Given $T^2 = T$ find all eigenvalues of $T$…
$begingroup$
Given a linear operator $T$ on a finite-dimensional vector space $V$, satisfying $T^2 = T$ answer the following:
(a) Using the dimension theorem, show that $N(T) bigoplus R(T) = V$;
(b) Identify all eigenvalues of $T$ and the corresponding eigenspaces and show that $T$ is diagonalizable.
I have already proven part (a). Basically define $Tv = w$ and the rest follows.
Part (b) however I am honestly confused on where to start.
linear-algebra eigenvalues-eigenvectors
asked Jan 9 at 0:21
Taylor McMillanTaylor McMillan
695
$endgroup$
add a comment |
$begingroup$
Given a linear operator $T$ on a finite-dimensional vector space $V$, satisfying $T^2 = T$ answer the following:
(a) Using the dimension theorem, show that $N(T) bigoplus R(T) = V$;
(b) Identify all eigenvalues of $T$ and the corresponding eigenspaces and show that $T$ is diagonalizable.
I have already proven part (a). Basically define $Tv = w$ and the rest follows.
Part (b) however I am honestly confused on where to start.
linear-algebra eigenvalues-eigenvectors
asked Jan 9 at 0:21
Taylor McMillanTaylor McMillan
695
$endgroup$
6
$begingroup$
Hint: how are the eigenvalues of $T$ and $T^2$ related?
$endgroup$
– user3482749
Jan 9 at 0:22
1
$begingroup$
I know that they must be the same. Since our vector space is finite-dimensional, then we are guaranteed the existence of an eigenvalue. That is, there exists $lambda in mathbb{F}$ such that $Tv = lambda v$. Since $T^2 = T$, we then have that $T^2v = lambda v$. It then follows that $T(Tv) = lambda Tv = lambda^2 v = lambda v$. Therefore, any eigenvalue of $T$ must satisfy $lambda^2 = lambda$, so our eigenvalues are 0 and 1. Is this correct? But then I am confused again...wouldn't this mean that we do not have a basis of eigenvectors, therefore, it is not diagonalizable?
$endgroup$
– Taylor McMillan
Jan 9 at 0:34
1
$begingroup$
Hint: what are the possible minimal polynomials given that $T^2-T=0$?
$endgroup$
– amd
Jan 9 at 2:10
add a comment |
$begingroup$
Given a linear operator $T$ on a finite-dimensional vector space $V$, satisfying $T^2 = T$ answer the following:
(a) Using the dimension theorem, show that $N(T) bigoplus R(T) = V$;
(b) Identify all eigenvalues of $T$ and the corresponding eigenspaces and show that $T$ is diagonalizable.
I have already proven part (a). Basically define $Tv = w$ and the rest follows.
Part (b) however I am honestly confused on where to start.
linear-algebra eigenvalues-eigenvectors
asked Jan 9 at 0:21
Taylor McMillanTaylor McMillan
695
$endgroup$
Given a linear operator $T$ on a finite-dimensional vector space $V$, satisfying $T^2 = T$ answer the following:
(a) Using the dimension theorem, show that $N(T) bigoplus R(T) = V$;
(b) Identify all eigenvalues of $T$ and the corresponding eigenspaces and show that $T$ is diagonalizable.
I have already proven part (a). Basically define $Tv = w$ and the rest follows.
Part (b) however I am honestly confused on where to start.
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Jan 9 at 0:21
Taylor McMillanTaylor McMillan
695
asked Jan 9 at 0:21
Taylor McMillanTaylor McMillan
695
asked Jan 9 at 0:21
Taylor McMillanTaylor McMillan
695
asked Jan 9 at 0:21
Taylor McMillanTaylor McMillan
695
asked Jan 9 at 0:21
Taylor McMillanTaylor McMillan
695
695
6
$begingroup$
Hint: how are the eigenvalues of $T$ and $T^2$ related?
$endgroup$
– user3482749
Jan 9 at 0:22
1
$begingroup$
I know that they must be the same. Since our vector space is finite-dimensional, then we are guaranteed the existence of an eigenvalue. That is, there exists $lambda in mathbb{F}$ such that $Tv = lambda v$. Since $T^2 = T$, we then have that $T^2v = lambda v$. It then follows that $T(Tv) = lambda Tv = lambda^2 v = lambda v$. Therefore, any eigenvalue of $T$ must satisfy $lambda^2 = lambda$, so our eigenvalues are 0 and 1. Is this correct? But then I am confused again...wouldn't this mean that we do not have a basis of eigenvectors, therefore, it is not diagonalizable?
$endgroup$
– Taylor McMillan
Jan 9 at 0:34
1
$begingroup$
Hint: what are the possible minimal polynomials given that $T^2-T=0$?
$endgroup$
– amd
Jan 9 at 2:10
add a comment |
6
$begingroup$
Hint: how are the eigenvalues of $T$ and $T^2$ related?
$endgroup$
– user3482749
Jan 9 at 0:22
1
$begingroup$
I know that they must be the same. Since our vector space is finite-dimensional, then we are guaranteed the existence of an eigenvalue. That is, there exists $lambda in mathbb{F}$ such that $Tv = lambda v$. Since $T^2 = T$, we then have that $T^2v = lambda v$. It then follows that $T(Tv) = lambda Tv = lambda^2 v = lambda v$. Therefore, any eigenvalue of $T$ must satisfy $lambda^2 = lambda$, so our eigenvalues are 0 and 1. Is this correct? But then I am confused again...wouldn't this mean that we do not have a basis of eigenvectors, therefore, it is not diagonalizable?
$endgroup$
– Taylor McMillan
Jan 9 at 0:34
1
$begingroup$
Hint: what are the possible minimal polynomials given that $T^2-T=0$?
$endgroup$
– amd
Jan 9 at 2:10
6
6
$begingroup$
Hint: how are the eigenvalues of $T$ and $T^2$ related?
$endgroup$
– user3482749
Jan 9 at 0:22
$begingroup$
Hint: how are the eigenvalues of $T$ and $T^2$ related?
$endgroup$
– user3482749
Jan 9 at 0:22
1
1
$begingroup$
I know that they must be the same. Since our vector space is finite-dimensional, then we are guaranteed the existence of an eigenvalue. That is, there exists $lambda in mathbb{F}$ such that $Tv = lambda v$. Since $T^2 = T$, we then have that $T^2v = lambda v$. It then follows that $T(Tv) = lambda Tv = lambda^2 v = lambda v$. Therefore, any eigenvalue of $T$ must satisfy $lambda^2 = lambda$, so our eigenvalues are 0 and 1. Is this correct? But then I am confused again...wouldn't this mean that we do not have a basis of eigenvectors, therefore, it is not diagonalizable?
$endgroup$
– Taylor McMillan
Jan 9 at 0:34
$begingroup$
I know that they must be the same. Since our vector space is finite-dimensional, then we are guaranteed the existence of an eigenvalue. That is, there exists $lambda in mathbb{F}$ such that $Tv = lambda v$. Since $T^2 = T$, we then have that $T^2v = lambda v$. It then follows that $T(Tv) = lambda Tv = lambda^2 v = lambda v$. Therefore, any eigenvalue of $T$ must satisfy $lambda^2 = lambda$, so our eigenvalues are 0 and 1. Is this correct? But then I am confused again...wouldn't this mean that we do not have a basis of eigenvectors, therefore, it is not diagonalizable?
$endgroup$
– Taylor McMillan
Jan 9 at 0:34
1
1
$begingroup$
Hint: what are the possible minimal polynomials given that $T^2-T=0$?
$endgroup$
– amd
Jan 9 at 2:10
$begingroup$
Hint: what are the possible minimal polynomials given that $T^2-T=0$?
$endgroup$
– amd
Jan 9 at 2:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By definition:
$$Tv = lambda v$$
Multiply both sides by $T$:
$$T^2 v = lambda T v.$$
Recall that $T^2 = T$:
$$Tv = lambda Tv.$$
Since $Tv = lambda v$, then:
$$lambda v = lambda cdot lambda v = lambda^2 v.$$
That is:
$$lambda = lambda^2.$$
This equation has only two solutions: $lambda = 0$ and $lambda = 1$.
For $lambda = 0$, the eigenspace corresponds to the null space of $T$.
For $lambda = 1$, the eigenspace is given by all vectors $v$ such that:
$$(T-I)v = 0,$$
where $I$ is the identity matrix.
answered Jan 9 at 0:37
the_candymanthe_candyman
8,80122045
$endgroup$
$begingroup$
Funny, I just commented this on the response above. Oh, I get it now. These are the ONLY eigenvalues, therefore it is diagonalizable!
$endgroup$
– Taylor McMillan
Jan 9 at 0:38
$begingroup$
The eigenspace for the unit eigenvalues is actually the range of $T$, since if $boldsymbol vin V$, then $T (Tboldsymbol v) =1cdot (Tboldsymbol v)$, so $Tboldsymbol v$ is an eigenvector.
$endgroup$
– nathan.j.mcdougall
Jan 9 at 0:40
1
$begingroup$
To be precise, $0$ and $1$ are the potential eigenvalues. The operator $T$ can have $0$ as the unique eigenvalue, in which case $T-1$ is invertible and from the equation you get $T=0$. It can also have $1$ as the unique eigenvalue in which case $T$ is invertible and $T=1$.
$endgroup$
– mouthetics
Jan 9 at 0:43
1
$begingroup$
@mouthetics I did not say that both $0$ and $1$ are eigenvalues of $T$. I just said that they are solutions of $lambda = lambda^2$.
$endgroup$
– the_candyman
Jan 9 at 0:44
add a comment |
$begingroup$
Hints:
- If $v$ is an eigenvector, then $lambda v=Tv=T^2v=lambda^2v$, so $lambda=lambda^2$.
$R(T)$ coincides with the eigenspace of $lambda=1$, then use part (a).
answered Jan 9 at 0:38
BerciBerci
60k23672
$endgroup$
add a comment |
Your Answer
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
By definition:
$$Tv = lambda v$$
Multiply both sides by $T$:
$$T^2 v = lambda T v.$$
Recall that $T^2 = T$:
$$Tv = lambda Tv.$$
Since $Tv = lambda v$, then:
$$lambda v = lambda cdot lambda v = lambda^2 v.$$
That is:
$$lambda = lambda^2.$$
This equation has only two solutions: $lambda = 0$ and $lambda = 1$.
For $lambda = 0$, the eigenspace corresponds to the null space of $T$.
For $lambda = 1$, the eigenspace is given by all vectors $v$ such that:
$$(T-I)v = 0,$$
where $I$ is the identity matrix.
answered Jan 9 at 0:37
the_candymanthe_candyman
8,80122045
$endgroup$
$begingroup$
Funny, I just commented this on the response above. Oh, I get it now. These are the ONLY eigenvalues, therefore it is diagonalizable!
$endgroup$
– Taylor McMillan
Jan 9 at 0:38
$begingroup$
The eigenspace for the unit eigenvalues is actually the range of $T$, since if $boldsymbol vin V$, then $T (Tboldsymbol v) =1cdot (Tboldsymbol v)$, so $Tboldsymbol v$ is an eigenvector.
$endgroup$
– nathan.j.mcdougall
Jan 9 at 0:40
1
$begingroup$
To be precise, $0$ and $1$ are the potential eigenvalues. The operator $T$ can have $0$ as the unique eigenvalue, in which case $T-1$ is invertible and from the equation you get $T=0$. It can also have $1$ as the unique eigenvalue in which case $T$ is invertible and $T=1$.
$endgroup$
– mouthetics
Jan 9 at 0:43
1
$begingroup$
@mouthetics I did not say that both $0$ and $1$ are eigenvalues of $T$. I just said that they are solutions of $lambda = lambda^2$.
$endgroup$
– the_candyman
Jan 9 at 0:44
add a comment |
$begingroup$
By definition:
$$Tv = lambda v$$
Multiply both sides by $T$:
$$T^2 v = lambda T v.$$
Recall that $T^2 = T$:
$$Tv = lambda Tv.$$
Since $Tv = lambda v$, then:
$$lambda v = lambda cdot lambda v = lambda^2 v.$$
That is:
$$lambda = lambda^2.$$
This equation has only two solutions: $lambda = 0$ and $lambda = 1$.
For $lambda = 0$, the eigenspace corresponds to the null space of $T$.
For $lambda = 1$, the eigenspace is given by all vectors $v$ such that:
$$(T-I)v = 0,$$
where $I$ is the identity matrix.
answered Jan 9 at 0:37
the_candymanthe_candyman
8,80122045
$endgroup$
$begingroup$
Funny, I just commented this on the response above. Oh, I get it now. These are the ONLY eigenvalues, therefore it is diagonalizable!
$endgroup$
– Taylor McMillan
Jan 9 at 0:38
$begingroup$
The eigenspace for the unit eigenvalues is actually the range of $T$, since if $boldsymbol vin V$, then $T (Tboldsymbol v) =1cdot (Tboldsymbol v)$, so $Tboldsymbol v$ is an eigenvector.
$endgroup$
– nathan.j.mcdougall
Jan 9 at 0:40
1
$begingroup$
To be precise, $0$ and $1$ are the potential eigenvalues. The operator $T$ can have $0$ as the unique eigenvalue, in which case $T-1$ is invertible and from the equation you get $T=0$. It can also have $1$ as the unique eigenvalue in which case $T$ is invertible and $T=1$.
$endgroup$
– mouthetics
Jan 9 at 0:43
1
$begingroup$
@mouthetics I did not say that both $0$ and $1$ are eigenvalues of $T$. I just said that they are solutions of $lambda = lambda^2$.
$endgroup$
– the_candyman
Jan 9 at 0:44
add a comment |
$begingroup$
By definition:
$$Tv = lambda v$$
Multiply both sides by $T$:
$$T^2 v = lambda T v.$$
Recall that $T^2 = T$:
$$Tv = lambda Tv.$$
Since $Tv = lambda v$, then:
$$lambda v = lambda cdot lambda v = lambda^2 v.$$
That is:
$$lambda = lambda^2.$$
This equation has only two solutions: $lambda = 0$ and $lambda = 1$.
For $lambda = 0$, the eigenspace corresponds to the null space of $T$.
For $lambda = 1$, the eigenspace is given by all vectors $v$ such that:
$$(T-I)v = 0,$$
where $I$ is the identity matrix.
answered Jan 9 at 0:37
the_candymanthe_candyman
8,80122045
$endgroup$
By definition:
$$Tv = lambda v$$
Multiply both sides by $T$:
$$T^2 v = lambda T v.$$
Recall that $T^2 = T$:
$$Tv = lambda Tv.$$
Since $Tv = lambda v$, then:
$$lambda v = lambda cdot lambda v = lambda^2 v.$$
That is:
$$lambda = lambda^2.$$
This equation has only two solutions: $lambda = 0$ and $lambda = 1$.
For $lambda = 0$, the eigenspace corresponds to the null space of $T$.
For $lambda = 1$, the eigenspace is given by all vectors $v$ such that:
$$(T-I)v = 0,$$
where $I$ is the identity matrix.
answered Jan 9 at 0:37
the_candymanthe_candyman
8,80122045
answered Jan 9 at 0:37
the_candymanthe_candyman
8,80122045
answered Jan 9 at 0:37
the_candymanthe_candyman
8,80122045
answered Jan 9 at 0:37
the_candymanthe_candyman
8,80122045
8,80122045
$begingroup$
Funny, I just commented this on the response above. Oh, I get it now. These are the ONLY eigenvalues, therefore it is diagonalizable!
$endgroup$
– Taylor McMillan
Jan 9 at 0:38
$begingroup$
The eigenspace for the unit eigenvalues is actually the range of $T$, since if $boldsymbol vin V$, then $T (Tboldsymbol v) =1cdot (Tboldsymbol v)$, so $Tboldsymbol v$ is an eigenvector.
$endgroup$
– nathan.j.mcdougall
Jan 9 at 0:40
1
$begingroup$
To be precise, $0$ and $1$ are the potential eigenvalues. The operator $T$ can have $0$ as the unique eigenvalue, in which case $T-1$ is invertible and from the equation you get $T=0$. It can also have $1$ as the unique eigenvalue in which case $T$ is invertible and $T=1$.
$endgroup$
– mouthetics
Jan 9 at 0:43
1
$begingroup$
@mouthetics I did not say that both $0$ and $1$ are eigenvalues of $T$. I just said that they are solutions of $lambda = lambda^2$.
$endgroup$
– the_candyman
Jan 9 at 0:44
add a comment |
$begingroup$
Funny, I just commented this on the response above. Oh, I get it now. These are the ONLY eigenvalues, therefore it is diagonalizable!
$endgroup$
– Taylor McMillan
Jan 9 at 0:38
$begingroup$
The eigenspace for the unit eigenvalues is actually the range of $T$, since if $boldsymbol vin V$, then $T (Tboldsymbol v) =1cdot (Tboldsymbol v)$, so $Tboldsymbol v$ is an eigenvector.
$endgroup$
– nathan.j.mcdougall
Jan 9 at 0:40
1
$begingroup$
To be precise, $0$ and $1$ are the potential eigenvalues. The operator $T$ can have $0$ as the unique eigenvalue, in which case $T-1$ is invertible and from the equation you get $T=0$. It can also have $1$ as the unique eigenvalue in which case $T$ is invertible and $T=1$.
$endgroup$
– mouthetics
Jan 9 at 0:43
1
$begingroup$
@mouthetics I did not say that both $0$ and $1$ are eigenvalues of $T$. I just said that they are solutions of $lambda = lambda^2$.
$endgroup$
– the_candyman
Jan 9 at 0:44
$begingroup$
Funny, I just commented this on the response above. Oh, I get it now. These are the ONLY eigenvalues, therefore it is diagonalizable!
$endgroup$
– Taylor McMillan
Jan 9 at 0:38
$begingroup$
Funny, I just commented this on the response above. Oh, I get it now. These are the ONLY eigenvalues, therefore it is diagonalizable!
$endgroup$
– Taylor McMillan
Jan 9 at 0:38
$begingroup$
The eigenspace for the unit eigenvalues is actually the range of $T$, since if $boldsymbol vin V$, then $T (Tboldsymbol v) =1cdot (Tboldsymbol v)$, so $Tboldsymbol v$ is an eigenvector.
$endgroup$
– nathan.j.mcdougall
Jan 9 at 0:40
$begingroup$
The eigenspace for the unit eigenvalues is actually the range of $T$, since if $boldsymbol vin V$, then $T (Tboldsymbol v) =1cdot (Tboldsymbol v)$, so $Tboldsymbol v$ is an eigenvector.
$endgroup$
– nathan.j.mcdougall
Jan 9 at 0:40
1
1
$begingroup$
To be precise, $0$ and $1$ are the potential eigenvalues. The operator $T$ can have $0$ as the unique eigenvalue, in which case $T-1$ is invertible and from the equation you get $T=0$. It can also have $1$ as the unique eigenvalue in which case $T$ is invertible and $T=1$.
$endgroup$
– mouthetics
Jan 9 at 0:43
$begingroup$
To be precise, $0$ and $1$ are the potential eigenvalues. The operator $T$ can have $0$ as the unique eigenvalue, in which case $T-1$ is invertible and from the equation you get $T=0$. It can also have $1$ as the unique eigenvalue in which case $T$ is invertible and $T=1$.
$endgroup$
– mouthetics
Jan 9 at 0:43
1
1
$begingroup$
@mouthetics I did not say that both $0$ and $1$ are eigenvalues of $T$. I just said that they are solutions of $lambda = lambda^2$.
$endgroup$
– the_candyman
Jan 9 at 0:44
$begingroup$
@mouthetics I did not say that both $0$ and $1$ are eigenvalues of $T$. I just said that they are solutions of $lambda = lambda^2$.
$endgroup$
– the_candyman
Jan 9 at 0:44
add a comment |
$begingroup$
Hints:
- If $v$ is an eigenvector, then $lambda v=Tv=T^2v=lambda^2v$, so $lambda=lambda^2$.
$R(T)$ coincides with the eigenspace of $lambda=1$, then use part (a).
answered Jan 9 at 0:38
BerciBerci
60k23672
$endgroup$
add a comment |
$begingroup$
Hints:
- If $v$ is an eigenvector, then $lambda v=Tv=T^2v=lambda^2v$, so $lambda=lambda^2$.
$R(T)$ coincides with the eigenspace of $lambda=1$, then use part (a).
answered Jan 9 at 0:38
BerciBerci
60k23672
$endgroup$
add a comment |
$begingroup$
Hints:
- If $v$ is an eigenvector, then $lambda v=Tv=T^2v=lambda^2v$, so $lambda=lambda^2$.
$R(T)$ coincides with the eigenspace of $lambda=1$, then use part (a).
answered Jan 9 at 0:38
BerciBerci
60k23672
$endgroup$
Hints:
- If $v$ is an eigenvector, then $lambda v=Tv=T^2v=lambda^2v$, so $lambda=lambda^2$.
$R(T)$ coincides with the eigenspace of $lambda=1$, then use part (a).
answered Jan 9 at 0:38
BerciBerci
60k23672
answered Jan 9 at 0:38
BerciBerci
60k23672
answered Jan 9 at 0:38
BerciBerci
60k23672
answered Jan 9 at 0:38
BerciBerci
60k23672
60k23672
add a comment |
add a comment |
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6
$begingroup$
Hint: how are the eigenvalues of $T$ and $T^2$ related?
$endgroup$
– user3482749
Jan 9 at 0:22
1
$begingroup$
I know that they must be the same. Since our vector space is finite-dimensional, then we are guaranteed the existence of an eigenvalue. That is, there exists $lambda in mathbb{F}$ such that $Tv = lambda v$. Since $T^2 = T$, we then have that $T^2v = lambda v$. It then follows that $T(Tv) = lambda Tv = lambda^2 v = lambda v$. Therefore, any eigenvalue of $T$ must satisfy $lambda^2 = lambda$, so our eigenvalues are 0 and 1. Is this correct? But then I am confused again...wouldn't this mean that we do not have a basis of eigenvectors, therefore, it is not diagonalizable?
$endgroup$
– Taylor McMillan
Jan 9 at 0:34
1
$begingroup$
Hint: what are the possible minimal polynomials given that $T^2-T=0$?
$endgroup$
– amd
Jan 9 at 2:10