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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.

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7.5 Change of Basis

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  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.

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