Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real...












1














Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real number, compute A(xv) in terms of x and v.



A(xv) = (?)



Not sure where to go with this question, is there a rule or condition that I am missing?



So far I thought it would be something using the following;



det(A - xI) = 0 and Ax = vx and solving for A but any simplification using these ideas have been wrong so far. Any suggestions?










share|cite|improve this question


















  • 2




    It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
    – DonAntonio
    Nov 20 at 0:54










  • I mistakenly thought x=23. Thank you!
    – S. Snake
    Nov 20 at 0:55
















1














Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real number, compute A(xv) in terms of x and v.



A(xv) = (?)



Not sure where to go with this question, is there a rule or condition that I am missing?



So far I thought it would be something using the following;



det(A - xI) = 0 and Ax = vx and solving for A but any simplification using these ideas have been wrong so far. Any suggestions?










share|cite|improve this question


















  • 2




    It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
    – DonAntonio
    Nov 20 at 0:54










  • I mistakenly thought x=23. Thank you!
    – S. Snake
    Nov 20 at 0:55














1












1








1







Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real number, compute A(xv) in terms of x and v.



A(xv) = (?)



Not sure where to go with this question, is there a rule or condition that I am missing?



So far I thought it would be something using the following;



det(A - xI) = 0 and Ax = vx and solving for A but any simplification using these ideas have been wrong so far. Any suggestions?










share|cite|improve this question













Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real number, compute A(xv) in terms of x and v.



A(xv) = (?)



Not sure where to go with this question, is there a rule or condition that I am missing?



So far I thought it would be something using the following;



det(A - xI) = 0 and Ax = vx and solving for A but any simplification using these ideas have been wrong so far. Any suggestions?







linear-algebra matrices eigenvalues-eigenvectors vectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 0:52









S. Snake

485




485








  • 2




    It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
    – DonAntonio
    Nov 20 at 0:54










  • I mistakenly thought x=23. Thank you!
    – S. Snake
    Nov 20 at 0:55














  • 2




    It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
    – DonAntonio
    Nov 20 at 0:54










  • I mistakenly thought x=23. Thank you!
    – S. Snake
    Nov 20 at 0:55








2




2




It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
– DonAntonio
Nov 20 at 0:54




It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
– DonAntonio
Nov 20 at 0:54












I mistakenly thought x=23. Thank you!
– S. Snake
Nov 20 at 0:55




I mistakenly thought x=23. Thank you!
– S. Snake
Nov 20 at 0:55















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