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Math 217: Coordinates and B-Matrices Section 5 Professor Karen Smith ************************************************** Math 217: Coordinates and B-matrices Section 5 Professor Karen Smith (c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. n n T n Definition. Let B = f~v1; : : : ; ~vng be a basis for R . Let R −! R be a linear transformation. The B-matrix of T is the n × n matrix B = [T (~v1)]B T (~v2)]B :::T (~vn)]B] : Computing the B-matrix: Compute one column at a time. For the j-th column: 1. Find T (~vj ). 2. Express T (~vj ) as a linear combination of the basis elements f~v1; : : : ; ~vng. 3. Write these B-coordinates of T (~vj ) as a column vector: this is the j-th column of the B-matrix. Do not forget this last step of converting back into B-coordinates! It is easy to n forget since the vectors T (~vj ) 2 R are columns already in standard coordinates. n For any vector ~v 2 R , we can understand T entirely in B-coordinates as follows: [T (~v)]B = B · [~v]B where B is the B-matrix of T . n T n Theorem. Let R −! R be a linear transformation (so T is multiplication by some matrix A). Then the B-matrix and the standard matrix A of T are similar: B = S−1AS where S = ~v1 ~v2 : : : ~vn : The matrix S is the change of basis matrix from B to the standard basis. The columns of S are the basis vectors of B expressed in the standard basis. ******************************************************************************* x 3x + 4y A. Let T : 2 ! 2 be the map 7! : R R y 4x − 3y 2 1. Prove that B = f~v1;~v2g = {−~e1 + 2~e2; 2~e1 + ~e2g is a basis for R . −1 2 Solution note: The square matrix has rank two, hence its columns are a 2 1 basis. −2 2. Compute and [~e + 3~e ] and 7 ~v + π~v . 4 1 2 B 13 1 2 B B 2 1 17=3 Solution note: In order, these are ; , and 0 1 π 3. Find a matrix S which \changes B-coordinates to standard coordinates." That is, S should satisfy S · [~v]B = [~v]f~e1:~e2g 2 for all vectors ~v 2 R . Verify explicitly that your S transforms the B-coordinates you found in (3) back into standard coordinates. −1 2 Solution note: S = is the change of basis matrix form B to the standard 2 1 basis. We can verify: −1 2 2 −2 −1 2 1 1 −1 2 17=3 −17=3 + 2π · = ; · = ; and · = ; 2 1 0 4 2 1 1 3 2 1 π 34=3 + π 17 which are the standard coordinates of 3 ~v1 + π~v2. 4. Find a matrix P which does the opposite|P should transform standard coordinates into B coordinates. That is, P should satisfy P · [~v]f~e1:~e2g = [~v]B 2 for all vectors ~v 2 R . Check that it works for the vectors in (2). What is the relationship between S and P ? Solution note: P does the reverse of S, so it is given by the inverse matrix: P = −1 2 −1 −1 1 −2 5 5 S = 5 = 2 1 : −2 −1 5 5 5. Find the B-matrix of T . Use to compute the image of ~v1 + ~v2. −5 0 1 Solution note: [T ] = : In B-coordinates, ~v + ~v is written . So its B 0 5 1 2 1 −5 0 1 −5 image in B-coordinates is · = . This is the vector −5~v + 5~v . 0 5 1 5 1 2 6. Find the standard matrix of T . Call it A. 3 4 Solution note: A = : 4 −3 7. What is the relationship between [T ]B and the standard matrix for T . Use this to compute [T ]B. Do you get the same answer? Alternatively, how could we compute the standard matrix from the B-matrix? −1 Solution note: [T ]B = S AS by the Theorem (memorize this carefully!). So we can compute −1 1 −2 3 4 −1 2 [T ] = : B 5 −2 −1 4 −3 2 1 Doing the multiplication we get the same answer. Alternatively, we can rearrange the statement in the theorem by multiplication on the left by S and on the right by S−1. Then we see −1 A = S[T ]BS ; so we can compute A by multiplying these three matrices together. Of course, we get the same answer, but this is a lot more tedious then finding its columns by seeing where ~e1 and ~e2 go under T . 4 B. Let B = f~e1;~e2;~e3 + ~e2;~e4 + ~e1g ⊂ R : 4 1. Prove B is a basis for R . Solution note: There are four vectors, so by Theorem 3.3.4, they are a basis for a 4-dimensional space if they span it. But their span includes all the standard vectors, since ~e1;~e2;~e3 = (~e3 + ~e2) − ~e2 and ~e4 = (~e4 + ~e1) − ~e1 are in B. T 2. Write ~x = [1 1 1 1] as a linear combination of the elements in B. What is [x]B? T Solution note: ~x = 1(~e3 + ~e2) + 1(~e4 + ~e1). So [x]B = [0 0 1 1] : 4 4 3. Consider the linear transformation T : R ! R whose matrix with respect the basis B is 23 0 0 03 60 −1 0 07 B = 6 7. Find the matrix of T in the standard basis (call it A). 40 0 a 05 0 0 0 b Solution note: The columns of the standard matrix will be the T (~ei) (expressed in T the standard basis). We know T (~e1) = 3~e1, so the first column is 3 0 0 0 . T Likewise T (~e2) = −~e2, so the second column is 0 −1 0 0 . The third column will be T (~e3) which is a tiny bit harder: we know ~e3 = (~e3 + ~e2) − ~e2, so T (~e3) = T (~e3 + ~e2) − T (~e2) = a(~e3 + ~e2) − (−~e2) = (a + 1)~e2 + a~e3. Note that in computing this, we used the B-matrix B to compute T (~e3 + ~e2). Thus the third column is 0 a + 1 a 0 : Similarly, T (~e4) = T (~e4 + ~e1) − T (~e1) = b(~e4 + ~e1) − 3(~e1), giving 23 0 0 b − 33 T 60 −1 a + 1 0 7 us b − 3 0 0 b for the fourth column. So A = 6 7. 40 0 a 0 5 0 0 0 b 213 2 2 3 3 3 C. Let T : R ! R be projection onto the plane Λ spanned by 415 and 4−15. 1 1 82 3 2 39 < 1 2 = 1. Is the set 415 ; 4−15 a basis for Λ? Explain. : 1 1 ; 3 2. Can the set in (1) be extended to a basis for all of R ? Explain. 3 3. Find a vector ~v 2 R perpendicular to Λ. 8213 2 2 39 < = S 3 4. Is the set B = 415 ; 4−15 f~vg a basis for R ? Explain. : 1 1 ; 5. Find the B-matrix of T . 6. Express the matrix for T in standard coordinates in terms of matrices you have already computed here. Solution note: (1) Yes, the plane has dimension two, and these are two vectors that are linearly independent, since neither is a scalar multiple of the other. (2) Yes, pick any vector not in the span of these. Then we have three vectors 3 linearly independent in the 3d space R . 2x3 (3) We need a vector 4y5 which has zero dot product with both given vectors. z This amounts to solving the linear system x + y + z = 0 2x − y + z = 0: Since the coefficient matrix is rank 2, by the rank-nullity theorem, this system has a one-dimensional kernel, so any non-zero vector will span it. (You can also solve it 2 2 3 directly). By inspection, we see 4 1 5 is a solution, hence a vector perpendicular to −3 the plane. (4) Yes, these are a basis, since ~v is not in the span of the previous vectors. 21 0 03 (5) [T ]B = 40 1 05 : 0 0 0 −1 (6) The standard matrix is A = S · [T ]B · S . 3 D. Let V be the subspace of R given by x1 + x2 − 2x3 = 0. 213 1 1. Find a basis of V in which the vector 1 has coordinates . 4 5 −1 1 213 223 233 2. Let ~v1 = 415, ~v2 = 405, ~v3 = 415. Find the coordinates of ~v1 in the basis (~v2;~v3), of ~v2 in 1 1 2 the basis (~v1;~v3) and of ~v3 in the basis (~v1;~v2)..
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