Fall 97 Exam 2 Solutions

(1)

• True, as 0 = det(A^10) = (det A)^10 implies det A = 0 and so 0 must be an eigenvalue of A.
• False, for example

   
  0 1 
  0 0 

• True, since at least one eigenvalue is zero and the trace is the sum of the two eigenvalues.
• True (envision doing A A^T by dotting rows of A with cols of A^T)
• False, the projection cleary collapses every angle between vectors to zero
• True, as there is only one pattern of 1's with six inversions (an even number)

(2)

(3)

• x_1 = t, x_2 = 1-t, x_3 = t.
• squareroot of 6.

(4)

• eigenvalue are 1,3 with algebraic mutlilpicity 1,2 respectively.
• dim(E_1)=1, dim(E_2)=1< algebraic multiplicity of 2. So there is no eigenbasis for A.

(5)

• 3 > or = dim(Im(A)) > or = dim(Im(AB))= 3, so rank(A)=dim(Im(A))=3 rank(B)=dim(Im(B))=3- dim(ker(B))=3-0=3
• If Ax=0, BAx=B(Ax)=B0=0, the corresponding eigenvalue is 0.
• If x=By, then BA(x)= BABy=B(AB)y=BIy=By=x, the corresponding eigenvalue is 1.
• Ker(A)(dim=2) contained in E_0, Im(B)(dim=3) contained in E_1, 5=2+3=dim(Ker(A))+dim(Im(B))< or = dim(E_0)+dim(E_1)< or = 5 so Ker(A)=E_0, Im(B)=E_1, they spand the totall space. eigenvalues of BA are 0,1 with geometric multiplicity 2,3 respectively.