Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real...
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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
asked Nov 20 at 0:52
S. Snake
324
add a comment |
up vote
1
down vote
favorite
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
asked Nov 20 at 0:52
S. Snake
324
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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 20 at 0:52
S. Snake
324
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
linear-algebra matrices eigenvalues-eigenvectors vectors
asked Nov 20 at 0:52
S. Snake
324
asked Nov 20 at 0:52
S. Snake
324
asked Nov 20 at 0:52
S. Snake
324
asked Nov 20 at 0:52
S. Snake
324
asked Nov 20 at 0:52
S. Snake
324
324
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
add a comment |
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
add a comment |
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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