Unitary Matrix and its product
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Let $A$ $in$ $mathbb{M_{3x3}(R)}$ be unitary. Prove that there exists $x$ $in mathbb{R^3}$ such that $A^2x$=x.
How should I proceed with this question? I just want a vague idea of how to do it.
linear-algebra matrices vector-spaces
asked Nov 15 at 9:54
Jimmy
14312
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Let $A$ $in$ $mathbb{M_{3x3}(R)}$ be unitary. Prove that there exists $x$ $in mathbb{R^3}$ such that $A^2x$=x.
How should I proceed with this question? I just want a vague idea of how to do it.
linear-algebra matrices vector-spaces
asked Nov 15 at 9:54
Jimmy
14312
1
$x=0$ is a trivial answer but you can also get nonzero $x$. Show that $1$ is an eigen value of $A^{2}$ by first looking at eigen values of $A$. A third degree polynomial with real coefficients cannot have all there roots non-real.
– Kavi Rama Murthy
Nov 15 at 9:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A$ $in$ $mathbb{M_{3x3}(R)}$ be unitary. Prove that there exists $x$ $in mathbb{R^3}$ such that $A^2x$=x.
How should I proceed with this question? I just want a vague idea of how to do it.
linear-algebra matrices vector-spaces
asked Nov 15 at 9:54
Jimmy
14312
Let $A$ $in$ $mathbb{M_{3x3}(R)}$ be unitary. Prove that there exists $x$ $in mathbb{R^3}$ such that $A^2x$=x.
How should I proceed with this question? I just want a vague idea of how to do it.
linear-algebra matrices vector-spaces
linear-algebra matrices vector-spaces
asked Nov 15 at 9:54
Jimmy
14312
asked Nov 15 at 9:54
Jimmy
14312
asked Nov 15 at 9:54
Jimmy
14312
asked Nov 15 at 9:54
Jimmy
14312
asked Nov 15 at 9:54
Jimmy
14312
14312
1
$x=0$ is a trivial answer but you can also get nonzero $x$. Show that $1$ is an eigen value of $A^{2}$ by first looking at eigen values of $A$. A third degree polynomial with real coefficients cannot have all there roots non-real.
– Kavi Rama Murthy
Nov 15 at 9:55
add a comment |
1
$x=0$ is a trivial answer but you can also get nonzero $x$. Show that $1$ is an eigen value of $A^{2}$ by first looking at eigen values of $A$. A third degree polynomial with real coefficients cannot have all there roots non-real.
– Kavi Rama Murthy
Nov 15 at 9:55
1
1
$x=0$ is a trivial answer but you can also get nonzero $x$. Show that $1$ is an eigen value of $A^{2}$ by first looking at eigen values of $A$. A third degree polynomial with real coefficients cannot have all there roots non-real.
– Kavi Rama Murthy
Nov 15 at 9:55
$x=0$ is a trivial answer but you can also get nonzero $x$. Show that $1$ is an eigen value of $A^{2}$ by first looking at eigen values of $A$. A third degree polynomial with real coefficients cannot have all there roots non-real.
– Kavi Rama Murthy
Nov 15 at 9:55
add a comment |
1 Answer
1
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up vote
4
down vote
accepted
Your problem is equivalent to showing that $A^{2}$ has 1 as an eigenvalue(assuming you are not actually satisfied with x=0 as a solution, cf the above comment).
In order to prove this you need to know/prove a few preliminary steps:
The eigenvalues of a unitary matrix are all complex numbers of unit modulus.
Since your matrix is real, its characteristic poly has real coefficients and degree 3. The roots of the char poly are the eigenvalues and since the coefficients of the poly are real, the roots come in pairs (a,a*) of complex conjugates.
Since the char poly has 3 roots they cannot all be complex non-real(since they come in pairs by 2.), just by counting. Therefore your matrix has a real eigenvalues (of modulus 1!). This eigenvalue can therefore only be +1 or -1. Eitherway, 1 is an eigenvalue of $A^{2}$
answered Nov 15 at 10:09
Sorin Tirc
61210
Can you help me by describing your thought process? I mean what you thought when you saw the problem and how you came to the conclusion that eigenvalues might be needed for this question?
– Jimmy
Nov 15 at 12:42
1
Hi @Jimmy. The definition of an eigenvalue $lambda$ for a matrix M over the complex numbers is: a complex number such that there exists a column vector x such that Mx = $lambda$x so as you can see your problem simply **is** a problem about the eigenvalues of $A^{2}$. x is called an eigenvector The rest of my thought process was merely remembering some facts about unitary matrices. Maybe you should brush up on some definitions and basic properties in linear algebra and matrices.
– Sorin Tirc
Nov 15 at 12:45
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Your problem is equivalent to showing that $A^{2}$ has 1 as an eigenvalue(assuming you are not actually satisfied with x=0 as a solution, cf the above comment).
In order to prove this you need to know/prove a few preliminary steps:
The eigenvalues of a unitary matrix are all complex numbers of unit modulus.
Since your matrix is real, its characteristic poly has real coefficients and degree 3. The roots of the char poly are the eigenvalues and since the coefficients of the poly are real, the roots come in pairs (a,a*) of complex conjugates.
Since the char poly has 3 roots they cannot all be complex non-real(since they come in pairs by 2.), just by counting. Therefore your matrix has a real eigenvalues (of modulus 1!). This eigenvalue can therefore only be +1 or -1. Eitherway, 1 is an eigenvalue of $A^{2}$
answered Nov 15 at 10:09
Sorin Tirc
61210
Can you help me by describing your thought process? I mean what you thought when you saw the problem and how you came to the conclusion that eigenvalues might be needed for this question?
– Jimmy
Nov 15 at 12:42
1
Hi @Jimmy. The definition of an eigenvalue $lambda$ for a matrix M over the complex numbers is: a complex number such that there exists a column vector x such that Mx = $lambda$x so as you can see your problem simply **is** a problem about the eigenvalues of $A^{2}$. x is called an eigenvector The rest of my thought process was merely remembering some facts about unitary matrices. Maybe you should brush up on some definitions and basic properties in linear algebra and matrices.
– Sorin Tirc
Nov 15 at 12:45
add a comment |
up vote
4
down vote
accepted
Your problem is equivalent to showing that $A^{2}$ has 1 as an eigenvalue(assuming you are not actually satisfied with x=0 as a solution, cf the above comment).
In order to prove this you need to know/prove a few preliminary steps:
The eigenvalues of a unitary matrix are all complex numbers of unit modulus.
Since your matrix is real, its characteristic poly has real coefficients and degree 3. The roots of the char poly are the eigenvalues and since the coefficients of the poly are real, the roots come in pairs (a,a*) of complex conjugates.
Since the char poly has 3 roots they cannot all be complex non-real(since they come in pairs by 2.), just by counting. Therefore your matrix has a real eigenvalues (of modulus 1!). This eigenvalue can therefore only be +1 or -1. Eitherway, 1 is an eigenvalue of $A^{2}$
answered Nov 15 at 10:09
Sorin Tirc
61210
Can you help me by describing your thought process? I mean what you thought when you saw the problem and how you came to the conclusion that eigenvalues might be needed for this question?
– Jimmy
Nov 15 at 12:42
1
Hi @Jimmy. The definition of an eigenvalue $lambda$ for a matrix M over the complex numbers is: a complex number such that there exists a column vector x such that Mx = $lambda$x so as you can see your problem simply **is** a problem about the eigenvalues of $A^{2}$. x is called an eigenvector The rest of my thought process was merely remembering some facts about unitary matrices. Maybe you should brush up on some definitions and basic properties in linear algebra and matrices.
– Sorin Tirc
Nov 15 at 12:45
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Your problem is equivalent to showing that $A^{2}$ has 1 as an eigenvalue(assuming you are not actually satisfied with x=0 as a solution, cf the above comment).
In order to prove this you need to know/prove a few preliminary steps:
The eigenvalues of a unitary matrix are all complex numbers of unit modulus.
Since your matrix is real, its characteristic poly has real coefficients and degree 3. The roots of the char poly are the eigenvalues and since the coefficients of the poly are real, the roots come in pairs (a,a*) of complex conjugates.
Since the char poly has 3 roots they cannot all be complex non-real(since they come in pairs by 2.), just by counting. Therefore your matrix has a real eigenvalues (of modulus 1!). This eigenvalue can therefore only be +1 or -1. Eitherway, 1 is an eigenvalue of $A^{2}$
answered Nov 15 at 10:09
Sorin Tirc
61210
Your problem is equivalent to showing that $A^{2}$ has 1 as an eigenvalue(assuming you are not actually satisfied with x=0 as a solution, cf the above comment).
In order to prove this you need to know/prove a few preliminary steps:
The eigenvalues of a unitary matrix are all complex numbers of unit modulus.
Since your matrix is real, its characteristic poly has real coefficients and degree 3. The roots of the char poly are the eigenvalues and since the coefficients of the poly are real, the roots come in pairs (a,a*) of complex conjugates.
Since the char poly has 3 roots they cannot all be complex non-real(since they come in pairs by 2.), just by counting. Therefore your matrix has a real eigenvalues (of modulus 1!). This eigenvalue can therefore only be +1 or -1. Eitherway, 1 is an eigenvalue of $A^{2}$
answered Nov 15 at 10:09
Sorin Tirc
61210
edited Nov 15 at 10:28
answered Nov 15 at 10:09
Sorin Tirc
61210
answered Nov 15 at 10:09
Sorin Tirc
61210
answered Nov 15 at 10:09
Sorin Tirc
61210
61210
Can you help me by describing your thought process? I mean what you thought when you saw the problem and how you came to the conclusion that eigenvalues might be needed for this question?
– Jimmy
Nov 15 at 12:42
1
Hi @Jimmy. The definition of an eigenvalue $lambda$ for a matrix M over the complex numbers is: a complex number such that there exists a column vector x such that Mx = $lambda$x so as you can see your problem simply **is** a problem about the eigenvalues of $A^{2}$. x is called an eigenvector The rest of my thought process was merely remembering some facts about unitary matrices. Maybe you should brush up on some definitions and basic properties in linear algebra and matrices.
– Sorin Tirc
Nov 15 at 12:45
add a comment |
Can you help me by describing your thought process? I mean what you thought when you saw the problem and how you came to the conclusion that eigenvalues might be needed for this question?
– Jimmy
Nov 15 at 12:42
1
Hi @Jimmy. The definition of an eigenvalue $lambda$ for a matrix M over the complex numbers is: a complex number such that there exists a column vector x such that Mx = $lambda$x so as you can see your problem simply **is** a problem about the eigenvalues of $A^{2}$. x is called an eigenvector The rest of my thought process was merely remembering some facts about unitary matrices. Maybe you should brush up on some definitions and basic properties in linear algebra and matrices.
– Sorin Tirc
Nov 15 at 12:45
Can you help me by describing your thought process? I mean what you thought when you saw the problem and how you came to the conclusion that eigenvalues might be needed for this question?
– Jimmy
Nov 15 at 12:42
Can you help me by describing your thought process? I mean what you thought when you saw the problem and how you came to the conclusion that eigenvalues might be needed for this question?
– Jimmy
Nov 15 at 12:42
1
1
Hi @Jimmy. The definition of an eigenvalue $lambda$ for a matrix M over the complex numbers is: a complex number such that there exists a column vector x such that Mx = $lambda$x so as you can see your problem simply **is** a problem about the eigenvalues of $A^{2}$. x is called an eigenvector The rest of my thought process was merely remembering some facts about unitary matrices. Maybe you should brush up on some definitions and basic properties in linear algebra and matrices.
– Sorin Tirc
Nov 15 at 12:45
Hi @Jimmy. The definition of an eigenvalue $lambda$ for a matrix M over the complex numbers is: a complex number such that there exists a column vector x such that Mx = $lambda$x so as you can see your problem simply **is** a problem about the eigenvalues of $A^{2}$. x is called an eigenvector The rest of my thought process was merely remembering some facts about unitary matrices. Maybe you should brush up on some definitions and basic properties in linear algebra and matrices.
– Sorin Tirc
Nov 15 at 12:45
add a comment |
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$x=0$ is a trivial answer but you can also get nonzero $x$. Show that $1$ is an eigen value of $A^{2}$ by first looking at eigen values of $A$. A third degree polynomial with real coefficients cannot have all there roots non-real.
– Kavi Rama Murthy
Nov 15 at 9:55