Proof on characteristic values The 2019 Stack Overflow Developer Survey Results Are InShow the...
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Proof on characteristic values
The 2019 Stack Overflow Developer Survey Results Are InShow the existence of a polynomial $f$ such that $T^{-1} = f(T)$ for an invertible linear operator $T.$Prove that $Scirc T$ and $Tcirc S$ have the same characteristic polynomial.The dimension of a linear space is a multiple of 3 if it has an endomorphism of a certain featureQuestion about linear transformation proofcharacteristic polynomial of semisimple matrixLinear operator of infinite dimensionLinear operator proof.Characteristic polynomial of AB and BA.Every invertible finite order matrix is semi-simpleCharacteristic Polynomial Independent From the Choice of Basis
$begingroup$
If $V$ is a vectorial space of finite dimension over a field $F$ and T is a invertible linear operator over the field $V$, also, say that $lambda in F^*$
Prove that $lambda$ is a characteristic value of T if and only if $frac{1}{lambda} $ is a characteristic value of $T^{-1}$
For the first direction I assume that $lambda$ is a characteristic value of T, i.e, $exists v in V- ${$0$} such that $Tv = lambda v$
What we have to prove is that $exists w in V- ${$0$} such that $T^{-1}w = frac{1}{lambda} w$
I guess the other direction is similar, but I have no clue how to prove this, how can I use the inversible hypothesis?
linear-algebra
asked Mar 21 at 5:35
Juju9704Juju9704
34511
$endgroup$
add a comment |
$begingroup$
If $V$ is a vectorial space of finite dimension over a field $F$ and T is a invertible linear operator over the field $V$, also, say that $lambda in F^*$
Prove that $lambda$ is a characteristic value of T if and only if $frac{1}{lambda} $ is a characteristic value of $T^{-1}$
For the first direction I assume that $lambda$ is a characteristic value of T, i.e, $exists v in V- ${$0$} such that $Tv = lambda v$
What we have to prove is that $exists w in V- ${$0$} such that $T^{-1}w = frac{1}{lambda} w$
I guess the other direction is similar, but I have no clue how to prove this, how can I use the inversible hypothesis?
linear-algebra
asked Mar 21 at 5:35
Juju9704Juju9704
34511
$endgroup$
$begingroup$
"What we have to prove is that [..]" -- Should it not be $T^{-1}$ at the end of this line?
$endgroup$
– Eevee Trainer
Mar 21 at 5:42
$begingroup$
You're right! My bad
$endgroup$
– Juju9704
Mar 21 at 5:44
1
$begingroup$
For the converse, you want to assume that $lambda^{-1}$ is an eigenvalue (characteristic value) of $T^{-1}$, and show that $lambda$ is an eigenvalue of $T$. Try writing down the equation you get from the assumption here, and then apply $T$ to both sides and rearrange.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:47
add a comment |
$begingroup$
If $V$ is a vectorial space of finite dimension over a field $F$ and T is a invertible linear operator over the field $V$, also, say that $lambda in F^*$
Prove that $lambda$ is a characteristic value of T if and only if $frac{1}{lambda} $ is a characteristic value of $T^{-1}$
For the first direction I assume that $lambda$ is a characteristic value of T, i.e, $exists v in V- ${$0$} such that $Tv = lambda v$
What we have to prove is that $exists w in V- ${$0$} such that $T^{-1}w = frac{1}{lambda} w$
I guess the other direction is similar, but I have no clue how to prove this, how can I use the inversible hypothesis?
linear-algebra
asked Mar 21 at 5:35
Juju9704Juju9704
34511
$endgroup$
If $V$ is a vectorial space of finite dimension over a field $F$ and T is a invertible linear operator over the field $V$, also, say that $lambda in F^*$
Prove that $lambda$ is a characteristic value of T if and only if $frac{1}{lambda} $ is a characteristic value of $T^{-1}$
For the first direction I assume that $lambda$ is a characteristic value of T, i.e, $exists v in V- ${$0$} such that $Tv = lambda v$
What we have to prove is that $exists w in V- ${$0$} such that $T^{-1}w = frac{1}{lambda} w$
I guess the other direction is similar, but I have no clue how to prove this, how can I use the inversible hypothesis?
linear-algebra
linear-algebra
asked Mar 21 at 5:35
Juju9704Juju9704
34511
asked Mar 21 at 5:35
Juju9704Juju9704
34511
edited Mar 21 at 5:44
Juju9704
asked Mar 21 at 5:35
Juju9704Juju9704
34511
asked Mar 21 at 5:35
Juju9704Juju9704
34511
asked Mar 21 at 5:35
Juju9704Juju9704
34511
34511
$begingroup$
"What we have to prove is that [..]" -- Should it not be $T^{-1}$ at the end of this line?
$endgroup$
– Eevee Trainer
Mar 21 at 5:42
$begingroup$
You're right! My bad
$endgroup$
– Juju9704
Mar 21 at 5:44
1
$begingroup$
For the converse, you want to assume that $lambda^{-1}$ is an eigenvalue (characteristic value) of $T^{-1}$, and show that $lambda$ is an eigenvalue of $T$. Try writing down the equation you get from the assumption here, and then apply $T$ to both sides and rearrange.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:47
add a comment |
$begingroup$
"What we have to prove is that [..]" -- Should it not be $T^{-1}$ at the end of this line?
$endgroup$
– Eevee Trainer
Mar 21 at 5:42
$begingroup$
You're right! My bad
$endgroup$
– Juju9704
Mar 21 at 5:44
1
$begingroup$
For the converse, you want to assume that $lambda^{-1}$ is an eigenvalue (characteristic value) of $T^{-1}$, and show that $lambda$ is an eigenvalue of $T$. Try writing down the equation you get from the assumption here, and then apply $T$ to both sides and rearrange.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:47
$begingroup$
"What we have to prove is that [..]" -- Should it not be $T^{-1}$ at the end of this line?
$endgroup$
– Eevee Trainer
Mar 21 at 5:42
$begingroup$
"What we have to prove is that [..]" -- Should it not be $T^{-1}$ at the end of this line?
$endgroup$
– Eevee Trainer
Mar 21 at 5:42
$begingroup$
You're right! My bad
$endgroup$
– Juju9704
Mar 21 at 5:44
$begingroup$
You're right! My bad
$endgroup$
– Juju9704
Mar 21 at 5:44
1
1
$begingroup$
For the converse, you want to assume that $lambda^{-1}$ is an eigenvalue (characteristic value) of $T^{-1}$, and show that $lambda$ is an eigenvalue of $T$. Try writing down the equation you get from the assumption here, and then apply $T$ to both sides and rearrange.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:47
$begingroup$
For the converse, you want to assume that $lambda^{-1}$ is an eigenvalue (characteristic value) of $T^{-1}$, and show that $lambda$ is an eigenvalue of $T$. Try writing down the equation you get from the assumption here, and then apply $T$ to both sides and rearrange.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$Tv = lambda v iff v=lambda T^{-1}v iff frac{1}{lambda}v=T^{-1}v $.
answered Mar 21 at 6:30
FredFred
48.5k11849
$endgroup$
add a comment |
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$begingroup$
$Tv = lambda v iff v=lambda T^{-1}v iff frac{1}{lambda}v=T^{-1}v $.
answered Mar 21 at 6:30
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
$Tv = lambda v iff v=lambda T^{-1}v iff frac{1}{lambda}v=T^{-1}v $.
answered Mar 21 at 6:30
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
$Tv = lambda v iff v=lambda T^{-1}v iff frac{1}{lambda}v=T^{-1}v $.
answered Mar 21 at 6:30
FredFred
48.5k11849
$endgroup$
$Tv = lambda v iff v=lambda T^{-1}v iff frac{1}{lambda}v=T^{-1}v $.
answered Mar 21 at 6:30
FredFred
48.5k11849
answered Mar 21 at 6:30
FredFred
48.5k11849
answered Mar 21 at 6:30
FredFred
48.5k11849
answered Mar 21 at 6:30
FredFred
48.5k11849
48.5k11849
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$begingroup$
"What we have to prove is that [..]" -- Should it not be $T^{-1}$ at the end of this line?
$endgroup$
– Eevee Trainer
Mar 21 at 5:42
$begingroup$
You're right! My bad
$endgroup$
– Juju9704
Mar 21 at 5:44
1
$begingroup$
For the converse, you want to assume that $lambda^{-1}$ is an eigenvalue (characteristic value) of $T^{-1}$, and show that $lambda$ is an eigenvalue of $T$. Try writing down the equation you get from the assumption here, and then apply $T$ to both sides and rearrange.
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:47