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Proving that $0$ is an eigenvalue of this transformation
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I'm currently self-studying linear algebra and am interested in proving the following:
Let $T:mathbbR^n rightarrow mathbbR^n$, where $ operatornamerank(T)<n$, prove that $0$ is an eigenvalue of the transformation.
What I'd like to do is use the inveritable matrix theorem, where since $ operatornamerank(T)$ doesn't equal $n$, I can use every part of theorem negated: http://mathworld.wolfram.com/InvertibleMatrixTheorem.html
One such part is: "18. 0 fails to be an eigenvalue of A."
Negated: $0$ is an eigenvalue of A, in this case, T.
Thus proving $0$ is an eigenvalue of T.
What do you think of this proof? If I had the wrong idea, could you go over one way to prove it?
linear-algebra proof-verification proof-writing eigenvalues-eigenvectors
edited Jul 18 at 21:38
Bernard
110k635103
asked Jul 18 at 21:34
HistorySweets
141
 |Â
show 4 more comments
up vote
2
down vote
favorite
I'm currently self-studying linear algebra and am interested in proving the following:
Let $T:mathbbR^n rightarrow mathbbR^n$, where $ operatornamerank(T)<n$, prove that $0$ is an eigenvalue of the transformation.
What I'd like to do is use the inveritable matrix theorem, where since $ operatornamerank(T)$ doesn't equal $n$, I can use every part of theorem negated: http://mathworld.wolfram.com/InvertibleMatrixTheorem.html
One such part is: "18. 0 fails to be an eigenvalue of A."
Negated: $0$ is an eigenvalue of A, in this case, T.
Thus proving $0$ is an eigenvalue of T.
What do you think of this proof? If I had the wrong idea, could you go over one way to prove it?
linear-algebra proof-verification proof-writing eigenvalues-eigenvectors
edited Jul 18 at 21:38
Bernard
110k635103
asked Jul 18 at 21:34
HistorySweets
141
1
This is not really a proof, but rather just an application of the theorem of invertible matrices. Are you actually trying to prove this implication without use of the invertible matrix theorem (or at least this part of the theorem)?
– Dave
Jul 18 at 21:38
1
The most recent chapters discussed applications of the theorem so I assumed it was permitted. In the case where it isnt really a proof, is it at least a correct application of the theorem?
– HistorySweets
Jul 18 at 21:39
@HistorySweets An application of a theorem is definitely a proof.
– user577471
Jul 18 at 21:40
So, is the proof okay? Sorry, proofs are still pretty new to me. At this point, i'm just wrapping my head around theorems and such.
– HistorySweets
Jul 18 at 21:41
This application is certainly correct: if a matrix is not invertible it has zero as an eigenvalue (in this case the matrix associated to $T$). But if the question asks for a proof of this implication, then I don't think what you have really does this; a proof more relies on use of the definitions, and one cannot assume this result from the invertible matrix theorem is true a priori.
– Dave
Jul 18 at 21:42
 |Â
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm currently self-studying linear algebra and am interested in proving the following:
Let $T:mathbbR^n rightarrow mathbbR^n$, where $ operatornamerank(T)<n$, prove that $0$ is an eigenvalue of the transformation.
What I'd like to do is use the inveritable matrix theorem, where since $ operatornamerank(T)$ doesn't equal $n$, I can use every part of theorem negated: http://mathworld.wolfram.com/InvertibleMatrixTheorem.html
One such part is: "18. 0 fails to be an eigenvalue of A."
Negated: $0$ is an eigenvalue of A, in this case, T.
Thus proving $0$ is an eigenvalue of T.
What do you think of this proof? If I had the wrong idea, could you go over one way to prove it?
linear-algebra proof-verification proof-writing eigenvalues-eigenvectors
edited Jul 18 at 21:38
Bernard
110k635103
asked Jul 18 at 21:34
HistorySweets
141
I'm currently self-studying linear algebra and am interested in proving the following:
Let $T:mathbbR^n rightarrow mathbbR^n$, where $ operatornamerank(T)<n$, prove that $0$ is an eigenvalue of the transformation.
What I'd like to do is use the inveritable matrix theorem, where since $ operatornamerank(T)$ doesn't equal $n$, I can use every part of theorem negated: http://mathworld.wolfram.com/InvertibleMatrixTheorem.html
One such part is: "18. 0 fails to be an eigenvalue of A."
Negated: $0$ is an eigenvalue of A, in this case, T.
Thus proving $0$ is an eigenvalue of T.
What do you think of this proof? If I had the wrong idea, could you go over one way to prove it?
linear-algebra proof-verification proof-writing eigenvalues-eigenvectors
edited Jul 18 at 21:38
Bernard
110k635103
asked Jul 18 at 21:34
HistorySweets
141
edited Jul 18 at 21:38
Bernard
110k635103
edited Jul 18 at 21:38
Bernard
110k635103
edited Jul 18 at 21:38
Bernard
110k635103
110k635103
asked Jul 18 at 21:34
HistorySweets
141
asked Jul 18 at 21:34
HistorySweets
141
asked Jul 18 at 21:34
HistorySweets
141
141
1
This is not really a proof, but rather just an application of the theorem of invertible matrices. Are you actually trying to prove this implication without use of the invertible matrix theorem (or at least this part of the theorem)?
– Dave
Jul 18 at 21:38
1
The most recent chapters discussed applications of the theorem so I assumed it was permitted. In the case where it isnt really a proof, is it at least a correct application of the theorem?
– HistorySweets
Jul 18 at 21:39
@HistorySweets An application of a theorem is definitely a proof.
– user577471
Jul 18 at 21:40
So, is the proof okay? Sorry, proofs are still pretty new to me. At this point, i'm just wrapping my head around theorems and such.
– HistorySweets
Jul 18 at 21:41
This application is certainly correct: if a matrix is not invertible it has zero as an eigenvalue (in this case the matrix associated to $T$). But if the question asks for a proof of this implication, then I don't think what you have really does this; a proof more relies on use of the definitions, and one cannot assume this result from the invertible matrix theorem is true a priori.
– Dave
Jul 18 at 21:42
 |Â
show 4 more comments
1
This is not really a proof, but rather just an application of the theorem of invertible matrices. Are you actually trying to prove this implication without use of the invertible matrix theorem (or at least this part of the theorem)?
– Dave
Jul 18 at 21:38
1
The most recent chapters discussed applications of the theorem so I assumed it was permitted. In the case where it isnt really a proof, is it at least a correct application of the theorem?
– HistorySweets
Jul 18 at 21:39
@HistorySweets An application of a theorem is definitely a proof.
– user577471
Jul 18 at 21:40
So, is the proof okay? Sorry, proofs are still pretty new to me. At this point, i'm just wrapping my head around theorems and such.
– HistorySweets
Jul 18 at 21:41
This application is certainly correct: if a matrix is not invertible it has zero as an eigenvalue (in this case the matrix associated to $T$). But if the question asks for a proof of this implication, then I don't think what you have really does this; a proof more relies on use of the definitions, and one cannot assume this result from the invertible matrix theorem is true a priori.
– Dave
Jul 18 at 21:42
1
1
This is not really a proof, but rather just an application of the theorem of invertible matrices. Are you actually trying to prove this implication without use of the invertible matrix theorem (or at least this part of the theorem)?
– Dave
Jul 18 at 21:38
This is not really a proof, but rather just an application of the theorem of invertible matrices. Are you actually trying to prove this implication without use of the invertible matrix theorem (or at least this part of the theorem)?
– Dave
Jul 18 at 21:38
1
1
The most recent chapters discussed applications of the theorem so I assumed it was permitted. In the case where it isnt really a proof, is it at least a correct application of the theorem?
– HistorySweets
Jul 18 at 21:39
The most recent chapters discussed applications of the theorem so I assumed it was permitted. In the case where it isnt really a proof, is it at least a correct application of the theorem?
– HistorySweets
Jul 18 at 21:39
@HistorySweets An application of a theorem is definitely a proof.
– user577471
Jul 18 at 21:40
@HistorySweets An application of a theorem is definitely a proof.
– user577471
Jul 18 at 21:40
So, is the proof okay? Sorry, proofs are still pretty new to me. At this point, i'm just wrapping my head around theorems and such.
– HistorySweets
Jul 18 at 21:41
So, is the proof okay? Sorry, proofs are still pretty new to me. At this point, i'm just wrapping my head around theorems and such.
– HistorySweets
Jul 18 at 21:41
This application is certainly correct: if a matrix is not invertible it has zero as an eigenvalue (in this case the matrix associated to $T$). But if the question asks for a proof of this implication, then I don't think what you have really does this; a proof more relies on use of the definitions, and one cannot assume this result from the invertible matrix theorem is true a priori.
– Dave
Jul 18 at 21:42
This application is certainly correct: if a matrix is not invertible it has zero as an eigenvalue (in this case the matrix associated to $T$). But if the question asks for a proof of this implication, then I don't think what you have really does this; a proof more relies on use of the definitions, and one cannot assume this result from the invertible matrix theorem is true a priori.
– Dave
Jul 18 at 21:42
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
An eigenvector for the value $0$ is just a non-trivial member of the null space or kernel of $T$: a vector $x neq 0$ such that $Tx=0$.
And the nullity rank theorem says that $dim(operatornameker(T)) + operatornamerank(T) = n (= dim(mathbbR^n))$, so if the rank is $< n$ we
know $dim(operatornameker(T)) > 0$ and so take any non-zero vector in it.
answered Jul 18 at 21:40
Henno Brandsma
91.6k342100
add a comment |Â
up vote
1
down vote
We have the Kernel of T is non empty, since the Rank of T is less than the dimension $n$.
That is $Tv=0$ for some $vne 0$
$Tv=0=0v$ implies that $0$ is an eigenvalue with eigenvector $v$
answered Jul 18 at 21:52
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
An eigenvector for the value $0$ is just a non-trivial member of the null space or kernel of $T$: a vector $x neq 0$ such that $Tx=0$.
And the nullity rank theorem says that $dim(operatornameker(T)) + operatornamerank(T) = n (= dim(mathbbR^n))$, so if the rank is $< n$ we
know $dim(operatornameker(T)) > 0$ and so take any non-zero vector in it.
answered Jul 18 at 21:40
Henno Brandsma
91.6k342100
add a comment |Â
up vote
1
down vote
An eigenvector for the value $0$ is just a non-trivial member of the null space or kernel of $T$: a vector $x neq 0$ such that $Tx=0$.
And the nullity rank theorem says that $dim(operatornameker(T)) + operatornamerank(T) = n (= dim(mathbbR^n))$, so if the rank is $< n$ we
know $dim(operatornameker(T)) > 0$ and so take any non-zero vector in it.
answered Jul 18 at 21:40
Henno Brandsma
91.6k342100
add a comment |Â
up vote
1
down vote
up vote
1
down vote
An eigenvector for the value $0$ is just a non-trivial member of the null space or kernel of $T$: a vector $x neq 0$ such that $Tx=0$.
And the nullity rank theorem says that $dim(operatornameker(T)) + operatornamerank(T) = n (= dim(mathbbR^n))$, so if the rank is $< n$ we
know $dim(operatornameker(T)) > 0$ and so take any non-zero vector in it.
answered Jul 18 at 21:40
Henno Brandsma
91.6k342100
An eigenvector for the value $0$ is just a non-trivial member of the null space or kernel of $T$: a vector $x neq 0$ such that $Tx=0$.
And the nullity rank theorem says that $dim(operatornameker(T)) + operatornamerank(T) = n (= dim(mathbbR^n))$, so if the rank is $< n$ we
know $dim(operatornameker(T)) > 0$ and so take any non-zero vector in it.
answered Jul 18 at 21:40
Henno Brandsma
91.6k342100
answered Jul 18 at 21:40
Henno Brandsma
91.6k342100
answered Jul 18 at 21:40
Henno Brandsma
91.6k342100
answered Jul 18 at 21:40
Henno Brandsma
91.6k342100
91.6k342100
add a comment |Â
add a comment |Â
up vote
1
down vote
We have the Kernel of T is non empty, since the Rank of T is less than the dimension $n$.
That is $Tv=0$ for some $vne 0$
$Tv=0=0v$ implies that $0$ is an eigenvalue with eigenvector $v$
answered Jul 18 at 21:52
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
1
down vote
We have the Kernel of T is non empty, since the Rank of T is less than the dimension $n$.
That is $Tv=0$ for some $vne 0$
$Tv=0=0v$ implies that $0$ is an eigenvalue with eigenvector $v$
answered Jul 18 at 21:52
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have the Kernel of T is non empty, since the Rank of T is less than the dimension $n$.
That is $Tv=0$ for some $vne 0$
$Tv=0=0v$ implies that $0$ is an eigenvalue with eigenvector $v$
answered Jul 18 at 21:52
Mohammad Riazi-Kermani
27.5k41852
We have the Kernel of T is non empty, since the Rank of T is less than the dimension $n$.
That is $Tv=0$ for some $vne 0$
$Tv=0=0v$ implies that $0$ is an eigenvalue with eigenvector $v$
answered Jul 18 at 21:52
Mohammad Riazi-Kermani
27.5k41852
answered Jul 18 at 21:52
Mohammad Riazi-Kermani
27.5k41852
answered Jul 18 at 21:52
Mohammad Riazi-Kermani
27.5k41852
answered Jul 18 at 21:52
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
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1
This is not really a proof, but rather just an application of the theorem of invertible matrices. Are you actually trying to prove this implication without use of the invertible matrix theorem (or at least this part of the theorem)?
– Dave
Jul 18 at 21:38
1
The most recent chapters discussed applications of the theorem so I assumed it was permitted. In the case where it isnt really a proof, is it at least a correct application of the theorem?
– HistorySweets
Jul 18 at 21:39
@HistorySweets An application of a theorem is definitely a proof.
– user577471
Jul 18 at 21:40
So, is the proof okay? Sorry, proofs are still pretty new to me. At this point, i'm just wrapping my head around theorems and such.
– HistorySweets
Jul 18 at 21:41
This application is certainly correct: if a matrix is not invertible it has zero as an eigenvalue (in this case the matrix associated to $T$). But if the question asks for a proof of this implication, then I don't think what you have really does this; a proof more relies on use of the definitions, and one cannot assume this result from the invertible matrix theorem is true a priori.
– Dave
Jul 18 at 21:42