If $A$ is hermitian matrix s.t. $A^99=I_n$, show that $A=I_n$.
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Let $A in M_n(mathbbC)$ be a hermitian matrix that satisfies the property $A^99=I_n$. I want to show that $A=I_n$.
By definition, $A$ is a hermitian matrix if $a_ij=overlinea_ji$.
From the fact that $A^99=I_n$, we deduce that $A$ is invertible and its inverse is $A^98$. Does that help to get that $A=I_n$ ? Or do we use somehow the definition of a hermitian matrix?
linear-algebra matrices
edited Sep 2 at 12:42
Shaun
7,44792972
asked Sep 2 at 12:39
Evinda
552412
 |Â
show 11 more comments
up vote
2
down vote
favorite
Let $A in M_n(mathbbC)$ be a hermitian matrix that satisfies the property $A^99=I_n$. I want to show that $A=I_n$.
By definition, $A$ is a hermitian matrix if $a_ij=overlinea_ji$.
From the fact that $A^99=I_n$, we deduce that $A$ is invertible and its inverse is $A^98$. Does that help to get that $A=I_n$ ? Or do we use somehow the definition of a hermitian matrix?
linear-algebra matrices
edited Sep 2 at 12:42
Shaun
7,44792972
asked Sep 2 at 12:39
Evinda
552412
3
Hint: Since $A$ is hermitian, it can be diagonalized. It is easy to write down the entries of a power of a diagonal matrix. And remember that the eigenvalues of $A$ are real.
– Henning Makholm
Sep 2 at 12:42
So since $A$ is hermitian, there exists an invertible matrix $P$ such that $P^-1AP$ is a diagonal matrix. How does this help? @HenningMakholm
– Evinda
Sep 2 at 12:55
x @Evinda: Simplify $A^99 = (PDP^-1)^99$.
– Henning Makholm
Sep 2 at 13:00
1
$D$ contains the eigenvalues, right? So from $D^99=I_n$, do we get that $lambda^99=1$ for each eigenvalue $lambda$? Thus there is only one eigenvalue, $lambda=1$. Thus $D=I$ and so $A=PP^-1=I_n$. Am I right? @HenningMakholm
– Evinda
Sep 2 at 14:17
1
@Evinda Yes, but you must be sure first that the eigenvalues of $A$ are real; otherwise $lambda^99=1$ woul not imply $lambda=1$. This happens because $A$ is hermitian
– Jose Brox
Sep 3 at 18:02
 |Â
show 11 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $A in M_n(mathbbC)$ be a hermitian matrix that satisfies the property $A^99=I_n$. I want to show that $A=I_n$.
By definition, $A$ is a hermitian matrix if $a_ij=overlinea_ji$.
From the fact that $A^99=I_n$, we deduce that $A$ is invertible and its inverse is $A^98$. Does that help to get that $A=I_n$ ? Or do we use somehow the definition of a hermitian matrix?
linear-algebra matrices
edited Sep 2 at 12:42
Shaun
7,44792972
asked Sep 2 at 12:39
Evinda
552412
Let $A in M_n(mathbbC)$ be a hermitian matrix that satisfies the property $A^99=I_n$. I want to show that $A=I_n$.
By definition, $A$ is a hermitian matrix if $a_ij=overlinea_ji$.
From the fact that $A^99=I_n$, we deduce that $A$ is invertible and its inverse is $A^98$. Does that help to get that $A=I_n$ ? Or do we use somehow the definition of a hermitian matrix?
linear-algebra matrices
linear-algebra matrices
edited Sep 2 at 12:42
Shaun
7,44792972
asked Sep 2 at 12:39
Evinda
552412
edited Sep 2 at 12:42
Shaun
7,44792972
asked Sep 2 at 12:39
Evinda
552412
edited Sep 2 at 12:42
Shaun
7,44792972
edited Sep 2 at 12:42
Shaun
7,44792972
edited Sep 2 at 12:42
Shaun
7,44792972
7,44792972
asked Sep 2 at 12:39
Evinda
552412
asked Sep 2 at 12:39
Evinda
552412
asked Sep 2 at 12:39
Evinda
552412
552412
3
Hint: Since $A$ is hermitian, it can be diagonalized. It is easy to write down the entries of a power of a diagonal matrix. And remember that the eigenvalues of $A$ are real.
– Henning Makholm
Sep 2 at 12:42
So since $A$ is hermitian, there exists an invertible matrix $P$ such that $P^-1AP$ is a diagonal matrix. How does this help? @HenningMakholm
– Evinda
Sep 2 at 12:55
x @Evinda: Simplify $A^99 = (PDP^-1)^99$.
– Henning Makholm
Sep 2 at 13:00
1
$D$ contains the eigenvalues, right? So from $D^99=I_n$, do we get that $lambda^99=1$ for each eigenvalue $lambda$? Thus there is only one eigenvalue, $lambda=1$. Thus $D=I$ and so $A=PP^-1=I_n$. Am I right? @HenningMakholm
– Evinda
Sep 2 at 14:17
1
@Evinda Yes, but you must be sure first that the eigenvalues of $A$ are real; otherwise $lambda^99=1$ woul not imply $lambda=1$. This happens because $A$ is hermitian
– Jose Brox
Sep 3 at 18:02
 |Â
show 11 more comments
3
Hint: Since $A$ is hermitian, it can be diagonalized. It is easy to write down the entries of a power of a diagonal matrix. And remember that the eigenvalues of $A$ are real.
– Henning Makholm
Sep 2 at 12:42
So since $A$ is hermitian, there exists an invertible matrix $P$ such that $P^-1AP$ is a diagonal matrix. How does this help? @HenningMakholm
– Evinda
Sep 2 at 12:55
x @Evinda: Simplify $A^99 = (PDP^-1)^99$.
– Henning Makholm
Sep 2 at 13:00
1
$D$ contains the eigenvalues, right? So from $D^99=I_n$, do we get that $lambda^99=1$ for each eigenvalue $lambda$? Thus there is only one eigenvalue, $lambda=1$. Thus $D=I$ and so $A=PP^-1=I_n$. Am I right? @HenningMakholm
– Evinda
Sep 2 at 14:17
1
@Evinda Yes, but you must be sure first that the eigenvalues of $A$ are real; otherwise $lambda^99=1$ woul not imply $lambda=1$. This happens because $A$ is hermitian
– Jose Brox
Sep 3 at 18:02
3
3
Hint: Since $A$ is hermitian, it can be diagonalized. It is easy to write down the entries of a power of a diagonal matrix. And remember that the eigenvalues of $A$ are real.
– Henning Makholm
Sep 2 at 12:42
Hint: Since $A$ is hermitian, it can be diagonalized. It is easy to write down the entries of a power of a diagonal matrix. And remember that the eigenvalues of $A$ are real.
– Henning Makholm
Sep 2 at 12:42
So since $A$ is hermitian, there exists an invertible matrix $P$ such that $P^-1AP$ is a diagonal matrix. How does this help? @HenningMakholm
– Evinda
Sep 2 at 12:55
So since $A$ is hermitian, there exists an invertible matrix $P$ such that $P^-1AP$ is a diagonal matrix. How does this help? @HenningMakholm
– Evinda
Sep 2 at 12:55
x @Evinda: Simplify $A^99 = (PDP^-1)^99$.
– Henning Makholm
Sep 2 at 13:00
x @Evinda: Simplify $A^99 = (PDP^-1)^99$.
– Henning Makholm
Sep 2 at 13:00
1
1
$D$ contains the eigenvalues, right? So from $D^99=I_n$, do we get that $lambda^99=1$ for each eigenvalue $lambda$? Thus there is only one eigenvalue, $lambda=1$. Thus $D=I$ and so $A=PP^-1=I_n$. Am I right? @HenningMakholm
– Evinda
Sep 2 at 14:17
$D$ contains the eigenvalues, right? So from $D^99=I_n$, do we get that $lambda^99=1$ for each eigenvalue $lambda$? Thus there is only one eigenvalue, $lambda=1$. Thus $D=I$ and so $A=PP^-1=I_n$. Am I right? @HenningMakholm
– Evinda
Sep 2 at 14:17
1
1
@Evinda Yes, but you must be sure first that the eigenvalues of $A$ are real; otherwise $lambda^99=1$ woul not imply $lambda=1$. This happens because $A$ is hermitian
– Jose Brox
Sep 3 at 18:02
@Evinda Yes, but you must be sure first that the eigenvalues of $A$ are real; otherwise $lambda^99=1$ woul not imply $lambda=1$. This happens because $A$ is hermitian
– Jose Brox
Sep 3 at 18:02
 |Â
show 11 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
$A$ is hermitian so $sigma(A) subseteq mathbbR$. On the other hand, the polynomial $z^99 - 1$ annihilates $A$ so
$$sigma(A) subseteq textzeroes of z^99-1 cap mathbbR = lefte^ifrac2kpi99 : k in 0, ldots, 98right cap mathbbR = 1$$
so $sigma(A) = 1$.
Since $A$ is diagonalizable, it diagonalizes to the indentity matrix so $A = I$.
answered Sep 2 at 15:35
mechanodroid
24.1k52244
Which is the $sigma$-function?
– Evinda
Sep 3 at 18:12
@Evinda $sigma(A)$ is the spectrum of $A$, i.e. the set of all eigenvalues of $A$.
– mechanodroid
Sep 3 at 19:20
And why does it hold that $sigma(A) subseteq textzeroes of z^99-1 cap mathbbR$ ?
– Evinda
Sep 3 at 20:16
@Evinda $A$ is hermitian so $sigma(A) subseteq mathbbR$. The polynomial $z^99 - 1$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides $z^99 - 1$. Therefore $$sigma(A) = textzeroes of m_A subseteq textzeroes of z^99-1$$ Hence, $sigma(A)$ is contained in the intersection.
– mechanodroid
Sep 3 at 20:35
1
@Evinda Yes, precisely. Since $A$ is diagonalizable, there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^-1$. The diagonal matrix $D$ has eigenvalues of $A$ along the diagonal, which must all be $1$. Therefore $D = I$ so $A = PIP^-1 = PP^-1 = I$.
– mechanodroid
Sep 4 at 8:35
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$A$ is hermitian so $sigma(A) subseteq mathbbR$. On the other hand, the polynomial $z^99 - 1$ annihilates $A$ so
$$sigma(A) subseteq textzeroes of z^99-1 cap mathbbR = lefte^ifrac2kpi99 : k in 0, ldots, 98right cap mathbbR = 1$$
so $sigma(A) = 1$.
Since $A$ is diagonalizable, it diagonalizes to the indentity matrix so $A = I$.
answered Sep 2 at 15:35
mechanodroid
24.1k52244
Which is the $sigma$-function?
– Evinda
Sep 3 at 18:12
@Evinda $sigma(A)$ is the spectrum of $A$, i.e. the set of all eigenvalues of $A$.
– mechanodroid
Sep 3 at 19:20
And why does it hold that $sigma(A) subseteq textzeroes of z^99-1 cap mathbbR$ ?
– Evinda
Sep 3 at 20:16
@Evinda $A$ is hermitian so $sigma(A) subseteq mathbbR$. The polynomial $z^99 - 1$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides $z^99 - 1$. Therefore $$sigma(A) = textzeroes of m_A subseteq textzeroes of z^99-1$$ Hence, $sigma(A)$ is contained in the intersection.
– mechanodroid
Sep 3 at 20:35
1
@Evinda Yes, precisely. Since $A$ is diagonalizable, there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^-1$. The diagonal matrix $D$ has eigenvalues of $A$ along the diagonal, which must all be $1$. Therefore $D = I$ so $A = PIP^-1 = PP^-1 = I$.
– mechanodroid
Sep 4 at 8:35
 |Â
show 2 more comments
up vote
3
down vote
accepted
$A$ is hermitian so $sigma(A) subseteq mathbbR$. On the other hand, the polynomial $z^99 - 1$ annihilates $A$ so
$$sigma(A) subseteq textzeroes of z^99-1 cap mathbbR = lefte^ifrac2kpi99 : k in 0, ldots, 98right cap mathbbR = 1$$
so $sigma(A) = 1$.
Since $A$ is diagonalizable, it diagonalizes to the indentity matrix so $A = I$.
answered Sep 2 at 15:35
mechanodroid
24.1k52244
Which is the $sigma$-function?
– Evinda
Sep 3 at 18:12
@Evinda $sigma(A)$ is the spectrum of $A$, i.e. the set of all eigenvalues of $A$.
– mechanodroid
Sep 3 at 19:20
And why does it hold that $sigma(A) subseteq textzeroes of z^99-1 cap mathbbR$ ?
– Evinda
Sep 3 at 20:16
@Evinda $A$ is hermitian so $sigma(A) subseteq mathbbR$. The polynomial $z^99 - 1$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides $z^99 - 1$. Therefore $$sigma(A) = textzeroes of m_A subseteq textzeroes of z^99-1$$ Hence, $sigma(A)$ is contained in the intersection.
– mechanodroid
Sep 3 at 20:35
1
@Evinda Yes, precisely. Since $A$ is diagonalizable, there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^-1$. The diagonal matrix $D$ has eigenvalues of $A$ along the diagonal, which must all be $1$. Therefore $D = I$ so $A = PIP^-1 = PP^-1 = I$.
– mechanodroid
Sep 4 at 8:35
 |Â
show 2 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$A$ is hermitian so $sigma(A) subseteq mathbbR$. On the other hand, the polynomial $z^99 - 1$ annihilates $A$ so
$$sigma(A) subseteq textzeroes of z^99-1 cap mathbbR = lefte^ifrac2kpi99 : k in 0, ldots, 98right cap mathbbR = 1$$
so $sigma(A) = 1$.
Since $A$ is diagonalizable, it diagonalizes to the indentity matrix so $A = I$.
answered Sep 2 at 15:35
mechanodroid
24.1k52244
$A$ is hermitian so $sigma(A) subseteq mathbbR$. On the other hand, the polynomial $z^99 - 1$ annihilates $A$ so
$$sigma(A) subseteq textzeroes of z^99-1 cap mathbbR = lefte^ifrac2kpi99 : k in 0, ldots, 98right cap mathbbR = 1$$
so $sigma(A) = 1$.
Since $A$ is diagonalizable, it diagonalizes to the indentity matrix so $A = I$.
answered Sep 2 at 15:35
mechanodroid
24.1k52244
answered Sep 2 at 15:35
mechanodroid
24.1k52244
answered Sep 2 at 15:35
mechanodroid
24.1k52244
answered Sep 2 at 15:35
mechanodroid
24.1k52244
24.1k52244
Which is the $sigma$-function?
– Evinda
Sep 3 at 18:12
@Evinda $sigma(A)$ is the spectrum of $A$, i.e. the set of all eigenvalues of $A$.
– mechanodroid
Sep 3 at 19:20
And why does it hold that $sigma(A) subseteq textzeroes of z^99-1 cap mathbbR$ ?
– Evinda
Sep 3 at 20:16
@Evinda $A$ is hermitian so $sigma(A) subseteq mathbbR$. The polynomial $z^99 - 1$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides $z^99 - 1$. Therefore $$sigma(A) = textzeroes of m_A subseteq textzeroes of z^99-1$$ Hence, $sigma(A)$ is contained in the intersection.
– mechanodroid
Sep 3 at 20:35
1
@Evinda Yes, precisely. Since $A$ is diagonalizable, there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^-1$. The diagonal matrix $D$ has eigenvalues of $A$ along the diagonal, which must all be $1$. Therefore $D = I$ so $A = PIP^-1 = PP^-1 = I$.
– mechanodroid
Sep 4 at 8:35
 |Â
show 2 more comments
Which is the $sigma$-function?
– Evinda
Sep 3 at 18:12
@Evinda $sigma(A)$ is the spectrum of $A$, i.e. the set of all eigenvalues of $A$.
– mechanodroid
Sep 3 at 19:20
And why does it hold that $sigma(A) subseteq textzeroes of z^99-1 cap mathbbR$ ?
– Evinda
Sep 3 at 20:16
@Evinda $A$ is hermitian so $sigma(A) subseteq mathbbR$. The polynomial $z^99 - 1$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides $z^99 - 1$. Therefore $$sigma(A) = textzeroes of m_A subseteq textzeroes of z^99-1$$ Hence, $sigma(A)$ is contained in the intersection.
– mechanodroid
Sep 3 at 20:35
1
@Evinda Yes, precisely. Since $A$ is diagonalizable, there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^-1$. The diagonal matrix $D$ has eigenvalues of $A$ along the diagonal, which must all be $1$. Therefore $D = I$ so $A = PIP^-1 = PP^-1 = I$.
– mechanodroid
Sep 4 at 8:35
Which is the $sigma$-function?
– Evinda
Sep 3 at 18:12
Which is the $sigma$-function?
– Evinda
Sep 3 at 18:12
@Evinda $sigma(A)$ is the spectrum of $A$, i.e. the set of all eigenvalues of $A$.
– mechanodroid
Sep 3 at 19:20
@Evinda $sigma(A)$ is the spectrum of $A$, i.e. the set of all eigenvalues of $A$.
– mechanodroid
Sep 3 at 19:20
And why does it hold that $sigma(A) subseteq textzeroes of z^99-1 cap mathbbR$ ?
– Evinda
Sep 3 at 20:16
And why does it hold that $sigma(A) subseteq textzeroes of z^99-1 cap mathbbR$ ?
– Evinda
Sep 3 at 20:16
@Evinda $A$ is hermitian so $sigma(A) subseteq mathbbR$. The polynomial $z^99 - 1$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides $z^99 - 1$. Therefore $$sigma(A) = textzeroes of m_A subseteq textzeroes of z^99-1$$ Hence, $sigma(A)$ is contained in the intersection.
– mechanodroid
Sep 3 at 20:35
@Evinda $A$ is hermitian so $sigma(A) subseteq mathbbR$. The polynomial $z^99 - 1$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides $z^99 - 1$. Therefore $$sigma(A) = textzeroes of m_A subseteq textzeroes of z^99-1$$ Hence, $sigma(A)$ is contained in the intersection.
– mechanodroid
Sep 3 at 20:35
1
1
@Evinda Yes, precisely. Since $A$ is diagonalizable, there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^-1$. The diagonal matrix $D$ has eigenvalues of $A$ along the diagonal, which must all be $1$. Therefore $D = I$ so $A = PIP^-1 = PP^-1 = I$.
– mechanodroid
Sep 4 at 8:35
@Evinda Yes, precisely. Since $A$ is diagonalizable, there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^-1$. The diagonal matrix $D$ has eigenvalues of $A$ along the diagonal, which must all be $1$. Therefore $D = I$ so $A = PIP^-1 = PP^-1 = I$.
– mechanodroid
Sep 4 at 8:35
 |Â
show 2 more comments
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3
Hint: Since $A$ is hermitian, it can be diagonalized. It is easy to write down the entries of a power of a diagonal matrix. And remember that the eigenvalues of $A$ are real.
– Henning Makholm
Sep 2 at 12:42
So since $A$ is hermitian, there exists an invertible matrix $P$ such that $P^-1AP$ is a diagonal matrix. How does this help? @HenningMakholm
– Evinda
Sep 2 at 12:55
x @Evinda: Simplify $A^99 = (PDP^-1)^99$.
– Henning Makholm
Sep 2 at 13:00
1
$D$ contains the eigenvalues, right? So from $D^99=I_n$, do we get that $lambda^99=1$ for each eigenvalue $lambda$? Thus there is only one eigenvalue, $lambda=1$. Thus $D=I$ and so $A=PP^-1=I_n$. Am I right? @HenningMakholm
– Evinda
Sep 2 at 14:17
1
@Evinda Yes, but you must be sure first that the eigenvalues of $A$ are real; otherwise $lambda^99=1$ woul not imply $lambda=1$. This happens because $A$ is hermitian
– Jose Brox
Sep 3 at 18:02