Transcription of SIMILAR MATRICES Similar Matrices - Mathematics
1 SIMILAR MatricesFix a linear transformationT:Rn Rnand an ordered basis,B= (~v1,..,~vn), ofRn. The standard matrix [T] and thematrix ofTwith respect toB, [T]B, are related by[T]S=S[T]Band [T] =S[T]BS 1and [T]B=S 1[T] hereS=[~v1| |~vn]is the change of basis matrix of the basis. In order to understand this relationshipbetter, it is convenient to take it as a definition and then study it twon nmatrices,AandB, we say thatAissimilartoBif there existsan invertiblen nmatrix,S, so thatAS= , this is equivalently, toA=SBS 1orB=S : IfAis SIMILAR toIn, thenA= :A=[2 31 2]is similarB=[100 1]. To see this, letS=[3 11 1],and computeAS=[3 11 1]= : IfAis SIMILAR toB, thenA2is SIMILAR toB2. To see this observe,that, by definition,A=SBS 1for some invertibleS.
2 Hence,A2= (SBS 1)(SBS 1) =SB(S 1S)BS 1=SBInBS 1=SB2S 1and soA2is SIMILAR : IfAis SIMILAR toBand one is invertible, then both are andA 1is SIMILAR toB 1. Indeed, supposeBis invertible, thenA=SBS 1for invertibleSand soAis also invertible as it is the product of three invertible MATRICES . Asimilar argument shows thatBis invertible ifAis. Finally, one hasA 1= (SBS 1) 1= (S 1) 1B 1S 1=SB 1S 1,which completes the proof. In general,Atis SIMILAR toBtfor any integert(needA,Binvertible whent <0).12 SIMILAR MATRICESEXAMPLE:A=[2 31 2]hasA100=I2. This is becauseAis SIMILAR toB=[100 1]. AsBis diagonal,B100=[110000( 1)100]=I2,and soAis SIMILAR toI2and hence is equal toI2. Similarly,A101=SB101S 1=SBS 1= of SIMILAR matricesBeing SIMILAR is aequivalence relation.
3 That is:(1) (Reflexive):Ais SIMILAR toA.(2) (Symmetric):Ais SIMILAR toB Bis SIMILAR toA(3) (Transitive): ifAis SIMILAR toBandBis SIMILAR toC, thenAis is relatively straightforward to check:(1) Follows by takingS=In.(2) Follows by observingAS=SBmeansBS 1=S 1A, =S AforS =S 1.(3) Follows by observingAS=SBandBT=TCmeans thatAST=SBT=STCand soAis SIMILAR toCby usingS =ST. Here we used thatS isthe product of two invertible MATRICES and so is is not easy, in general, to tell whether two MATRICES are SIMILAR and this is aquestion we will return to later in the class. It can be easy to tell when they SIMILAR , thennull(A) = null(B)(and sorank(A) =rank(B)). SIMILAR toB, we haveAS=SBfor some invertibleS.
4 Now supposethat~z ker(B). It is clear thatS~z ker(A). Indeed,A(S~z) = (AS)~z= (SB)~z=S(B~z) =S~0 =~ , if~z1,..,~zpare a basis of ker(B), thenS~z1,..,S~zpare linearly inde-pendent. Indeed,c1S~z1+ +cpS~zp=~0 S(c1~z1+ cp~zp)Hence,c1~z1+ cp~zp ker(S) c1~z1+ cp~zp=~ must be a trivial linear relation (asz1,..,zpare a basis) which shows theonly linear relations amongS~z1,..,S~zpis the trivial one which proves the , we haveplinearly independent vectors in ker(A) and sonull(A) = dim ker(A) p= dim(ker(B)) = null(B).As this argument is symmetric inAandBwe conclude also that null(B) null(A)which proves the result. Finally, by the rank-nullity theoremrank(A) =n null(A) =n null(B) = rank(B)wherenis the number of columns inAandB.
5 SIMILAR MATRICES3 EXAMPLE:[1 12 2]is not SIMILAR to[120 1]. By inspection, the first matrixhas rank = 1 and second has rank = MatricesA matrix isdiagonalif its only non-zero entries are on the diagonal. For instance,B= k1000k2000k3 ,is a 3 3 diagonal matrix. Geometrically, a diagonal matrix acts by stretching each of the standard vectors. Algebraically, this meansBis diagonal if and only ifB~ei=ki~eifor each standard vector~ei. As we have seen diagonal MATRICES and MATRICES thatare SIMILAR to diagonal MATRICES are extremely useful for computing large powersof the matrix. As such, it is natural to ask when a given matrix is SIMILAR to adiagonal have the following complete answer:Theorem matrixAis SIMILAR to a diagonal matrix if and only if there isan ordered basisB= (~v1.)
6 ,~vn)so thatA~vi=ki~vifor someki isAstretches the~viby a factorki. It is worth mentioning that the~viareexamples ofeigenvectors ofA(a topic we will study later). first thatAis SIMILAR to a diagonal matrixB. That is,AS=SBfor a diagonal matrixBand invertible matrixS. AsBis diagonalB~ei=ki~eiwherekiis theith entry on the ~vi=S~ei, thenB= (~v1,..,~vn) is an ordered basis. Moreover,A~vi=AS~ei=SB~ei=S(B~ei) =S(ki~ei) =kiS~ei=ki~ verifies one direction of the , if one has the ordered basisB= (~v1,..,~vn) so thatA~vi=ki~vi, andSis the change of basis matrix ofB, then(S 1AS)~ei=S 1(A(S~ei)) =S 1(A(~vi)) =S 1(ki~vi) =kiS 1~vi=ki~ ,B=S 1 ASis diagonal and is SIMILAR toAby definition. This may be rephrased in terms of linear transformations as follows:Theorem linear transformationT:Rn Rngiven byT(~x) =A~xhas abasisB= (~v1.
7 ,~vn)so that[T]Bis diagonal if and only ifT(~vi) = i~ [T] andB= [T]Band apply the preceding theorem. EXAMPLE: LetR 2:R2 R2be rotation counter clockwise by 90 . For any~v6=~0,R 2(~v) is perpendicular to~v, so there are no non-zero vectors that getstretched. Hence, there is no basisBfor which [R 2]Bis MATRICESEXAMPLE: LetP:R2 R2be projection onto the liney=x. This meansP([11])=[11]andP([1 1])=[00].Hence, ifB=([11],[1 1]),then[P]B=[1 00 0].As a consequence, ifS=[111 1]is the change of basis matrix ofB, then[P] =S[P]BS 1=S[1 00 0]S 1.
