Math 217: Coordinates and B-Matrices Section 5 Professor Karen Smith **************************************************

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Math 217: Coordinates and B-matrices

n n T n Deﬁnition. Let B = {~v1, . . . , ~vn} 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 {~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 n forget since the vectors T (~vj ) ∈ R are columns already in standard coordinates.

n For any vector ~v ∈ 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 = {~v1,~v2} = {−~e1 + 2~e2, 2~e1 + ~e2} 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]{~e1.~e2} 2 for all vectors ~v ∈ 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]{~e1.~e2} = [~v]B

2 for all vectors ~v ∈ 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 = {~e1,~e2,~e3 + ~e2,~e4 + ~e1} ⊂ 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 3 0 0 0 0 −1 0 0 B =  . Find the matrix of T in the standard basis (call it A). 0 0 a 0 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 3 0 0 b − 3 T 0 −1 a + 1 0  us b − 3 0 0 b for the fourth column. So A =  . 0 0 a 0  0 0 0 b

1  2  3 3 C. Let T : R → R be projection onto the plane Λ spanned by 1 and −1. 1 1      1 2  1. Is the set 1 , −1 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 ∈ R perpendicular to Λ. 1  2    S 3 4. Is the set B = 1 , −1 {~v} 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 . 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 are a basis, since ~v is not in the span of the previous vectors. 1 0 0 (5) [T ]B = 0 1 0 . 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. 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).