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If there are basis $B$ and $B'$ s.t. $[T]_BB=[T^-1]_B'B'$ does $Spec(T)cap lmid neqemptysetiff $
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Let $V$ a $mathbb C-$vector space of finite dimension. Let $Tin mathcal L(V)$ invertible. If there exist basis $B$ and $B'$ of $V$ s.t. $$[T]_BB=[T^-1]_B'B'$$ does $$Spec(T)cap lambda mid neqemptysetiff Spec(T)caplambda neqemptyset,$$
where $Spect(T)$ is the set of eigenvalues.
I know that $[T^-1]_B'B'=[T^-1]_B'B'$, and thus $$[T]_BB[T]_B'B'=1.$$
Therefore, as matrices, $[T]_B'B'$ is the inverse of $[T]_BB$. Moreover, if $lambda in Spect(T)$ then $frac1lambda in Spect(T^-1)$. Therefore, $$lambda v=[T]_BBv=[T^-1]_B'B'v=frac1lambda vimplies (lambda ^2-1)v=0implies lambda ^2=1$$
since eigenvectors are $neq 0$. We conclude that $lambda =pm 1$ what is a contradiction.
I'm not sure that I'm doing well. Any help would be appreciated.
linear-algebra
asked Jul 28 at 10:40
MSE
1,471315
add a comment |Â
up vote
1
down vote
favorite
Let $V$ a $mathbb C-$vector space of finite dimension. Let $Tin mathcal L(V)$ invertible. If there exist basis $B$ and $B'$ of $V$ s.t. $$[T]_BB=[T^-1]_B'B'$$ does $$Spec(T)cap lambda mid neqemptysetiff Spec(T)caplambda neqemptyset,$$
where $Spect(T)$ is the set of eigenvalues.
I know that $[T^-1]_B'B'=[T^-1]_B'B'$, and thus $$[T]_BB[T]_B'B'=1.$$
Therefore, as matrices, $[T]_B'B'$ is the inverse of $[T]_BB$. Moreover, if $lambda in Spect(T)$ then $frac1lambda in Spect(T^-1)$. Therefore, $$lambda v=[T]_BBv=[T^-1]_B'B'v=frac1lambda vimplies (lambda ^2-1)v=0implies lambda ^2=1$$
since eigenvectors are $neq 0$. We conclude that $lambda =pm 1$ what is a contradiction.
I'm not sure that I'm doing well. Any help would be appreciated.
linear-algebra
asked Jul 28 at 10:40
MSE
1,471315
Be careful, writing $[T]_BBv=[T]_B'B'v$ really don't make sense.
– Surb
Jul 28 at 12:05
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $V$ a $mathbb C-$vector space of finite dimension. Let $Tin mathcal L(V)$ invertible. If there exist basis $B$ and $B'$ of $V$ s.t. $$[T]_BB=[T^-1]_B'B'$$ does $$Spec(T)cap lambda mid neqemptysetiff Spec(T)caplambda neqemptyset,$$
where $Spect(T)$ is the set of eigenvalues.
I know that $[T^-1]_B'B'=[T^-1]_B'B'$, and thus $$[T]_BB[T]_B'B'=1.$$
Therefore, as matrices, $[T]_B'B'$ is the inverse of $[T]_BB$. Moreover, if $lambda in Spect(T)$ then $frac1lambda in Spect(T^-1)$. Therefore, $$lambda v=[T]_BBv=[T^-1]_B'B'v=frac1lambda vimplies (lambda ^2-1)v=0implies lambda ^2=1$$
since eigenvectors are $neq 0$. We conclude that $lambda =pm 1$ what is a contradiction.
I'm not sure that I'm doing well. Any help would be appreciated.
linear-algebra
asked Jul 28 at 10:40
MSE
1,471315
Let $V$ a $mathbb C-$vector space of finite dimension. Let $Tin mathcal L(V)$ invertible. If there exist basis $B$ and $B'$ of $V$ s.t. $$[T]_BB=[T^-1]_B'B'$$ does $$Spec(T)cap lambda mid neqemptysetiff Spec(T)caplambda neqemptyset,$$
where $Spect(T)$ is the set of eigenvalues.
I know that $[T^-1]_B'B'=[T^-1]_B'B'$, and thus $$[T]_BB[T]_B'B'=1.$$
Therefore, as matrices, $[T]_B'B'$ is the inverse of $[T]_BB$. Moreover, if $lambda in Spect(T)$ then $frac1lambda in Spect(T^-1)$. Therefore, $$lambda v=[T]_BBv=[T^-1]_B'B'v=frac1lambda vimplies (lambda ^2-1)v=0implies lambda ^2=1$$
since eigenvectors are $neq 0$. We conclude that $lambda =pm 1$ what is a contradiction.
I'm not sure that I'm doing well. Any help would be appreciated.
linear-algebra
asked Jul 28 at 10:40
MSE
1,471315
edited Jul 28 at 12:03
asked Jul 28 at 10:40
MSE
1,471315
asked Jul 28 at 10:40
MSE
1,471315
asked Jul 28 at 10:40
MSE
1,471315
1,471315
Be careful, writing $[T]_BBv=[T]_B'B'v$ really don't make sense.
– Surb
Jul 28 at 12:05
add a comment |Â
Be careful, writing $[T]_BBv=[T]_B'B'v$ really don't make sense.
– Surb
Jul 28 at 12:05
Be careful, writing $[T]_BBv=[T]_B'B'v$ really don't make sense.
– Surb
Jul 28 at 12:05
Be careful, writing $[T]_BBv=[T]_B'B'v$ really don't make sense.
– Surb
Jul 28 at 12:05
add a comment |Â
2 Answers
2
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oldest
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1
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If $[mathcal T]_B,B = [mathcal T^-1 ]_B',B'$, then $[mathcal T]_B,B$ is similar to $[mathcal T^-1]_B,B$, hence $mathrm Spec (mathcal T) = mathrm Spec (mathcal T^-1)$. Since $lambda in mathrm Spec (mathcal T) iff lambda ^-1 in mathrm Spec (mathcal T^-1) = mathrm Spec(mathcal T)$, the assertion is correct.
In different bases, the matrix of eigenvectors are different.
answered Jul 28 at 11:03
xbh
1,0457
add a comment |Â
up vote
0
down vote
Careful, the eigenvalues are invariant, but not the eigenvectors.
I would solve it like this: The eigenvalues are independent of the basis, and hence, by your observation, in our situation the spectrum is closed under inversion.
answered Jul 28 at 10:52
AlgebraicsAnonymous
66611
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
accepted
If $[mathcal T]_B,B = [mathcal T^-1 ]_B',B'$, then $[mathcal T]_B,B$ is similar to $[mathcal T^-1]_B,B$, hence $mathrm Spec (mathcal T) = mathrm Spec (mathcal T^-1)$. Since $lambda in mathrm Spec (mathcal T) iff lambda ^-1 in mathrm Spec (mathcal T^-1) = mathrm Spec(mathcal T)$, the assertion is correct.
In different bases, the matrix of eigenvectors are different.
answered Jul 28 at 11:03
xbh
1,0457
add a comment |Â
up vote
1
down vote
accepted
If $[mathcal T]_B,B = [mathcal T^-1 ]_B',B'$, then $[mathcal T]_B,B$ is similar to $[mathcal T^-1]_B,B$, hence $mathrm Spec (mathcal T) = mathrm Spec (mathcal T^-1)$. Since $lambda in mathrm Spec (mathcal T) iff lambda ^-1 in mathrm Spec (mathcal T^-1) = mathrm Spec(mathcal T)$, the assertion is correct.
In different bases, the matrix of eigenvectors are different.
answered Jul 28 at 11:03
xbh
1,0457
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $[mathcal T]_B,B = [mathcal T^-1 ]_B',B'$, then $[mathcal T]_B,B$ is similar to $[mathcal T^-1]_B,B$, hence $mathrm Spec (mathcal T) = mathrm Spec (mathcal T^-1)$. Since $lambda in mathrm Spec (mathcal T) iff lambda ^-1 in mathrm Spec (mathcal T^-1) = mathrm Spec(mathcal T)$, the assertion is correct.
In different bases, the matrix of eigenvectors are different.
answered Jul 28 at 11:03
xbh
1,0457
If $[mathcal T]_B,B = [mathcal T^-1 ]_B',B'$, then $[mathcal T]_B,B$ is similar to $[mathcal T^-1]_B,B$, hence $mathrm Spec (mathcal T) = mathrm Spec (mathcal T^-1)$. Since $lambda in mathrm Spec (mathcal T) iff lambda ^-1 in mathrm Spec (mathcal T^-1) = mathrm Spec(mathcal T)$, the assertion is correct.
In different bases, the matrix of eigenvectors are different.
answered Jul 28 at 11:03
xbh
1,0457
answered Jul 28 at 11:03
xbh
1,0457
answered Jul 28 at 11:03
xbh
1,0457
answered Jul 28 at 11:03
xbh
1,0457
1,0457
add a comment |Â
add a comment |Â
up vote
0
down vote
Careful, the eigenvalues are invariant, but not the eigenvectors.
I would solve it like this: The eigenvalues are independent of the basis, and hence, by your observation, in our situation the spectrum is closed under inversion.
answered Jul 28 at 10:52
AlgebraicsAnonymous
66611
add a comment |Â
up vote
0
down vote
Careful, the eigenvalues are invariant, but not the eigenvectors.
I would solve it like this: The eigenvalues are independent of the basis, and hence, by your observation, in our situation the spectrum is closed under inversion.
answered Jul 28 at 10:52
AlgebraicsAnonymous
66611
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Careful, the eigenvalues are invariant, but not the eigenvectors.
I would solve it like this: The eigenvalues are independent of the basis, and hence, by your observation, in our situation the spectrum is closed under inversion.
answered Jul 28 at 10:52
AlgebraicsAnonymous
66611
Careful, the eigenvalues are invariant, but not the eigenvectors.
I would solve it like this: The eigenvalues are independent of the basis, and hence, by your observation, in our situation the spectrum is closed under inversion.
answered Jul 28 at 10:52
AlgebraicsAnonymous
66611
answered Jul 28 at 10:52
AlgebraicsAnonymous
66611
answered Jul 28 at 10:52
AlgebraicsAnonymous
66611
answered Jul 28 at 10:52
AlgebraicsAnonymous
66611
66611
add a comment |Â
add a comment |Â
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Be careful, writing $[T]_BBv=[T]_B'B'v$ really don't make sense.
– Surb
Jul 28 at 12:05