國立政治大學112學年度轉學生招生考試
科目:線性代數系所別:應用數學系三年級
解答:ℓ is a subspace of R2⇒(0,0)∈ℓ⇒0=m⋅0+b⇒b=0b=0⇒ℓ={(x,y)∈R2:y=mx}Suppose {w1=(x1,y1)∈ℓw2=(x2,y2)∈ℓ⇒{y1=mx1y2=mx2⇒ay1+by2=(ax1+bx2)m⇒(ax1+bx2,ay1+by2)=aw1+bw2∈ℓ⇒ℓ is a subspace of R2QED解答:A=[2−1−1−12−1−1−12]⇒det(A−λI)=−λ(λ−3)2=0⇒λ=0,3λ1=0⇒(A−λ1I)v=0⇒[2−1−1−12−1−1−12][x1x2x3]=0⇒{x1=x3x2=x3⇒v=x3[111]⇒ we choose v1=[111]λ2=3⇒(A−λ3I)v=0⇒[−1−1−1−1−1−1−1−1−1][x1x2x3]=0⇒x1+x2+x3=0⇒v=[−x2−x3x2x3]=x2[−110]+x3[−101]⇒ we choose v2=[−110],v3=[−101]⇒P=[v1v2v3],D=[λ1000λ2000λ3]⇒A=[1−1−1110101][000030003][131313−1323−13−13−1323]
解答:A=[100020003]⇒det(A−λI)=−(λ−1)(λ−2)(λ−3)=0λ1=3⇒(A−λ1I)v=0⇒[−2000−10000][x1x2x3]=0⇒x1=x2=0⇒v=x3[001],we choose v1=[001]λ2=2⇒(A−λ2I)v=0⇒[−100000001][x1x2x3x4]=0⇒x1=x3=0⇒v=x2[010],we choose v2=[010]λ3=1⇒(A−λ3I)v=0⇒[000010002][x1x2x3]=0⇒x2=x3=0⇒v=x1[100],we choose v3=[100]⇒P=[v1v2v2]⇒Yes, they are similar. P=[001010100]
解答:(a)T:Rn→R⇒[T] is 1 by n⇒rank([T])=1⇒dim(N(T))=n−rank([T])=n−1QED(b)Suppose avi+bw=0,i=1−(n−1), then T(avi+bw)=0⇒aT(vi)+bT(w)=0⇒bT(w)=0⇒b=0(w∉N(T)⇒T(w)≠0)⇒avi+bw=avi=0⇒a=0⇒vi and w are linearly independent ⇒B′ is a basis of RnQED(c)From (b), we have u∈Rn⇒u∈span(B′)⇒u=a1v1+a2v2+⋯+an−1vn−1+bw We can choose v=a1v1+a2v2+⋯+an−1vn−1∈span(B)⇒v∈N(T) ,and choose b=T(u)T(w), where T(w)≠0. Then u=v+T(u)T(w)wQED
解答:(a) [A∣I]=[121100254010110001]R2−2R1→R2,R3−R1→R3→[121100012−2100−1−1−101]R1−2R2→R2,R3+R2→R3→[10−35−20012−210001−311]R1+3R3→R1,R2−2R3→R2→[100−4130104−1−2001−311]⇒A−1=[−4134−1−2−311](b) rref(A)=I3⇒Yes, they are independent.(c) A−1b=[−33−2]⇒b=−3A1+3A2−2A3
解答:A=[2−10103−1001100−103]⇒det(A−λI)=(λ−2)3(λ−3)=0λ1=2⇒(A−λ1I)v=0⇒[0−10101−1001−100−101][x1x2x3x4]=0⇒{x2=x4x2=x3⇒v=a[1000]+b[0111],we choose v1=[1000],v2=[0111]λ2=3⇒(A−λ2I)v=0⇒[−1−10100−1001−200−100][x1x2x3x4]=0⇒{x1=x4x2=x3=0⇒v=a[1001],we choose v3=[1001]A is 4 by 4 and has 3 eigenvectors, then J=[2100020000200003]
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