Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 21 September 2014

Lecture 9: Change of basis

Let V be a vector space. Let B={b1,b2,} be a basis of V. Let C={c1,c2,} be another basis of V.

The change of basis matrix from B to C is P=(pij)given by cj=p1jb1 +p2jb2+. The change of basis matrix from C to B is Q=(qij)given by bj=q1jc1 +q2j+. Let f:VV be a linear transformation. The matrix of f with respect to B is Bf=(fijB) given byf(bj)= f1jBb1+ f2jBb2+. The matrix of f with respect to C is Cf(fijC) given byf(cj)= f1jCc1+ f2jCc2+.

(a) P=Q-1.
(b) Bf=QCfQ-1.

My favourite vector space V=3= { (a1a231) |a1,a2,a3 } has basis B={b1,b2,b3} with b1=(100), b2=(010), b3=(001). The matrix Bf= ( 001 100 010 ) defines a linear transformation f:VV, f(b1)=b2, f(b2)=b3, f(b3)=b1 and f(3621) = f(3b1+6b2+21b3) = 3f(b1)+ 6f(b2)+ 21f(b3) = 3b2+6b3+21b1 = (2136). Another basis of V is C={c1,c2,c3} with c1=(111), c2=(1-1+3i2-1-3i2), c2=(1-1-3i2-1+3i2). Then c1 = b1+b2+b3, c2 = b1+(-1+3i2) b2+(-1-3i2)b3, c3 = b1+(-1-3i2) b2+(-1+3i2)b3, and b1 = 13(c1+c2+c3), b2 = 13 ( c1+ -1-3i2c2+ -1+3i2c3 ) , b3 = 13 ( c1+ -1+3i2c2+ -1-3i2c3 ) . Helpful: Let ζ=-1+3i2 and note that ζ2 = (-1+3i2)2= 1-23i-34= -1-3i2and ζ3 = (-1-3i)2 (-1+3i)2= 1+34=1, and 1+ζ+ζ2=0. So c1 = b1+b2+b3, c2 = b1+ζb2+ ζ2b3, c3 = b1+ζ2b2+ζb3 and 13(c1+c2+c3) = 13 ( (111)+ (1ζζ2)+ (1ζ2ζ) ) =13(300)= (100)=b1, 13(c1+ζ2c2+ζc3) = 13 ( (111)+ ζ2(1ζζ2)+ ζ(1ζ2ζ) ) =13 ( (111)+ (ζ21ζ)+ (ζ1ζ2) ) =13(030)=b2, 13(c1+ζc2+ζ2c3) = 13 ( (111)+ ζ(1ζζ2)+ ζ2(1ζ2ζ) ) =13 ( (111)+ (ζζ21)+ (ζ2ζ1) ) =13(003)=bb. The change of basis matrix from B to C is P= ( 111 1ζζ2 1ζ2ζ ) and the change of basis matrix from C to B is Q= ( 131313 1313ζ213ζ 1313ζ13ζ2 ) and PQ= ( 111 1ζζ2 1ζ2ζ ) ( 131313 1313ζ213ζ 1313ζ13ζ2 ) = ( 100 010 001 ) . So P-1=Q.

The matrix of f with respect to C: f(c1) = f(111) =(111) =c1, f(c2) = f(1ζζ2) =(ζ21ζ) =ζ2(1ζζ2) =ζ2c2, f(c3) = f(1ζ2ζ) =(ζ1ζ2) =ζ(1ζ2ζ) =ζc1. So the matrix of f with respect to C is Cf= ( 100 0ζ20 00ζ ) . Magic: QBfP = ( 131313 1313ζ213ζ 1313ζ13ζ2 ) ( 001 100 010 ) ( 111 1ζζ2 1ζ2ζ ) = 13 ( 111 1ζ2ζ 1ζζ2 ) ( 1ζ2ζ 111 1ζζ2 ) = 13 ( 100 03ζ20 003ζ ) = ( 100 0ζ20 00ζ ) = Cf.

Let f:VV be a linear transformation. Let B and C be bases of V. Let P be the change of basis matrix from B to C,
Q the change of basis matrix from C to B,
Bf the matrix of f with respect to B,
Cf the matrix of f with respect to C.
Then

(a) P=Q-1,
(b) Bf=QCfQ-1.

Proof.

(a)
To show:
(aa) PQ=1,
(ab) QP=1.
(aa)
We know: P = (pij)with cj=p1j b1+p2jb2+, Q = (qk)with b=q1c1+ q2c2+.
To show: PQ=1.
To show:
(aaa) (PQ)ii=1,
(aab) (PQ)ij=0 if ij.
(aaa) (PQ)ii= pi1q1i+ pi2q2i+ pi3q3i+ since bi = q1ic1+ q2ic2+ q3ic3+ = q1i ( p11b1+ p21b2+ p31b3+ ) + q2i ( p12b1+ p22b2+ p32b3+ ) + q3i ( p13b1+ p23b2+ p33b3+ ) = ( q1ip11+ q2ip12+ q3ip13+ ) b1+ ( q1ip21+ q2ip22+ q3ip23+ ) b2+. If we use sum notation bj = qjc = qj ( m pmbm ) = ,m pm qjbm. So pm qj=0 if mj and pj qj=1.

Notes and References

These are a typed copy of Lecture 9 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 12, 2011.

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