Answer Key for Exam #2
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1 . Use elimination on an augmented matrix: Answer Key for Exam # The fourth column has no pivot, so x 4 is a free variable. The corresponding system is x + x 4 =, x =, x x 4 = which we solve for the pivot variables: Therefore x = x 4 x = x = +x 4 x 4 = x 4 x x = + x x 4 x 4. We perform the eliminations A = 4 = R 4 A basis for the row space is the pivot rows of R, or of A. A basis for the column space is the pivot columns of A (but not of R). A basis for the nullspace can be found as in problem or by taking the negative of the upper right corner of R, putting a identity matrix below it, taking the three columns of that. So the only basis that requires more work is the left nullspace. To get it we transpose the pivot columns of A eliminate: ( ) Here we can solve the corresponding system, or throw away the identity on the left, negate the rest, put a identity under it. We also have another basis for the column space in the rows of the last matrix above. In conclusion A row space basis is or 4
2 A null space basis is A column space basis is A left null space basis is or The factored form of A that displays bases for all four is A = ) (. To see which vector to keep we start by computing all the dot products for the three vectors. If v = v = v = 4 then v v = + + = 4, v v = = 8, v v = = 4, v v = = 8, v v = = 8, v v = = Recall that the projection of b onto a is b a a a a. If we take a = v then the ratios will both be 4 8 =, so v seems like a good one to keep. Then the projection of v onto v is v v v = 4 =, v v 8 therefore v = = + e, where e is the error in the projection. We want to replace v by some multiple of e. We have 6 4 e = = = w, 4 so we throw away v replace it by w. Next we do the same thing with v. The projection of v onto v is v v v = 4 =, v v 8
3 therefore v = = 4 + e, where e is the error in the projection, again we want to replace v by some multiple of e. We have 4 6 e = 8 = = w 9, so we throw away v replace it by w. If we rename v as w, we now have w = w = w =, 9 where w w w w, but usually we would not have w w at this point. (You might, if you re lucky; in fact, if we had decided to keep v instead of v then we would have been lucky at this stage.) In this case w w = + + = 44, so they are not perpendicular. Since we also have w w = = 44 w w = = 9, it is a good idea to keep w change w. The projection of w onto w is therefore w w w = 44 =, w w 44 w = = + e = Replacing w by the simplest multiple of e, we now have, which are all perpendicular to each other, so we only have to fix the lengths. The dot product of the last vector with itself is, we did the others earlier, so we finally get that an orthonormal basis for the subspace of R 4 spanned by v, v v is If we had kept w changed w we would have instead
4 If we had kept v initially changed the others we would have wound up with the orthonormal basis , if we had kept v at the beginning changed the others we could have come out with or with Let A be the matrix with v v as columns. Then so A T A = 6 = 6 ( ) 6 6 = 6, 6 6 = 6 8 P = 4 = so we find that the projection matrix P onto the subspace S is We also have that 4, 4 9 ( P = = R = P I = = ),
5 is the reflection matrix through S. The projection of v = 4 onto S is 4 P v = = 8 4 =, the reflection of v through S is 6 R v = = = The projection is the average of the reflection v itself, this could have been used to avoid one of the last two matrix multiplications.. If P = 9 then P is symmetric P = = P Any matrix P which satisfies P = P = P T is a projection matrix. The trace of P is ( ) =, so the subspace T that P projects onto is -dimensional. Therefore any two rows or columns of P will be a basis for it as long as they are not multiples of each other. We can use the same trick on I P to get a basis for T. Since only one of the rows of P is very nice, though, let s eliminate: P = P projects onto its own column space, or its row space since P is symmetric. This gives us a nice basis for T, a basis for T comes from the null space of P : A basis for T is A basis for T is 6. We would like to find the line y = mx + b which best fits the four points (, ), (, ), (, ) (4, ) in the sense of least squares. If the line fit exactly we would have = m + b = m + b = m + b = 4m + b or m = b 4
6 This equation has no solution, but we can find the least squares solution by multiplying by the transpose of the matrix: 4 m 4 = b 4 which becomes ( 4 ) m = b Solving this by elimination on an augmented matrix we get Therefore b = m =, so m = b = the best line is y = x +.. If U is a 6-dimensional subspace of R 8 then the projection matrix onto U is P = A ( A T A ) A T, where A is any 8 6 matrix whose columns are a basis of A. Calculating P directly by h would not be fun since we would have to invert the 6 6 matrix A T A. According to the fundamental theorem of linear algebra, the left null space of A is an 8 6 = dimensional subspace of R 8 it is the orthogonal complement of the column space of A. In other words, U is a -dimensional subspace of R 8 we know how to find a basis for it. Thus an indirect way to find P is as follows: (i) find a basis { v, v } for the left null space of A (ii) let A be the matrix with v v as columns, calculate P = A ( A T A ) A T (iii) P is the projection matrix onto U, therefore I P is the projection matrix onto U. Step (i) takes some work, but in (ii) we only have to invert a matrix. Scores: For the exams which I have at this writing, the median is 9 the mean is about 88.. Score Frequency Score Frequency Score Frequency ??
Answer Key for Exam #2
Answer Key for Exam #. Use elimination on an augmented matrix: 8 6 7 7. The corresponding system is x 7x + x, x + x + x, x x which we solve for the pivot variables x, x x : x +7x x x x x x x x x x x Therefore
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