. Note that: cos(A) = /(||u||*||v||) = 2 / SQRT(6) = SQRT(2/3) = 0.81649658... and A = 35.2644 degrees (approx., or 0.6154797... radians). I forgot to take the square root when I punched it into the calculator! Here's a few more details I put together for you: Addendum: Details of Row Reductions In 2(a) we are asked to find matrix P, the transition matrix converting B-coordinates into B'-coordinates, which I outlined as: P = (G^-1) * M a two stage process, converting B-coordinates into standard coordinates (by M) and then standard coordinates into B'-coordinates (by G^-1). Thus: / 1 4 -2 | 1 2 -4 \ [ G | M ] = | 2 1 5 | 3 -5 2 | \-2 -4 8 | 4 2 -6 / may be reduced via elementary row operations to: / 1 0 0 | -11/16 -55/16 73/16 \ [ I | P ] = | 0 1 0 | 25/32 45/32 -83/32 | \ 0 0 1 | 23/32 3/32 -29/32 / In 2(b) we are asked to find matrix Q, the transition matrix converting B'-coordinates into B-coordinates, which I outlined as a similar (but reversed) procedure that calls for: Q = (M^-1) * G converting B'-coordinates to standard, and then to B-coordinates. Thus: / 1 2 -4 | 1 4 -2 \ [ M | G ] = | 3 -5 2 | 2 1 5 | \ 4 2 -6 |-2 -4 8 / may be reduced via elementary row operations to: / 1 0 0 | -33/13 -86/13 80/13 \ [ I | Q ] = | 0 1 0 | -37/13 -85/13 57/13 | \ 0 0 1 | -30/13 -77/13 55/13 / In 2(c) we asked to confirm that P and Q are matrix inverses. Since square matrices are involved, it suffices to take their product in either direction: /-11/16 -55/16 73/16 \ / -33/13 -86/13 80/13 \ P * Q = | 25/32 45/32 -83/32 | * | -37/13 -85/13 57/13 | \ 23/32 3/32 -29/32 / \ -30/13 -77/13 55/13 / /208/208 0 0 \ = | 0 208/208 0 | \ 0 0 208/208/ = I Finally we can compute now the answer to part 2(d) using matrix Q: / -33/13 -86/13 80/13 \ /-1 \ Q * [x]_B' = | -37/13 -85/13 57/13 | * | 0 | \ -30/13 -77/13 55/13 / \ 2 / / 193/13 \ = | 151/13 | \ 140/13 / which upon simplification to mixed fractions gives: [x]_B = ( 14 11/13, 11 8/13, 10 10/13)' in agreement with what was previously calculated in more direct fashion. regards, mathtalk-ga```
 alison28-ga rated this answer: `Very thorough and clear answers. Great job! Enjoy the \$50.00`