Math 217: Using Coordinate Space to Model Any Vector Space Section 3 Professor Karen Smith Fundamental Idea of Linear Algebra: L

Math 217: Using Coordinate Space to Model any Vector Space

Fundamental Idea of Linear Algebra: Let V be any vector space, say, of dimension d. Using Coordinates, we can

d 1. Model the vector space V as R , and T 2. Model any linear transformation V −→ V as multiplication by a d × d matrix.

Technique: Fix a basis B = (~v1, . . . ,~vd) for V . d Model each vector ~v ∈ V as its “column of B-coordinates” in R . d This identiﬁes V with R , which you should think of as the “B- coordinate space.” T d [T ]B d T d [T ]B d Then V −→ V is then modeled by R −→ R . Because V −→ V is linear, so is R −→ R . d [T ]B d The Crucial Theorem says that R −→ R is given by left multiplication by some d × d matrix. This matrix is called the B-matrix of T . We denote it by [T ]B.

A. Discuss this technique with your table, using diagrams and arrows, until you feel you can explain it in your own words.

2 3 4 B. Let P4 be the space of polynomials of degree 4 or less. Fix the basis B = (1, x, x , x , x ).

5 1. Describe the map f 7→ [f]B from P4 to the B-coordinate space R explicitly for an arbitrary element of P4.

0 2. Prove that the map D : P4 → P4 sending f 7→ f − f is a linear transformation. 3. Find the matrix of D with respect to the basis B.

4. Compute D(x3 +x2) in two different ways: first directly from the definition in (2), then using the B-matrix. Make sure your answers agree!

5. Use your matrix [D]B to ﬁnd the dimension of imD. 6. Is D is invertible? Explain.

−1 −1 7. Find the matrix of D with respect to B. That is, ﬁnd [D ]B. 8. Use your matrix to compute D−1(g) for the case g(x) = 1 + x. [Note that your answer should be a polynomial!] Verify directly that D ◦ D−1(g) = g in this case. Do the same for another 4 vector in P4, say g = x + x. 9. Use your answer to (7) to give a formula for D−1(g) for arbitrary g(x) = a+bx+cx2+dx3+ex4. [Note that your formula should produce an element in P4!]

−1 −1 10. True or False: [D ]B = [D]B . 4 3 2 11. Now let A = (x , x , x , x, 2). Find [D]A. Is [D]A = [D]B in general? Solution note: a b 5 2 3 4   1. The map is P4 → R sending f = a + bx + cx + dx + ex 7→ c.   d e

0 0 0 2. D(f + g) = (f + g) − (f + g) = f − f + g − g = D(f) + D(g) for all f, g ∈ P4. Also 0 0 D(cf) = (cf) − (cf) = c(f − f ) = cD(f) for all f ∈ P4 and all scalars c ∈ R. 1 −1 0 0 0  0 1 −2 0 0  3 2   3. [D(x + x )]B = [D]B = 0 0 1 −3 0    0 0 0 1 −4 0 0 0 0 1

4. Directly from the deﬁnition: D(x3 + x2) = (x3 + x2) − (x3 + x2)0 = x3 − 2x2 − 2x. Using the B-matrix: we convert everything into B-coordinates:

1 −1 0 0 0  0  0  0 1 −2 0 0  0 −2 3 2       [D]B[x + x ]B = 0 0 1 −3 0  1 = −2       0 0 0 1 −4 1  1  0 0 0 0 1 0 0

which represents the vector (polynomial) 2x − 2x2 + x3. They agree!

5. The dimension of the image is the rank of the matrix, which is obviously 5. This means the map is surjective.

6. Yes, D is invertible. By rank-nullity, the dimension of the kernel is 0, so the map is also injective. So it is a bijective linear transformation. 1 1 2 6 24 0 1 2 6 24   7. The matrix of D−1 is the inverse of the matrix of D. So it is 0 0 1 3 12   0 0 0 1 4  0 0 0 0 1

8. Compute D−1(x + 1) is computed by converting to B-coordinates and multiplying by the matrix of D−1 above (and then converting back to a polynomial). So

1 1 2 6 24 1 2 0 1 2 6 24 1 1 −1 −1       [D (x + 1)]B = [D ]B[(x + 1)]B = 0 0 1 3 12 0 = 0 ,       0 0 0 1 4  0 0 0 0 0 0 1 0 0

which represents the polynomial 2 + x. Solution note:

9. Compute

1 1 2 6 24 a a + b + 2c + 6d + 24e 0 1 2 6 24 b  b + 2c + 6d + 24e  −1 2 3 4       [D ]B[a + bx + cx + dx + ex ]B = 0 0 1 3 12 c =  c + 3d + 12e  ,       0 0 0 1 4  d  d + 4e  0 0 0 0 1 e e

so g−1(a + bx + cx2 + dx3 + ex4) = a + b + 2c + 6d + 24e + (b + 2c + 6d + 24e)x + (c + 3d + 12e)x2 + (d + 4e)x3 + ex4.

10. TRUE!

11. The matrix totally depends on the choice of basis! For the basis A, we have

 1 0 0 0 0 −4 1 0 0 0    0 −3 1 0 0    0 0 −2 1 0 0 0 0 −1 1 A handy technique to compute the B-matrix:

T Let B = {~v1, . . . ,~vn} be a basis for a vector space V . Let V −→ V 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 when V = R so the vectors T (~vj ) are already columns!

A handy technique to compute the Change of Basis Matrix:

Let B = (~v1, . . . ,~vn) andA = (~w1, . . . , ~wn) be two bases for V . The Change of Basis matrix from B to A is SB→A = [[~v1]A [~v2]A ... [~vn]A] .

Computing the change of basis matrix: To go FROM B TO A, compute one column at a time. For the j-th column:

1. Take the j-the basis element ~vj in B.

2. Express ~vj as a linear combination of the basis elements A = ( ~w1, . . . , ~wn).

3. Write these A-coordinates of ~vj as a column vector: this is the j-th column of the change of basis matrix SB→A.

Note that if A is the standard basis, or is otherwsie “simple,” it will be probably be easy to find the A-coordinates. In this case, SB→A is probably easier to find than SA→B, which you can find by taking the inverse of SB→A instead.

2×2 C. Consider the basis B = (E11,E12,E21,E22) for R . 2×2 4 1. Describe the map A 7→ [A]B from R to the B-coordinate space R explicitly for arbitrary 2×2 A ∈ R . 2×2 T 2×2 T 2. Prove that the map R −→ R sending A 7→ A − A is linear.

3. Find the matrix of T with respect to the basis B. What are the dimensions of [T ]B? What does Rank-Nullity tell us about the dimensions of the kernal and image of T ?

1 4. Use [T ]B to ﬁnd bases for the image and kernel of T .

2×2 5. Let A be some other (arbitrary) basis for R . What is the rank of [T ]A? Explain. 2×2 6. Now ﬁx A = (E11 + E12, 2E12,E21 − E22, 5E22). For arbitrary A ∈ R , ﬁnd [A]A.

1 2×2 2×2 Note that these bases are subsets of R —don’t give their B-coordinates but do give actual vectors in R . 4 4 7. Let R → R be the canonical map from the A-coordinate space to the B-coordinate space 2×2 of R which sends each [A]A 7→ [A]B for each matrix A. Without doing any computation, explain why this is an isomorphism.

8. Find the matrix of the map in (7). Explain why we denote this by SA→B, thinking about its source and target. Why do we call it the change of basis matrix from A to B?

9. What is the relationship between SB→A and SA→B? 10. Write down the source and target space for each space involved in the composition

SB→A ◦ [T ]B ◦ SA→B,

and explain why this composition is [T ]A. Solution note: a 2×2 2×2 a b b 1. R → R sends A = 7→  . c d c d

2. T (A + B) = (A + B)T − (A + B) = AT − A + BT − B = T (A) + T (B) and similarly T (cA) = (cA)T − (cA) = cAT − cA = c(T (A)).

0 0 0 0 0 −1 1 0 3. The matrix [T ]B is 4 × 4. It is [TB] =   . The image of [T ]B (and hence T ) 0 1 −1 0 0 0 0 0 has dimension 1, since this matrix is obviously rank 1. By Rank-Nullity, [T ]B (and hence T ) has kernel of dimension 3.  0  0 1  1  4. Basis for image of T is (which we get by interpreting the B-coordinate column   −1 0 −1 0 1 0 0 0 0 1 as a matrix). Basis for the kernel of T is ( , , ). 0 0 0 1 1 0

5. The rank [T ]A should be 1 in ANY basis because the image of T is one, regardless of the choice of basis we are using to model.  a  a b 2×2 (b − a)/2 6. For arbitrary A = ∈ R , we have [A]A =   . c d  c  (c + d)/5

4 2×2 4 7. This map is the composition of the maps R → R → R given by [A]A 7→ A 7→ [A]B. Since both are isomorphisms, so is the composition. 1 0 0 0 1 2 0 0 8. SA→B =  . This matrix converts A-coordinate columns to B-coordinate columns. 0 0 1 0 0 0 −1 5

9. SA→B and SB→A are inverse to eachother.

u1 2 D. Representing Linear Transformations in Non-Standard Bases. Let ~u = ∈ R be u2 a unit vector, and let L be the line it spans. Recall that the map “projection onto L” is a linear transformation 2 2 π : R → R 2 u1 u1u2 whose matrix is A = 2 . u1u2 u2

1. Let S = (~e1,~e2). Compute the S-matrix of π using the Handy Technique. Does your answer make sense? 2 2. Let ~v be another unit vector such that ~u · ~v = 0. Explain why B = (~u,~v) is a basis for R . Draw a picture. 3. Compute the kernel and the image of π by direct geometric considerations (and rank nullity). Find a basis for each. 4. Find the B matrix of π using the Handy Technique. Why might expressing π in this basis be more convenient than the standard basis?

2 5. Explain how to use the B matrix of π to compute π(~w) where ~w is any vector in R . 2 6. There is a canonical map SB→S sending B-coordinates to S-coordinates of R . What is the source and target? Find the matrix of this map by computing its columns. [This is NOT a B-matrix!]

7. Find SS→B by inverting SB→S . Why does this work? 8. What is the relationship between the B matrix and the S matrix of π.

Solution note: 2 u1 u1u2 1. [T ]S = 2 = A. u1u2 u2 2. The vectors are perpendicular, hence not scalar multiples of eachother. This means that they 2 are linearly independent, and since there are 2 of them and R is 2 dimensional, they are a basis.

3. The image consists of the line spanned by ~u. A basis for the image is {~u}. The kernel consists of the perpendicular line spanned by ~v. A basis for the kernel is {~v}.

4. Since π(~u) = ~u, the B-coordinates of π(~u) = [1 0]T . Since π(~v) = ~0, the B-coordinates of 1 0 π(~v) = [0 0]T . So [π] = . Lots of zeros makes computation fast and easy! Also we B 0 0 easily see that the rank is 1.

5. The technique is: write ~w in B-coordinates, then multiply by the B matrix. The result will be the column of B-coordinates of π(~w). You can rewrite these in standard coordinates if you want.

2 6. The canonical map SB→S has source the B-coordinate space R and target the S-coordinate 2 u1 −u2 space R . It is SB→S = , or the matrix whose columns are the basis B. u2 u1 u1 u2 7. SS→B = , which is the inverse of the matrix in the previous problem. −u2 u1 8. They are similar! if we represent a linear transformation in two diﬀerent bases, the matrices are always similar. Precisely

1 0 [π] = = S−1 [π] S = S−1AS B 0 0 B→S S B→S

where S is the matrix whose columns are the basis elements of B.