Selected Problems — Matrix Algebra Math 2300

1 Selected Problems Matrix AlgebraMath that ifAis nonsingular thenATis nonsingular and(AT) 1= (A 1) :Lets put into words what are we asked to show in this problem. First, we mustshow that if a Matrix is invertible, then so is its transpose. We must also show that theinverse of the transpose is the same as the transpose of the inverse. In other words, if wethink of inverting and transposing as processes we may perform on square matrices, then forinvertible matrices, these two processes performed in either order yield the same :LetAbe nonsingular. By definition, there existsA 1such thatA 1A=AA 1= we note thatIT=I(verify this).

2 Now we see thatI=IT= (A 1A)T,sinceA 1A=I=AT(A 1)T,since (AB)T= we haveAT(A 1)T=I(1)Similarly, we can show that(A 1)TAT=I.(2)ThusATand (A 1)Tmultiply to give the identity Matrix . HenceATis invertible. Further-more, (1) and (2) show that (A 1)Tis the inverse ofAT. In mathematical notation, thisstatement becomes the equation(A 1)T= (AT) (Note: is an abbreviation for the Latin phrase Quod erat demonstrandum whichmeans That which was to be demonstrated , signifying the end of the proof.) that ifAis a symmetric nonsingular Matrix , thenA 1is also :SinceAis symmetric,AT=A. SinceAis nonsingularA 1exists. Now, we needto examine the transpose ofA 1.

3 The result we have just proved in the previous problem isrelevant!(A 1)T= (AT) 1,by previous problem=A 1,sinceAT=AThusA 1is symmetric, since it is equal to its own ann nmatrix and letxandybe vectors inRn. Show that ifAx=Ayandx =y,then the matrixAmust be (by contradiction):We are given thatAx=Aywithx =y. We have to argue thatthis forcesAto be singular. Suppose, on the contrary, thatAis nonsingular. Then thereexistsA 1such thatA 1A=AA 1=I. Now,Ax=Ay= A 1(Ax) =A 1(Ay),left multiplying byA 1= (A 1A)x= (A 1A)y,since Matrix multiplication is associative= Ix=IysinceA 1A=I= x=ywhich contradictsx =y. HenceAmust be ann nmatrix and letB=A+ATandC=A AT(3)(a)Show thatBis symmetric andCis.

4 To showBis symmetric, we must show that it equals its own (A+AT)T,by (3)=AT+ (AT)T,since transpose of a sum is the sum of the transposes=AT+A,since taking the transpose twice yields the original Matrix =Bsince Matrix addition is commutativeThusB=BTand soBis showCis skew-symmetric, we must show thatC= (A AT)T,by (3)=AT (AT)T,since the transpose of a difference is the difference of the transposes=AT A,since taking the transpose twice yields the original Matrix = (A AT),factoring out the scalar 1= C,by (3)ThusCT= Cwhich impliesC= CTand we see thatCis (b)Show that everyn nmatrix can be represented as a sum of a symmetric Matrix and askew-symmetric : LetAbe ann nmatrix.

5 ThenATexists and is also ann nmatrix. By part(a),A+ATis symmetric andA ATis skew-symmetric. Now we notice that(A+AT) + (A AT) = 2A(4)since Matrix addition is associative and is close to what we want, but not exactly what we want. We have a symmetricmatrix and a skew-symmetric Matrix that add to give 2A, the matrixAtimes the scalar2. We fix the problem by multiplying both sides of (4) by 1 [(A+AT) + (A AT)] =12(2A)= 12(A+AT) +12(A AT) =Asince scalar multiplication distributes over Matrix , we note that multiplying a symmetric Matrix by a scalar yields a symmetricmatrix (the reader should verify this). Similarly, a scalar times a skew-symmetric matrixyields a skew-symmetric Matrix (verify).

6 Thus12(A+AT) and12(A AT) are symmetricand skew-symmetric respectively and we have expressedAas the sum of a symmetricmatrix and a skew-symmetric general, Matrix multiplication is not commutative ( ,AB =BA). However, in certainspecial cases the commutative property does hold. Show that:(a)IfAandBaren ndiagonal matrices, thenAB= We may think of this productrow-wise: theith row ofABis a linear combination of therows ofBwith scalars coming from theith row ofA. Thus in determining the first rowofAB, for instance, we see that the first row ofBgets scaled bya11while all the otherrows get scaled by 0. So the first row ofABhas first entrya11b11, all the other entriesbeing 0.

7 The attentive reader should verify thatAB= 3 Similarly,BA= Since multiplication of scalars is commutative,aiibii=biiaiifori= 1.. n. ThusAB=BA.(b)IfAis ann nmatrix andB=a0I+a1A+a2A2+..+akAk(5)wherea0, a1, .. , akare scalars, thenAB= :AB=A(a0I+a1A+a2A2+..+akAk) by (5)=a0A+a1A2+a2A3+..+akAk+1since Matrix multiplication distributesover Matrix addition= (a0I+a1A+a2A2+..+akAk)Afactoring out A on theright=BAby (5) by Niloufer Mackey and Todd ThomasFirst created: February Updated: Friday 18thJanuary, 2013, 10:514