Mixing
![Which branch of software development matches the situation? [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgjbpfN9tAutmK93bJRC3ZoROZzi2TJDms5n8_qJuhgE0a9b52OOHayv3NGT8igAdFL7byXNst-_1DZK5SjrIJ28_6RQPUpBROqMs5s6jo-ZsjX8kjDwfxJufIitH3TaQRXWaGSQKRQib-f/s72-c/1.jpg)
Problem related to linear transformation
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Problem
Assume that a linear transformation $T$ has two eigenvectors $x$ and $y$ belonging to two distinct eigenvalue
$lambda$ and $mu$ . If $ax+by$ is an eigenvector of$T$,prove that $a=0$ or $b=0$.
Attempt
$T(x)=lambda x$ and $T(y) = mu y$.
$T(ax+by) = omega(ax+by)$,where $omega$ is an eigenvalue for $ax+by$.
$a(lambda -omega)=0$ and $b(mu-omega) =0$ since $x$ and$ y$ are independent( distinct eigenvalues)
.
Problem
How to arrive at result after that ?
linear-algebra linear-transformations
asked Sep 1 at 6:16
blue boy
1,085513
add a comment |Â
up vote
2
down vote
favorite
Problem
Assume that a linear transformation $T$ has two eigenvectors $x$ and $y$ belonging to two distinct eigenvalue
$lambda$ and $mu$ . If $ax+by$ is an eigenvector of$T$,prove that $a=0$ or $b=0$.
Attempt
$T(x)=lambda x$ and $T(y) = mu y$.
$T(ax+by) = omega(ax+by)$,where $omega$ is an eigenvalue for $ax+by$.
$a(lambda -omega)=0$ and $b(mu-omega) =0$ since $x$ and$ y$ are independent( distinct eigenvalues)
.
Problem
How to arrive at result after that ?
linear-algebra linear-transformations
asked Sep 1 at 6:16
blue boy
1,085513
show that if $a$ is not zero then $b=0$
– Lozenges
Sep 1 at 6:36
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem
Assume that a linear transformation $T$ has two eigenvectors $x$ and $y$ belonging to two distinct eigenvalue
$lambda$ and $mu$ . If $ax+by$ is an eigenvector of$T$,prove that $a=0$ or $b=0$.
Attempt
$T(x)=lambda x$ and $T(y) = mu y$.
$T(ax+by) = omega(ax+by)$,where $omega$ is an eigenvalue for $ax+by$.
$a(lambda -omega)=0$ and $b(mu-omega) =0$ since $x$ and$ y$ are independent( distinct eigenvalues)
.
Problem
How to arrive at result after that ?
linear-algebra linear-transformations
asked Sep 1 at 6:16
blue boy
1,085513
Problem
Assume that a linear transformation $T$ has two eigenvectors $x$ and $y$ belonging to two distinct eigenvalue
$lambda$ and $mu$ . If $ax+by$ is an eigenvector of$T$,prove that $a=0$ or $b=0$.
Attempt
$T(x)=lambda x$ and $T(y) = mu y$.
$T(ax+by) = omega(ax+by)$,where $omega$ is an eigenvalue for $ax+by$.
$a(lambda -omega)=0$ and $b(mu-omega) =0$ since $x$ and$ y$ are independent( distinct eigenvalues)
.
Problem
How to arrive at result after that ?
linear-algebra linear-transformations
asked Sep 1 at 6:16
blue boy
1,085513
asked Sep 1 at 6:16
blue boy
1,085513
asked Sep 1 at 6:16
blue boy
1,085513
asked Sep 1 at 6:16
blue boy
1,085513
1,085513
show that if $a$ is not zero then $b=0$
– Lozenges
Sep 1 at 6:36
add a comment |Â
show that if $a$ is not zero then $b=0$
– Lozenges
Sep 1 at 6:36
show that if $a$ is not zero then $b=0$
– Lozenges
Sep 1 at 6:36
show that if $a$ is not zero then $b=0$
– Lozenges
Sep 1 at 6:36
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
$T(ax+by) = omega(ax+by)$, where $omega$ is an eigenvalue for $ax+by$.
As noted, by linearity $,omega(ax+by) = T(ax+by) = a T(x) + bT(y) = alambda x + b mu y,$, and by linear independence $,a lambda = a omega,$ and $,bmu = b omega,$. Multiplying the first equality by $b$ and the second one by $a$ then subtracting gives $requirecancel,balambda-abmu=baomega-abomega=0 iff ab(lambda-mu)=0,$, and since $,lambda ne mu;ldots$
answered Sep 1 at 6:42
dxiv
55.9k64798
add a comment |Â
up vote
2
down vote
Your approach is correct. Now that $lambda ne mu$, there can be no $omega$ that simultaneously satisfies both the equations (assuming a and b are not equal to zero). So the possible solutions are,
a or b = 0 and $omega$ taking the appropriate value to satisfy the other equation.
a, b both are 0, which means the vector becomes a 0 vector and its a trivial case.
So if ax+by is an eigenvector of T, then either a=0 or b=0.
answered Sep 1 at 6:37
Lokesh Kumar
1363
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$T(ax+by) = omega(ax+by)$, where $omega$ is an eigenvalue for $ax+by$.
As noted, by linearity $,omega(ax+by) = T(ax+by) = a T(x) + bT(y) = alambda x + b mu y,$, and by linear independence $,a lambda = a omega,$ and $,bmu = b omega,$. Multiplying the first equality by $b$ and the second one by $a$ then subtracting gives $requirecancel,balambda-abmu=baomega-abomega=0 iff ab(lambda-mu)=0,$, and since $,lambda ne mu;ldots$
answered Sep 1 at 6:42
dxiv
55.9k64798
add a comment |Â
up vote
3
down vote
accepted
$T(ax+by) = omega(ax+by)$, where $omega$ is an eigenvalue for $ax+by$.
As noted, by linearity $,omega(ax+by) = T(ax+by) = a T(x) + bT(y) = alambda x + b mu y,$, and by linear independence $,a lambda = a omega,$ and $,bmu = b omega,$. Multiplying the first equality by $b$ and the second one by $a$ then subtracting gives $requirecancel,balambda-abmu=baomega-abomega=0 iff ab(lambda-mu)=0,$, and since $,lambda ne mu;ldots$
answered Sep 1 at 6:42
dxiv
55.9k64798
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$T(ax+by) = omega(ax+by)$, where $omega$ is an eigenvalue for $ax+by$.
As noted, by linearity $,omega(ax+by) = T(ax+by) = a T(x) + bT(y) = alambda x + b mu y,$, and by linear independence $,a lambda = a omega,$ and $,bmu = b omega,$. Multiplying the first equality by $b$ and the second one by $a$ then subtracting gives $requirecancel,balambda-abmu=baomega-abomega=0 iff ab(lambda-mu)=0,$, and since $,lambda ne mu;ldots$
answered Sep 1 at 6:42
dxiv
55.9k64798
$T(ax+by) = omega(ax+by)$, where $omega$ is an eigenvalue for $ax+by$.
As noted, by linearity $,omega(ax+by) = T(ax+by) = a T(x) + bT(y) = alambda x + b mu y,$, and by linear independence $,a lambda = a omega,$ and $,bmu = b omega,$. Multiplying the first equality by $b$ and the second one by $a$ then subtracting gives $requirecancel,balambda-abmu=baomega-abomega=0 iff ab(lambda-mu)=0,$, and since $,lambda ne mu;ldots$
answered Sep 1 at 6:42
dxiv
55.9k64798
answered Sep 1 at 6:42
dxiv
55.9k64798
answered Sep 1 at 6:42
dxiv
55.9k64798
answered Sep 1 at 6:42
dxiv
55.9k64798
55.9k64798
add a comment |Â
add a comment |Â
up vote
2
down vote
Your approach is correct. Now that $lambda ne mu$, there can be no $omega$ that simultaneously satisfies both the equations (assuming a and b are not equal to zero). So the possible solutions are,
a or b = 0 and $omega$ taking the appropriate value to satisfy the other equation.
a, b both are 0, which means the vector becomes a 0 vector and its a trivial case.
So if ax+by is an eigenvector of T, then either a=0 or b=0.
answered Sep 1 at 6:37
Lokesh Kumar
1363
add a comment |Â
up vote
2
down vote
Your approach is correct. Now that $lambda ne mu$, there can be no $omega$ that simultaneously satisfies both the equations (assuming a and b are not equal to zero). So the possible solutions are,
a or b = 0 and $omega$ taking the appropriate value to satisfy the other equation.
a, b both are 0, which means the vector becomes a 0 vector and its a trivial case.
So if ax+by is an eigenvector of T, then either a=0 or b=0.
answered Sep 1 at 6:37
Lokesh Kumar
1363
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your approach is correct. Now that $lambda ne mu$, there can be no $omega$ that simultaneously satisfies both the equations (assuming a and b are not equal to zero). So the possible solutions are,
a or b = 0 and $omega$ taking the appropriate value to satisfy the other equation.
a, b both are 0, which means the vector becomes a 0 vector and its a trivial case.
So if ax+by is an eigenvector of T, then either a=0 or b=0.
answered Sep 1 at 6:37
Lokesh Kumar
1363
Your approach is correct. Now that $lambda ne mu$, there can be no $omega$ that simultaneously satisfies both the equations (assuming a and b are not equal to zero). So the possible solutions are,
a or b = 0 and $omega$ taking the appropriate value to satisfy the other equation.
a, b both are 0, which means the vector becomes a 0 vector and its a trivial case.
So if ax+by is an eigenvector of T, then either a=0 or b=0.
answered Sep 1 at 6:37
Lokesh Kumar
1363
answered Sep 1 at 6:37
Lokesh Kumar
1363
answered Sep 1 at 6:37
Lokesh Kumar
1363
answered Sep 1 at 6:37
Lokesh Kumar
1363
1363
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901422%2fproblem-related-to-linear-transformation%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
show that if $a$ is not zero then $b=0$
– Lozenges
Sep 1 at 6:36