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eigenvalues Linear Transformation proof

0

1

$begingroup$

I am given the linear transformation $T:Vto V$ that has $u$ as an eigenvector with $x$ as its eigenvalue, and $v$ as another eigenvector of $T$ with $g$ as its eigenvalue.
I need to prove that if $u+v$ is eigenvector of $T$ then $x=g$.

eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 27 at 18:13

J. W. Tanner

5,0551520

asked Mar 27 at 17:47

ga asga as

84

$endgroup$

add a comment | 

0

1

$begingroup$

I am given the linear transformation $T:Vto V$ that has $u$ as an eigenvector with $x$ as its eigenvalue, and $v$ as another eigenvector of $T$ with $g$ as its eigenvalue.
I need to prove that if $u+v$ is eigenvector of $T$ then $x=g$.

eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 27 at 18:13

J. W. Tanner

5,0551520

asked Mar 27 at 17:47

ga asga as

84

$endgroup$

add a comment | 

$begingroup$

I am given the linear transformation $T:Vto V$ that has $u$ as an eigenvector with $x$ as its eigenvalue, and $v$ as another eigenvector of $T$ with $g$ as its eigenvalue.
I need to prove that if $u+v$ is eigenvector of $T$ then $x=g$.

eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 27 at 18:13

J. W. Tanner

5,0551520

asked Mar 27 at 17:47

ga asga as

84

$endgroup$

share|cite|improve this question

edited Mar 27 at 18:13

J. W. Tanner

5,0551520

asked Mar 27 at 17:47

ga asga as

84

share|cite|improve this question

edited Mar 27 at 18:13

J. W. Tanner

5,0551520

asked Mar 27 at 17:47

ga asga as

84

edited Mar 27 at 18:13

J. W. Tanner

5,0551520

edited Mar 27 at 18:13

J. W. Tanner

5,0551520

asked Mar 27 at 17:47

ga asga as

84

asked Mar 27 at 17:47

ga asga as

84

1 Answer
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active

oldest

votes

1

$begingroup$

Hint You have $T(u)=xu$ and $T(v)=gv$. So if $u+v$ is an eigenvector of $T$ associated to say $lambda$ i.e. $T(u+v)=lambda(u+v)$ then

$$T(u+v)=lambda(u+v)=T(u)+T(v)=xu+gv$$
so
$$(x-lambda)u=(lambda-g)v$$
Now you should discus two possible cases:

• first case: $u$ and $v$ are colinear;


• second case: $u$ and $v$ are linearly independent.

share|cite|improve this answer

answered Mar 27 at 17:58

user296113user296113

7,015928

$endgroup$

add a comment | 

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1

$begingroup$

Hint You have $T(u)=xu$ and $T(v)=gv$. So if $u+v$ is an eigenvector of $T$ associated to say $lambda$ i.e. $T(u+v)=lambda(u+v)$ then

$$T(u+v)=lambda(u+v)=T(u)+T(v)=xu+gv$$
so
$$(x-lambda)u=(lambda-g)v$$
Now you should discus two possible cases:

• first case: $u$ and $v$ are colinear;


• second case: $u$ and $v$ are linearly independent.

share|cite|improve this answer

answered Mar 27 at 17:58

user296113user296113

7,015928

$endgroup$

add a comment | 

1

$begingroup$

Hint You have $T(u)=xu$ and $T(v)=gv$. So if $u+v$ is an eigenvector of $T$ associated to say $lambda$ i.e. $T(u+v)=lambda(u+v)$ then

$$T(u+v)=lambda(u+v)=T(u)+T(v)=xu+gv$$
so
$$(x-lambda)u=(lambda-g)v$$
Now you should discus two possible cases:

• first case: $u$ and $v$ are colinear;


• second case: $u$ and $v$ are linearly independent.

share|cite|improve this answer

answered Mar 27 at 17:58

user296113user296113

7,015928

$endgroup$

add a comment | 

$begingroup$

Hint You have $T(u)=xu$ and $T(v)=gv$. So if $u+v$ is an eigenvector of $T$ associated to say $lambda$ i.e. $T(u+v)=lambda(u+v)$ then

$$T(u+v)=lambda(u+v)=T(u)+T(v)=xu+gv$$
so
$$(x-lambda)u=(lambda-g)v$$
Now you should discus two possible cases:

• first case: $u$ and $v$ are colinear;


• second case: $u$ and $v$ are linearly independent.

share|cite|improve this answer

answered Mar 27 at 17:58

user296113user296113

7,015928

$endgroup$

share|cite|improve this answer

answered Mar 27 at 17:58

user296113user296113

7,015928

answered Mar 27 at 17:58

user296113user296113

7,015928

answered Mar 27 at 17:58

user296113user296113

7,015928