SokMATH2250S24PS5.p...RaneyLinear AlgebraSpring 2024Problem Set 51. Show that x=⎣⎡−211⎦⎤ is an eigenvector of A=⎣⎡011112−110⎦⎤ and find the corresponding eigenvalue.2. Show that λ=i=−1 is an eigenvalue of A=[01−10], and find a corresponding eigenvector.3. Let A=[2640]. Find the eigenvalues of A and bases for the corresponding eigenspaces.4. Let A be an idempotent matrix (that is, A2=A ). Show that λ=0 and λ=1 are the only possible eigenvalues of A.5. Let A=⎣⎡011112−110⎦⎤1 be the linear transformation defined byT(p(x))=xp′(x)(a) Which, if any, of the polynomialsp1(x)=1,p2(x)=x,p3(x)=x2are in A=⎣⎡011112−110⎦⎤2 ?(b) Which, if any, of the polynomialsp1(x)=1,p2(x)=x,p3(x)=x2are in A=⎣⎡011112−110⎦⎤3 ?6. Find the dimension of the vector space A=⎣⎡011112−110⎦⎤4 and give a basis for A=⎣⎡011112−110⎦⎤5.7. Define a linear transformation A=⎣⎡011112−110⎦⎤6 byT(A)=AB−BAwhere A=⎣⎡011112−110⎦⎤7. Find a basis for A=⎣⎡011112−110⎦⎤2.(D)DashboardCalendarA=⎣⎡011112−110⎦⎤9To-doNotificationsλ=i=−10Inbox
Q. SokMATH2250S24PS5.p...RaneyLinear AlgebraSpring 2024Problem Set 51. Show that x=⎣⎡−211⎦⎤ is an eigenvector of A=⎣⎡011112−110⎦⎤ and find the corresponding eigenvalue.2. Show that λ=i=−1 is an eigenvalue of A=[01−10], and find a corresponding eigenvector.3. Let A=[2640]. Find the eigenvalues of A and bases for the corresponding eigenspaces.4. Let A be an idempotent matrix (that is, A2=A ). Show that λ=0 and λ=1 are the only possible eigenvalues of A.5. Let A=⎣⎡011112−110⎦⎤1 be the linear transformation defined byT(p(x))=xp′(x)(a) Which, if any, of the polynomialsp1(x)=1,p2(x)=x,p3(x)=x2are in A=⎣⎡011112−110⎦⎤2 ?(b) Which, if any, of the polynomialsp1(x)=1,p2(x)=x,p3(x)=x2are in A=⎣⎡011112−110⎦⎤3 ?6. Find the dimension of the vector space A=⎣⎡011112−110⎦⎤4 and give a basis for A=⎣⎡011112−110⎦⎤5.7. Define a linear transformation A=⎣⎡011112−110⎦⎤6 byT(A)=AB−BAwhere A=⎣⎡011112−110⎦⎤7. Find a basis for A=⎣⎡011112−110⎦⎤2.(D)DashboardCalendarA=⎣⎡011112−110⎦⎤9To-doNotificationsλ=i=−10Inbox
Check Eigenvector: To show that x is an eigenvector of A, we need to check if Ax=λx for some scalar λ. Calculation: A⋅x=[01−1111120]⋅[−211] =[(0⋅−2)+(1⋅1)+(−1⋅1)(1⋅−2)+(1⋅1)+(1⋅1)(1⋅−2)+(2⋅1)+(0⋅1)] =[0+1−1−2+1+1−2+2+0] =[000] Since Ax=0, which is not a scalar multiple of x, x is not an eigenvector.