Invertible linear operator with diagonalizable power is diagonalizable
$begingroup$
This question is a follow-up to my other question here:
If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.
My problem is as follows:
Given a vector space $V$ over $mathbb{C}$, and a linear operator $T$ on $V$, show that if $T$ is invertible and $T^k$ is diagonalizable for some $kgeq2$, then $T$ is diagonalizable.
My attempt at a solution:
Because the vector space is over $mathbb{C}$, $T$ has an eigenvalue (by the fundamental theorem of algebra). Let the eigenvalues of $T$ be denoted by $lambda_1,lambda_2,cdots,lambda_n$, for some $ngeq1$. Then, observe that for some eigenvector $v$,
$$Tv=lambda_i vRightarrowlambda_i (Tv)=lambda_i(lambda_i v)Rightarrow T(lambda_i v)=lambda_i^2 vRightarrow T(Tv)=lambda_i^2 vRightarrow T^2v=lambda_i^2 v.$$
This argument can be inductively continued to show that the eigenvalues of $T^k$ are $lambda_1^k,lambda_2^k,cdots,lambda_n^k$. Because $T^k$ is diagonalizable, these eigenvalues are distinct (a proof that I'm omitting for brevity here). Because the $lambda_i^k$'s are distinct, it follows that the $lambda_i$'s are distinct. Therefore, $T$ is diagonalizable (again based on the proof that I'm omitting).
Here are my questions:
(a) First and foremost, is this a valid proof? I'm skeptical because it is so much simpler than the proof I've linked (which is in terms of matrices, not linear operators, but that shouldn't make a significant difference). I also haven't (explicitly) used the fact that $T$ is invertible, so I feel something is missing.
(b) It seems like this problem can be approached using Jordan canonical form, but I'm struggling to do so. Any suggestions would be appreciated.
linear-algebra proof-verification alternative-proof diagonalization
edited Dec 6 '18 at 1:41
Gerry Myerson
147k8149302
asked Dec 6 '18 at 0:44
AtsinaAtsina
820117
$endgroup$
|
show 4 more comments
$begingroup$
This question is a follow-up to my other question here:
If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.
My problem is as follows:
Given a vector space $V$ over $mathbb{C}$, and a linear operator $T$ on $V$, show that if $T$ is invertible and $T^k$ is diagonalizable for some $kgeq2$, then $T$ is diagonalizable.
My attempt at a solution:
Because the vector space is over $mathbb{C}$, $T$ has an eigenvalue (by the fundamental theorem of algebra). Let the eigenvalues of $T$ be denoted by $lambda_1,lambda_2,cdots,lambda_n$, for some $ngeq1$. Then, observe that for some eigenvector $v$,
$$Tv=lambda_i vRightarrowlambda_i (Tv)=lambda_i(lambda_i v)Rightarrow T(lambda_i v)=lambda_i^2 vRightarrow T(Tv)=lambda_i^2 vRightarrow T^2v=lambda_i^2 v.$$
This argument can be inductively continued to show that the eigenvalues of $T^k$ are $lambda_1^k,lambda_2^k,cdots,lambda_n^k$. Because $T^k$ is diagonalizable, these eigenvalues are distinct (a proof that I'm omitting for brevity here). Because the $lambda_i^k$'s are distinct, it follows that the $lambda_i$'s are distinct. Therefore, $T$ is diagonalizable (again based on the proof that I'm omitting).
Here are my questions:
(a) First and foremost, is this a valid proof? I'm skeptical because it is so much simpler than the proof I've linked (which is in terms of matrices, not linear operators, but that shouldn't make a significant difference). I also haven't (explicitly) used the fact that $T$ is invertible, so I feel something is missing.
(b) It seems like this problem can be approached using Jordan canonical form, but I'm struggling to do so. Any suggestions would be appreciated.
linear-algebra proof-verification alternative-proof diagonalization
edited Dec 6 '18 at 1:41
Gerry Myerson
147k8149302
asked Dec 6 '18 at 0:44
AtsinaAtsina
820117
$endgroup$
$begingroup$
Could cite precisely what result are you omitting? You should be using invertibility somewhere. For example, nilpotent matrices such as Jordan blocks of eigenvalue zero are nilpotent (and so for a sufficiently high power, diagonalizable) and not diagonzalizable.
$endgroup$
– Guido A.
Dec 6 '18 at 0:53
$begingroup$
@GuidoA. mathworld.wolfram.com/DiagonalizableMatrix.html The diagonalization theorem gives that an $ntimes n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors. These must correspond to $n$ distinct eigenvalues. And since an operator can be represented as a matrix...
$endgroup$
– Atsina
Dec 6 '18 at 0:56
1
$begingroup$
Could you clarify your statement? The identity matrix $I_n in M_n(mathbb{k})$ seems pretty diagonalizable to me and has a unique eigenvalue.
$endgroup$
– Guido A.
Dec 6 '18 at 0:59
$begingroup$
@GuidoA. Hmmm. It's probably my mistake...but this seems to me that MathWorld has an error. Certainly $mbox{dim }V$ eigenvalues implies diagonalizability, but the converse doesn't appear to be true, as suggested by MathWorld
$endgroup$
– Atsina
Dec 6 '18 at 1:03
1
$begingroup$
Are you aware of the following fact: A linear operator $T$ on a finite-dimensional $mathbb{C}$-vector space $V$ is diagonalizable if and only if there exists a polynomial $P$ satisfying $Pleft(Tright) = 0$ and $P^prime neq 0$ (this means that the derivative of $P$ is not identically $0$ as a polynomial)? If you know this, I suggest you think about how to get such a polynomial for $T$ from such a polynomial for $T^k$.
$endgroup$
– darij grinberg
Dec 6 '18 at 4:37
|
show 4 more comments
$begingroup$
This question is a follow-up to my other question here:
If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.
My problem is as follows:
Given a vector space $V$ over $mathbb{C}$, and a linear operator $T$ on $V$, show that if $T$ is invertible and $T^k$ is diagonalizable for some $kgeq2$, then $T$ is diagonalizable.
My attempt at a solution:
Because the vector space is over $mathbb{C}$, $T$ has an eigenvalue (by the fundamental theorem of algebra). Let the eigenvalues of $T$ be denoted by $lambda_1,lambda_2,cdots,lambda_n$, for some $ngeq1$. Then, observe that for some eigenvector $v$,
$$Tv=lambda_i vRightarrowlambda_i (Tv)=lambda_i(lambda_i v)Rightarrow T(lambda_i v)=lambda_i^2 vRightarrow T(Tv)=lambda_i^2 vRightarrow T^2v=lambda_i^2 v.$$
This argument can be inductively continued to show that the eigenvalues of $T^k$ are $lambda_1^k,lambda_2^k,cdots,lambda_n^k$. Because $T^k$ is diagonalizable, these eigenvalues are distinct (a proof that I'm omitting for brevity here). Because the $lambda_i^k$'s are distinct, it follows that the $lambda_i$'s are distinct. Therefore, $T$ is diagonalizable (again based on the proof that I'm omitting).
Here are my questions:
(a) First and foremost, is this a valid proof? I'm skeptical because it is so much simpler than the proof I've linked (which is in terms of matrices, not linear operators, but that shouldn't make a significant difference). I also haven't (explicitly) used the fact that $T$ is invertible, so I feel something is missing.
(b) It seems like this problem can be approached using Jordan canonical form, but I'm struggling to do so. Any suggestions would be appreciated.
linear-algebra proof-verification alternative-proof diagonalization
edited Dec 6 '18 at 1:41
Gerry Myerson
147k8149302
asked Dec 6 '18 at 0:44
AtsinaAtsina
820117
$endgroup$
This question is a follow-up to my other question here:
If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.
My problem is as follows:
Given a vector space $V$ over $mathbb{C}$, and a linear operator $T$ on $V$, show that if $T$ is invertible and $T^k$ is diagonalizable for some $kgeq2$, then $T$ is diagonalizable.
My attempt at a solution:
Because the vector space is over $mathbb{C}$, $T$ has an eigenvalue (by the fundamental theorem of algebra). Let the eigenvalues of $T$ be denoted by $lambda_1,lambda_2,cdots,lambda_n$, for some $ngeq1$. Then, observe that for some eigenvector $v$,
$$Tv=lambda_i vRightarrowlambda_i (Tv)=lambda_i(lambda_i v)Rightarrow T(lambda_i v)=lambda_i^2 vRightarrow T(Tv)=lambda_i^2 vRightarrow T^2v=lambda_i^2 v.$$
This argument can be inductively continued to show that the eigenvalues of $T^k$ are $lambda_1^k,lambda_2^k,cdots,lambda_n^k$. Because $T^k$ is diagonalizable, these eigenvalues are distinct (a proof that I'm omitting for brevity here). Because the $lambda_i^k$'s are distinct, it follows that the $lambda_i$'s are distinct. Therefore, $T$ is diagonalizable (again based on the proof that I'm omitting).
Here are my questions:
(a) First and foremost, is this a valid proof? I'm skeptical because it is so much simpler than the proof I've linked (which is in terms of matrices, not linear operators, but that shouldn't make a significant difference). I also haven't (explicitly) used the fact that $T$ is invertible, so I feel something is missing.
(b) It seems like this problem can be approached using Jordan canonical form, but I'm struggling to do so. Any suggestions would be appreciated.
linear-algebra proof-verification alternative-proof diagonalization
linear-algebra proof-verification alternative-proof diagonalization
edited Dec 6 '18 at 1:41
Gerry Myerson
147k8149302
asked Dec 6 '18 at 0:44
AtsinaAtsina
820117
edited Dec 6 '18 at 1:41
Gerry Myerson
147k8149302
asked Dec 6 '18 at 0:44
AtsinaAtsina
820117
edited Dec 6 '18 at 1:41
Gerry Myerson
147k8149302
edited Dec 6 '18 at 1:41
Gerry Myerson
147k8149302
edited Dec 6 '18 at 1:41
Gerry Myerson
147k8149302
147k8149302
asked Dec 6 '18 at 0:44
AtsinaAtsina
820117
asked Dec 6 '18 at 0:44
AtsinaAtsina
820117
asked Dec 6 '18 at 0:44
AtsinaAtsina
820117
820117
$begingroup$
Could cite precisely what result are you omitting? You should be using invertibility somewhere. For example, nilpotent matrices such as Jordan blocks of eigenvalue zero are nilpotent (and so for a sufficiently high power, diagonalizable) and not diagonzalizable.
$endgroup$
– Guido A.
Dec 6 '18 at 0:53
$begingroup$
@GuidoA. mathworld.wolfram.com/DiagonalizableMatrix.html The diagonalization theorem gives that an $ntimes n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors. These must correspond to $n$ distinct eigenvalues. And since an operator can be represented as a matrix...
$endgroup$
– Atsina
Dec 6 '18 at 0:56
1
$begingroup$
Could you clarify your statement? The identity matrix $I_n in M_n(mathbb{k})$ seems pretty diagonalizable to me and has a unique eigenvalue.
$endgroup$
– Guido A.
Dec 6 '18 at 0:59
$begingroup$
@GuidoA. Hmmm. It's probably my mistake...but this seems to me that MathWorld has an error. Certainly $mbox{dim }V$ eigenvalues implies diagonalizability, but the converse doesn't appear to be true, as suggested by MathWorld
$endgroup$
– Atsina
Dec 6 '18 at 1:03
1
$begingroup$
Are you aware of the following fact: A linear operator $T$ on a finite-dimensional $mathbb{C}$-vector space $V$ is diagonalizable if and only if there exists a polynomial $P$ satisfying $Pleft(Tright) = 0$ and $P^prime neq 0$ (this means that the derivative of $P$ is not identically $0$ as a polynomial)? If you know this, I suggest you think about how to get such a polynomial for $T$ from such a polynomial for $T^k$.
$endgroup$
– darij grinberg
Dec 6 '18 at 4:37
|
show 4 more comments
$begingroup$
Could cite precisely what result are you omitting? You should be using invertibility somewhere. For example, nilpotent matrices such as Jordan blocks of eigenvalue zero are nilpotent (and so for a sufficiently high power, diagonalizable) and not diagonzalizable.
$endgroup$
– Guido A.
Dec 6 '18 at 0:53
$begingroup$
@GuidoA. mathworld.wolfram.com/DiagonalizableMatrix.html The diagonalization theorem gives that an $ntimes n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors. These must correspond to $n$ distinct eigenvalues. And since an operator can be represented as a matrix...
$endgroup$
– Atsina
Dec 6 '18 at 0:56
1
$begingroup$
Could you clarify your statement? The identity matrix $I_n in M_n(mathbb{k})$ seems pretty diagonalizable to me and has a unique eigenvalue.
$endgroup$
– Guido A.
Dec 6 '18 at 0:59
$begingroup$
@GuidoA. Hmmm. It's probably my mistake...but this seems to me that MathWorld has an error. Certainly $mbox{dim }V$ eigenvalues implies diagonalizability, but the converse doesn't appear to be true, as suggested by MathWorld
$endgroup$
– Atsina
Dec 6 '18 at 1:03
1
$begingroup$
Are you aware of the following fact: A linear operator $T$ on a finite-dimensional $mathbb{C}$-vector space $V$ is diagonalizable if and only if there exists a polynomial $P$ satisfying $Pleft(Tright) = 0$ and $P^prime neq 0$ (this means that the derivative of $P$ is not identically $0$ as a polynomial)? If you know this, I suggest you think about how to get such a polynomial for $T$ from such a polynomial for $T^k$.
$endgroup$
– darij grinberg
Dec 6 '18 at 4:37
$begingroup$
Could cite precisely what result are you omitting? You should be using invertibility somewhere. For example, nilpotent matrices such as Jordan blocks of eigenvalue zero are nilpotent (and so for a sufficiently high power, diagonalizable) and not diagonzalizable.
$endgroup$
– Guido A.
Dec 6 '18 at 0:53
$begingroup$
Could cite precisely what result are you omitting? You should be using invertibility somewhere. For example, nilpotent matrices such as Jordan blocks of eigenvalue zero are nilpotent (and so for a sufficiently high power, diagonalizable) and not diagonzalizable.
$endgroup$
– Guido A.
Dec 6 '18 at 0:53
$begingroup$
@GuidoA. mathworld.wolfram.com/DiagonalizableMatrix.html The diagonalization theorem gives that an $ntimes n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors. These must correspond to $n$ distinct eigenvalues. And since an operator can be represented as a matrix...
$endgroup$
– Atsina
Dec 6 '18 at 0:56
$begingroup$
@GuidoA. mathworld.wolfram.com/DiagonalizableMatrix.html The diagonalization theorem gives that an $ntimes n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors. These must correspond to $n$ distinct eigenvalues. And since an operator can be represented as a matrix...
$endgroup$
– Atsina
Dec 6 '18 at 0:56
1
1
$begingroup$
Could you clarify your statement? The identity matrix $I_n in M_n(mathbb{k})$ seems pretty diagonalizable to me and has a unique eigenvalue.
$endgroup$
– Guido A.
Dec 6 '18 at 0:59
$begingroup$
Could you clarify your statement? The identity matrix $I_n in M_n(mathbb{k})$ seems pretty diagonalizable to me and has a unique eigenvalue.
$endgroup$
– Guido A.
Dec 6 '18 at 0:59
$begingroup$
@GuidoA. Hmmm. It's probably my mistake...but this seems to me that MathWorld has an error. Certainly $mbox{dim }V$ eigenvalues implies diagonalizability, but the converse doesn't appear to be true, as suggested by MathWorld
$endgroup$
– Atsina
Dec 6 '18 at 1:03
$begingroup$
@GuidoA. Hmmm. It's probably my mistake...but this seems to me that MathWorld has an error. Certainly $mbox{dim }V$ eigenvalues implies diagonalizability, but the converse doesn't appear to be true, as suggested by MathWorld
$endgroup$
– Atsina
Dec 6 '18 at 1:03
1
1
$begingroup$
Are you aware of the following fact: A linear operator $T$ on a finite-dimensional $mathbb{C}$-vector space $V$ is diagonalizable if and only if there exists a polynomial $P$ satisfying $Pleft(Tright) = 0$ and $P^prime neq 0$ (this means that the derivative of $P$ is not identically $0$ as a polynomial)? If you know this, I suggest you think about how to get such a polynomial for $T$ from such a polynomial for $T^k$.
$endgroup$
– darij grinberg
Dec 6 '18 at 4:37
$begingroup$
Are you aware of the following fact: A linear operator $T$ on a finite-dimensional $mathbb{C}$-vector space $V$ is diagonalizable if and only if there exists a polynomial $P$ satisfying $Pleft(Tright) = 0$ and $P^prime neq 0$ (this means that the derivative of $P$ is not identically $0$ as a polynomial)? If you know this, I suggest you think about how to get such a polynomial for $T$ from such a polynomial for $T^k$.
$endgroup$
– darij grinberg
Dec 6 '18 at 4:37
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Since (a) is sorted out by the comments, let me only comment on (b). The point here is that $T$ is invertible if and only if all Jordan blocks have a non-zero eigenvalue and it is diagonalizable if and only if all Jordan blocks have size one. If you bring $T$ to Jordan canonical form, then a matrix representation of $T^k$ is obtained from just taking the $k$-th powers of the Jordan blocks of $T$.
Hence you can prove your claim by showing that if $J$ is a Jordan-block of size at least two with non-zero eigenvalue, then for each $kinmathbb N$, the matrix $J^k$ is not diagonalizable. But this is easily verified directly, since $J^k$ is always upper triangular with (equal) non-zero entries on the main diagonal and also the entries in the diagonal right above the main one are non-zero.
answered Dec 6 '18 at 12:01
Andreas CapAndreas Cap
11.3k923
$endgroup$
add a comment |
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$begingroup$
Since (a) is sorted out by the comments, let me only comment on (b). The point here is that $T$ is invertible if and only if all Jordan blocks have a non-zero eigenvalue and it is diagonalizable if and only if all Jordan blocks have size one. If you bring $T$ to Jordan canonical form, then a matrix representation of $T^k$ is obtained from just taking the $k$-th powers of the Jordan blocks of $T$.
Hence you can prove your claim by showing that if $J$ is a Jordan-block of size at least two with non-zero eigenvalue, then for each $kinmathbb N$, the matrix $J^k$ is not diagonalizable. But this is easily verified directly, since $J^k$ is always upper triangular with (equal) non-zero entries on the main diagonal and also the entries in the diagonal right above the main one are non-zero.
answered Dec 6 '18 at 12:01
Andreas CapAndreas Cap
11.3k923
$endgroup$
add a comment |
$begingroup$
Since (a) is sorted out by the comments, let me only comment on (b). The point here is that $T$ is invertible if and only if all Jordan blocks have a non-zero eigenvalue and it is diagonalizable if and only if all Jordan blocks have size one. If you bring $T$ to Jordan canonical form, then a matrix representation of $T^k$ is obtained from just taking the $k$-th powers of the Jordan blocks of $T$.
Hence you can prove your claim by showing that if $J$ is a Jordan-block of size at least two with non-zero eigenvalue, then for each $kinmathbb N$, the matrix $J^k$ is not diagonalizable. But this is easily verified directly, since $J^k$ is always upper triangular with (equal) non-zero entries on the main diagonal and also the entries in the diagonal right above the main one are non-zero.
answered Dec 6 '18 at 12:01
Andreas CapAndreas Cap
11.3k923
$endgroup$
add a comment |
$begingroup$
Since (a) is sorted out by the comments, let me only comment on (b). The point here is that $T$ is invertible if and only if all Jordan blocks have a non-zero eigenvalue and it is diagonalizable if and only if all Jordan blocks have size one. If you bring $T$ to Jordan canonical form, then a matrix representation of $T^k$ is obtained from just taking the $k$-th powers of the Jordan blocks of $T$.
Hence you can prove your claim by showing that if $J$ is a Jordan-block of size at least two with non-zero eigenvalue, then for each $kinmathbb N$, the matrix $J^k$ is not diagonalizable. But this is easily verified directly, since $J^k$ is always upper triangular with (equal) non-zero entries on the main diagonal and also the entries in the diagonal right above the main one are non-zero.
answered Dec 6 '18 at 12:01
Andreas CapAndreas Cap
11.3k923
$endgroup$
Since (a) is sorted out by the comments, let me only comment on (b). The point here is that $T$ is invertible if and only if all Jordan blocks have a non-zero eigenvalue and it is diagonalizable if and only if all Jordan blocks have size one. If you bring $T$ to Jordan canonical form, then a matrix representation of $T^k$ is obtained from just taking the $k$-th powers of the Jordan blocks of $T$.
Hence you can prove your claim by showing that if $J$ is a Jordan-block of size at least two with non-zero eigenvalue, then for each $kinmathbb N$, the matrix $J^k$ is not diagonalizable. But this is easily verified directly, since $J^k$ is always upper triangular with (equal) non-zero entries on the main diagonal and also the entries in the diagonal right above the main one are non-zero.
answered Dec 6 '18 at 12:01
Andreas CapAndreas Cap
11.3k923
answered Dec 6 '18 at 12:01
Andreas CapAndreas Cap
11.3k923
answered Dec 6 '18 at 12:01
Andreas CapAndreas Cap
11.3k923
answered Dec 6 '18 at 12:01
Andreas CapAndreas Cap
11.3k923
11.3k923
add a comment |
add a comment |
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$begingroup$
Could cite precisely what result are you omitting? You should be using invertibility somewhere. For example, nilpotent matrices such as Jordan blocks of eigenvalue zero are nilpotent (and so for a sufficiently high power, diagonalizable) and not diagonzalizable.
$endgroup$
– Guido A.
Dec 6 '18 at 0:53
$begingroup$
@GuidoA. mathworld.wolfram.com/DiagonalizableMatrix.html The diagonalization theorem gives that an $ntimes n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors. These must correspond to $n$ distinct eigenvalues. And since an operator can be represented as a matrix...
$endgroup$
– Atsina
Dec 6 '18 at 0:56
1
$begingroup$
Could you clarify your statement? The identity matrix $I_n in M_n(mathbb{k})$ seems pretty diagonalizable to me and has a unique eigenvalue.
$endgroup$
– Guido A.
Dec 6 '18 at 0:59
$begingroup$
@GuidoA. Hmmm. It's probably my mistake...but this seems to me that MathWorld has an error. Certainly $mbox{dim }V$ eigenvalues implies diagonalizability, but the converse doesn't appear to be true, as suggested by MathWorld
$endgroup$
– Atsina
Dec 6 '18 at 1:03
1
$begingroup$
Are you aware of the following fact: A linear operator $T$ on a finite-dimensional $mathbb{C}$-vector space $V$ is diagonalizable if and only if there exists a polynomial $P$ satisfying $Pleft(Tright) = 0$ and $P^prime neq 0$ (this means that the derivative of $P$ is not identically $0$ as a polynomial)? If you know this, I suggest you think about how to get such a polynomial for $T$ from such a polynomial for $T^k$.
$endgroup$
– darij grinberg
Dec 6 '18 at 4:37