1 / 25

7.5 Change of Basis

7.5 Change of Basis. Background. Sometimes it is convenient for us to change our basis. For instance, it might be helpful to change the coordinate axes in  3 to better suit our situation. When we do, the vectors we have will also change.

clarke-chan
Download Presentation

7.5 Change of Basis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 7.5 Change of Basis

  2. Background Sometimes it is convenient for us to change our basis. For instance, it might be helpful to change the coordinate axes in 3 to better suit our situation. When we do, the vectors we have will also change. This section deals with those changes and changes to the matrix of a LT when we change basis.

  3. Dealing with the change If D={d1,d2} is our new basis and B={b1,b2} our old, we will want to know the relationship between the coordinate vectors CD(v) and CB(v). b1=rd1 + sd2 b2=td1 + ud2 Then any vector v in V can be written using original basis: v=v1b1 + v2b2 becomes v=v1(rd1+sd2) +v2(td1+ud2) = (rv1+tv2)d1 + (sv1 +uv2)d2 P=[CD(b1) CD(b2)]

  4. Definition If B={b1,…,bn} and D are any two ordered bases of V, we define the change of basis matrix (or just change matrix) PDB from B to D in block form by listing the columns: PDB = [CD(b1) …. CD(bn) ]

  5. Theorem 1 Let B={b1,…,bn} and D={d1,…,dn} be two ordered bases of a vector space V. If PDB is defined as on the last slide, then CD(v) = PDBCB(v) holds for every vector v in V Proof: already shown in two dimensions-general case is homework.

  6. Example In P2, find PDB if B = {1,x,x2} and D={1,(1-x),(1-x)2}. Then use it to express p = p(x) = a + bx+cx2 as a polynomial in powers of (1-x). Sol’n: 1) Find PDB =[CD(1) CD(x) CD(x2)] 2) CD(p) = PDBCB(p) where CB(p) = [a b c]T 3) write p(x) as linear combo of the new basis vectors.

  7. Theorem 2 Let B,D, and E be three ordered bases of an n-dimensional vector space V. Then: 1. PBB= I 2. PDB is invertible and (PDB)-1 = PBD 3. PED PDB = PEB Proof: Using thm 1 for part 3: (3) PED PDBCB(v) = PED CD(v) = CE(v) = PEB CB(v) Let v = bi for each i. CB(bi) is col i of In. So for any matrix A, A CB(bi) = col i of A. So we will get PED PDB = PEB

  8. Theorem 2 (1) is homework (2) from (3) if E = B, we have PBD PDB= PBB=I from 1 so this gives us that (PDB)-1 = PBD (and could go other way) 

  9. Definition Given linear operator T: VV and an ordered basis B={e1,…,en} of V, the matrix MBB(T) will be called the matrix of T with respect to B (MB(T)): MB(T) = [CB[T(e1)] … CB[T(en)]]

  10. Theorem 3 Let B0 and B be two ordered bases of a finite dimensional vector space V. If T: VV is any linear operator, the matrices MB(T) and MB0(T) of T wrt these bases are similar: MB(T) = P-1MB0(T)P where P = PB0B = [CB0(e1) … CB0(e0)] Where B={e1,…,en}

  11. Theorem 3-proof B={e1,…,en} and B0 are ordered bases of V. CB0(v) = PB0BCB(v) by thm 1 (Let PB0B = P) CB[T(v)] = MB(T)CB(v) from 7.4 (thm 2) So PMB(T)CB(v) = P CB[T(v)] = CB0[T(v)] = MB0(T)CB0(v) = MB0(T)PCB(v) Since CB(ej) = col j of I, we get PMB(T) = MB0(T)P P-1=PBB0 so MB(T) = P-1MB0(T)P 

  12. Example T: 3 3; T(a,b,c) = (2a-b,b+c,c-3a). If B0 denotes the standard basis of 3, and B = {(1,1,0),(1,0,1),(0,1,0)}, find an invertible matrix P such that P-1MB0(T)P = MB(T). Sol’n: Find MB0(T) = [CB0[T(e1)] CB0[T(e2)] CB0[T(e3)] = [CB0(2,0,-3) CB0(-1,1,0) CB0(0,1,1)] Then find MB(T) = [CB[T(b1)] CB[T(b2)] CB[T(b3)] = [CB(1,1,-3) CB(2,1,-2) CB(-1,1,0)] Then P=PB0B = [CB0(1,1,0) CB0(1,0,1) CB0(0,1,0)] 

  13. Diagonalization Recall working with similar matrices when we originally discussed diagonalization. We said that a matrix, A, was diagonalizable if it was similar to a real diagonal matrix (i.e. P-1AP is diagonal). Let (nxn)A = MB0(T) be the matrix for the transformation T: VV wrt an ordered basis B0. If we can find another ordered basis of V, B, such that MB(T)=D is diagonal then we can use thm 3 to find an invertible P such that P-1AP = D. We just have to find a basis B such that MB(T) is diagonal.

  14. Diagonalization-cont For any (nxn) A, we can say MB0(TA) = A where TA: nn; TA(X) = AX and B0 is the standard basis of n.

  15. Theorem 4 A (nxn), B0 is standard basis of n 1. If A’ = P-1AP, let B be the ordered basis of n consisting of the columns of P in order. Then MB0(TA) = A and MB(TA) = A’ 2. If B is any ordered basis of n, let P be the invertible matrix whose columns are the vectors of B in order. Then: MB(TA) = P-1AP

  16. Theorem 4 Proof: (1) let B be the ordered basis of n consisting of the columns of P in order. Show MB0(TA) = A and MB(TA) = A’ let X1,…,Xn be the columns of P. Then PB0B=[CB0(X1) … CB0(Xn)] = [X1 … Xn]=P Thm 3 says that MB(T)=P-1MB0(T)P So MB0(TA)=[TA(E1) … TA(En)] = [AE1 … AEn] = A (2) P = PB0B (since P is just the cols of B in order) Thm 3 says: MB(TA)=P-1MB0(TA)P=P-1AP

  17. Example Verify that P-1AP = D, and then find basis B of 2 st MB(TA)=D Sol’n: P-1AP = D  AP = PD (w/ P invertible) easy to show. Then, we let B consist of the columns of P: MB(TA) = P-1AP = D (by thm 4):

  18. A little summary So now to find P such that P-1AP is a particular type of matrix (in this case, diagonal), we just need to find a basis B such that MB(TA) = D (of desired type-here diagonal). Then P will be the matrix with the vectors in B as its columns.

  19. Example If T: VV is an operator w/ V finite dimensional, show that TST = T for some invertible operator S: VV Sol’n: Let B = {e1,…,er,er+1,…,en} be a basis of V w/ ker T=span {er+1,…,en} Then {T(e1),…,T(er)} is independent (7.2, thm 5) w/ all its vectors in V. Let {T(e1),…,T(er),fr+1,…,fn} be a basis of V. Let S: VV; S[T(ei)]=ei for 1 ≤ i ≤ r S(fj) = ej for r < j ≤ n

  20. Example (cont) S[T(ei)]=ei for 1 ≤ i ≤ r S(fj) = ej for r < j ≤ n TST(v)=TST(v1e1+…+vnen)=T(v1ST(e1)+…+vnST(en)) =T(v1e1+…+vner)+T(vr+1S(0)+…+vnS(0)) = T(v1e1+…+vnen) + T(0+…+0) = T(v1e1+…+vnen) + T(vr+1er+1 +…+vnen) =T(v)

  21. Example A (nxn) show that AUA = A for some invertible U. Solution: Let T =TA: n n. Let TST = T where S: n n is an isomorphism. If B0 is the standard basis of n, then A = MB0(T) (thm 4) If U = MB0(S), then U is invertible (7.4 thm 4) and: AUA = MB0(T)MB0(S)MB0(T) = MB0(TST) = MB0(T) = A (used thm 3 of 7.4)

  22. Recall similarity invariants Recall that trace, det, rank were all similarity invariants (ie all similar matrices will have same det, trace, rank). If T: VV is a linear operator, then all of the matrices of T with respect to the different bases of V are similar (by thm 3) and thus have the same rank, trace, and det. Generally, these similarity invariants are then a property of T and not just of its individual matrices. So, for instance, we say rank T = rank A where A is any mtx of T det T = det MB(T) where B is any basis of V tr T = tr MB(T)

  23. Example Let S and T be linear operators on finite dimensional space V. Show that det (ST) = det SdetT Choose a basis B of V and use 7.4 thm 3: det(ST) = det[MB(ST)] by argument on last slide = det[MB(S)MB(T)] by 7.4 thm 3 = det[MB(S)]det[MB(T)] by laws of determinants = det(S) det(T) by argument on last slide.

  24. Characteristic Polynomial The characteristic polynomial is a similarity invariant. T: VV is linear, and let A = MB(T), w/ B any basis of V cT(x) = cA(x)

  25. Example Find the characteristic polynomial cT(x) of T: P2P2 given by T(a+bx+cx2) = (b+c) + (a+c)x + (a+b)x2 Sol’n: Let B={1,x,x2} the standard basis of P2 Then MB(T) =[CB[T(1)] CB[T(x)] CB[T(x2)] ] =[CB(x+x2) CB(1+x2) CB(1+x)] Then find the characteristic polynomial.

More Related