TRUE/FALSE QUESTIONS FOR MIDTERM 2

1 MATH 54 TRUE/FALSE QUESTIONS FOR MIDTERM 2 SOLUTIONSPEYAM RYAN TABRIZIAN1. (a)TRUEIfAis diagonalizable, thenA3is diagonalizable.(A=PDP 1, soA3=PD3P= P D P 1, where P=Pand D=D3, which is diagonal)(b)TRUEIfAis a3 3matrix with3(linearly independent)eigenvectors, thenAis diagonalizable(This is one of the facts we talked about in lecture, the point isthat to figure out ifAis diagonalizable, look at the eigenvec-tors)(c)TRUEIfAis a3 3matrix with eigenvalues = 1,2,3,thenAis invertible(No eigenvalue which is0, so by the IMT,Ais invertible)(d)TRUEIfAis a3 3matrix with eigenvalues = 1,2,3,thenAis (always) diagonalizable(this is the useful test we ve been talking about in lecture,Aisdiagonalizable since it has3distinct eigenvalues)(e)FALSEIfAis a3 3matrix with eigenvalues = 1,2,2,thenAis (always) not diagonalizable(TakeA= 1 0 00 2 00 0 2 , it is diagonal, hence diagonalizable)Date: Monday, April 13th, RYAN TABRIZIAN(f)FALSEIf xis the orthogonal projection ofxonW, then xisorthogonal tox.

2 (Draw a picture)(g)FALSEIf uis the orthogonal projection ofuonSpan{v},then: u=(u vv v)u(It s u=(u vv v)v, it has to be a multiple ofv)(h)TRUEIfQis an orthogonal matrix, thenQis invertible.(Remember that in this course, orthogonal matrices are square)2. (a)FALSEIfAis diagonalizable, then it is example, takeA=[0 00 0]. It is diagonalizablebecause itis diagonal, but it is not invertible!(b)FALSEIfAis invertible, thenAis diagonalizableTakeA=[1 10 1](this is the magic counterexample we talkedabout in lecture). It is invertible becausedet(A) = 16= show it is not diagonalizable, let s find the eigenvalues andeigenvectors ofA:Eigenvalues:det( I A) = 1 10 1 = ( 1)2= 0 Which gives us = :Nul(I A) =Nul[0 10 0]MATH 54 TRUE/FALSE QUESTIONS FOR MIDTERM 2 SOLUTIONS3 Which gives y= 0, soy= 0, hence:Nul(I A) ={[x0]}=Span{[10]}Since there is only one (linearly independent) eigenvector,Aisnot diagonalizable! (30 points, 5 pts each)Label the following statements sure toJUSTIFY YOUR ANSWERS!

3 !!You may use anyfacts from the book or from lecture.(a) IfA={a1,a2,a3}andD={d1,d2,d3}are bases forV, andPis the matrix whoseith column is[di]A, then for allxinV,we have[x]D=P[x]AFALSEF irst of all,P=[[d1]A[d2]A[d3]A]=PA D(remember, you always evaluate with respect to the new, coolbasis, here it isA), so we should have:[x]A=PA D[x]D=P[x]DAnd not the opposite!(b) A3 3matrixAwith only one eigenvalue cannot be diagonal-izableSUPER false !!!!!!!!!!4 PEYAM RYAN TABRIZIANR emember that to check if a matrix is not diagonalizable, youreally have to look at the eigenvectors!For example,A= 2 0 00 2 00 0 2 has only eigenvalue2, but isdiagonalizable (it s diagonal!). Or you can chooseAto be theOmatrix, or the identity matrix, this also works!(c) Ifv1andv2are2eigenvectors ofAcorresponding to2differ-enteigenvalues 1and 2, thenv1andv2are linearly indepen-dent!TRUE(finally!)Note:The proof is a bit complicated, but I ve seen this on apast exam! I think at that point, the professor wanted to getrevenge on his students for not coming to lecture!

4 Remember that eigenvectors have to be nonzero!Now, assumeav1+bv2= applyAto this to get:A(av1+bv2) =A(0) =0 That is:aA(v1) +bA(v2) =0a 1v1+b 2v2=0 However, we can also multiply the original equation by 1toget:a 1v1+b 1v2=0 Subtracting this equation from the one preceding it, we get:b( 1 2)v2=0 SoMATH 54 TRUE/FALSE QUESTIONS FOR MIDTERM 2 SOLUTIONS5b( 1 2) =0 But 16= 2, so 1 26= 0, hence we getb= going back to the first equation, we get:av1=0 Soa= 0, and we re done!(d) If a matrixAhas orthogonal columns, then it is an that an orthogonalmatrix has to have orthonormalcolumns!(e) For every subspaceWand every vectory,y ProjWyisorthogonal toProjWy(proof by picture is ok here)TRUEDraw a picture!ProjWyis just another name for y.(f) Ifyis already inW, thenProjWy=yTRUEA gain, draw a picture!If you want a more mathematical proof, here it is:LetB={w1, wp}be an orthogonal basis forW(p=Dim(W)).6 PEYAM RYAN TABRIZIANT heny=(y w1w1 w1)w1+ +(y wpwp wp) then, by definition ofProjWy= y, we get: y=(y w1w1 w1)w1+ +(y wpwp wp)wp=ySo y=yin this (a) IfAis a3 3matrix with eigenvalues = 0,2,3, thenAmustbe diagonalizable!

5 TRUE(ann nmatrix with3distinct eigenvalues is diago-nalizable)(b) There does not exist a3 3matrixAwith eigenvalues =1, 1, 1 + (here we assumeAhas real entries; eigenvalues alwayscome in complex conjugate pairs, ifAhas eigenvalue 1+i, it must also have eigenvalue 1 i)(c) IfAis a symmetric matrix, then all its eigenvectors are : TakeAto be your favorite symmetric matrix, and,for example, takevto be one eigenvector, andwto be thesameeigenvector (or a different eigenvector corresponding toMATH 54 TRUE/FALSE QUESTIONS FOR MIDTERM 2 SOLUTIONS7the same eigenvalue). That s why we had to apply the GramSchmidt process to each eigenspace in the previous problem!(d) IfQis an orthogonaln nmatrix, thenRow(Q) =Col(Q).TRUE: (sinceQis orthogonal,QTQ=I, soQis invertible,henceRow(Q) =Col(Q) =Rn)(e) The equationAx=b, whereAis an nmatrix always has aunique least-squares : TakeAto be the zero matrix, andbto be the zerovector! This statement is true ifAhas rankn.

6 (f) IfAB=I, thenBA= : LetA=[1 0]andB=[10]. ThenAB=I, butBA=[1 00 0]!(g) IfAis a square matrix, thenRank(A) =Rank(A2) false : LetA=[0 10 0], thenRank(A) = 1, butA2=[0 00 0], soRank(A2) = RYAN TABRIZIAN(h) IfWis a subspace, andPyis the orthogonal projection ofyontoW, thenP2y=PyTRUE(draw a picture! If you orthogonally projectPy= yonW, you get y)(i) IfT:V W, wheredim(V) = 3anddim(W) = 2, thenTcannot be (by Rank-Nullity theorem,dim(Nul(T))+Rank(T) =3. ButRank(T)can only be at mostdim(W) = 2, sodim(Nul(T))>0, soNul(T)6={0})(j) IfAis similar toB, thendet(A) =det(B).TRUE(IfA=PBP 1, thendet(A) =det(B))