Mixing
Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real...
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 '18 at 0:52
S. Snake
485
add a comment |
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 '18 at 0:52
S. Snake
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 '18 at 0:54
I mistakenly thought x=23. Thank you!
– S. Snake
Nov 20 '18 at 0:55
add a comment |
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 '18 at 0:52
S. Snake
485
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 '18 at 0:52
S. Snake
485
asked Nov 20 '18 at 0:52
S. Snake
485
asked Nov 20 '18 at 0:52
S. Snake
485
asked Nov 20 '18 at 0:52
S. Snake
485
asked Nov 20 '18 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 '18 at 0:54
I mistakenly thought x=23. Thank you!
– S. Snake
Nov 20 '18 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 '18 at 0:54
I mistakenly thought x=23. Thank you!
– S. Snake
Nov 20 '18 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 '18 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 '18 at 0:54
I mistakenly thought x=23. Thank you!
– S. Snake
Nov 20 '18 at 0:55
I mistakenly thought x=23. Thank you!
– S. Snake
Nov 20 '18 at 0:55
add a comment |
1 Answer
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If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
– aleph_two
yesterday
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
– aleph_two
yesterday
add a comment |
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
– aleph_two
yesterday
add a comment |
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
answered yesterday
community wiki
aleph_two
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
– aleph_two
yesterday
add a comment |
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
– aleph_two
yesterday
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
– aleph_two
yesterday
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
– aleph_two
yesterday
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 '18 at 0:54
I mistakenly thought x=23. Thank you!
– S. Snake
Nov 20 '18 at 0:55