Coordinates and B-matrices

Math 217: Coordinates and B-matrices T Definition. Let B = {~v1 , . . . , ~vn } be a basis for Rn . Let Rn −→ Rn 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 {~v1 , . . . , ~vn }. 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 forget since the vectors T (~vj ) ∈ Rn are columns already in standard coordinates. For any vector ~v ∈ Rn , we can understand T entirely in B-coordinates as follows: [T (~v )]B = B · [~v ]B where B is the B-matrix of T . T Theorem. Let Rn −→ Rn 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 −1 AS   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 2 2 7→ . A. Let T : R → R be the map y 4x − 3y 1. Prove that B = {~v1 , ~v2 } = {−~e1 + 2~e2 , 2~e1 + ~e2 } is a basis for R2 .   −1 2 has rank two, hence its columns are a Solution note: The square matrix 2 1 basis.   7  −2 2. Compute and [~e1 + 3~e2 ]B and 13 ~v1 + π~v2 B . 4 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 ]{~e1 .~e2 } for all vectors ~v ∈ R2 . Verify explicitly that your S transforms the B-coordinates you found in (3) back into standard coordinates.  Solution note: basis.  −1 2 S=  −1 2 is the change of basis matrix form B to the standard 2 1 We can verify:               2 2 −2 −1 2 1 1 −1 2 17/3 −17/3 + 2π · = , · = , and · = , 1 0 4 2 1 1 3 2 1 π 34/3 + π which are the standard coordinates of 17 v1 3~ + 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 ]{~e1 .~e2 } = [~v ]B for all vectors ~v ∈ R2 . Check that it works for the vectors in (2). What is the relationship between S and P ? Solution note: P does reverse of S, so it is given by the inverse matrix: P =  the   −1 2 1 −2 = 52 51 . S −1 = −1 5 −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 ]B = . In B-coordinates, ~v1 + ~v2 is written . So its 0 5 1       −5 1 −5 0 . This is the vector −5~v1 + 5~v2 . = · image in B-coordinates is 5 1 0 5 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? Solution note: [T ]B = S −1 AS by the Theorem (memorize this carefully!). So we can compute     −1 1 −2 3 4 −1 2 [T ]B = . 2 1 5 −2 −1 4 −3 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 A = S[T ]B S −1 , 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 . B. Let B = {~e1 , ~e2 , ~e3 + ~e2 , ~e4 + ~e1 } ⊂ R4 . 1. Prove B is a basis for R4 . 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. 2. Write ~x = [1 1 1 1]T as a linear combination of the elements in B. What is [x]B ? Solution note: ~x = 1(~e3 + ~e2 ) + 1(~e4 + ~e1 ). So [x]B = [0 0 1 1]T . 3. Consider  3 0 B= 0 0 the 0 −1 0 0 lineartransformation T : R4 → R4 whose matrix with respect the basis B is 0 0 0 0 . Find the matrix of T in the standard basis (call it A). a 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   3 0 0 b−3 0 −1 a + 1  T 0  . us b − 3 0 0 b for the fourth column. So A =  0 0 a 0  0 0 0 b     1 2 C. Let T : R3 → R3 be projection onto the plane Λ spanned by 1 and −1. 1 1     2   1 1. Is the set 1 , −1 a basis for Λ? Explain.   1 1 2. Can the set in (1) be extended to a basis for all of R3 ? Explain. 3. Find a vector ~v ∈ R3 perpendicular to Λ.     2 S  1    1 , −1 4. Is the set B = {~v } a basis for R3 ? 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 . x (3) We need a vector y  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 directly). By inspection, we see  1  is a solution, hence a vector perpendicular to −3 the plane. (4) Yes, these basis, since ~v is not in the span of the previous vectors.  are a  1 0 0 (5) [T ]B = 0 1 0 . 0 0 0 (6) The standard matrix is A = S · [T ]B · S −1 . D. Let V be the subspace of R3 given by x1 + x2 − 2x3 = 0.     1 1   . 1. Find a basis of V in which the vector 1 has coordinates −1 1       1 2 3 2. Let ~v1 = 1, ~v2 = 0, ~v3 = 1. 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 ).