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Orthogonal and unitary operators

Why aren't all real self-adjoint operators diagonal?Self - adjoint and Unitary operatorOrthogonal Projectors and OperatorsProof that Every Positive Operator on V has a Unique Positive Square RootAre self-adjoint / Hermitian operators necessarily orthogonal / unitary?self-adjoint / orthonormal basis of eigenvectors$U,T$ are self-adjoint with $TU=UT$, then there is orthonormal basis of eigenvectors of $T$ and $U$.Unitary and Orthogonal OperatorsFind a “canonical form” for the linear operators that are both self-adjoint and unitary in a finite-dimensional complex inner product space.Proving a linear involution on a finite dimensional inner product space $V$ over $mathbb{C}$ is a self adjoint operator.

0

$begingroup$

1) Theorem:

Let $T$ be a linear operator on a finite-dimensional real inner product space $V$. Then $V$ has an orthonormal basis of eigenvectors of $T$ with corresponding eigenvalues of absolute value 1 if and only if $T$ is both self adjoint and orthogonal.

2) Theorem:

Let $T$ be a linear operator on a finite-dimensional complex inner product space $V$. Then $V$ has an orthonormal basis of eigenvectors with absolute value 1 if and only if $T$ is unitary.

So my question is why doesnt a unitary operator have to be self-adjoint to have orthonormal basis of eigenvectors with corresponding eigenvalues of absolute value 1 but orthogonal operators have to?

linear-algebra

share|cite|improve this question

edited Mar 14 at 20:41

Alan Muniz

2,61311030

asked Mar 14 at 20:30

StudentStudent

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Orthogonal operators could still be missing eigenvectors since they may have complex eigenvalues, for example a 90 degree rotation of a plane.
  $endgroup$
  – Joppy
  Mar 14 at 21:13
  

  
  
  
  

add a comment | 

0

$begingroup$

1) Theorem:

Let $T$ be a linear operator on a finite-dimensional real inner product space $V$. Then $V$ has an orthonormal basis of eigenvectors of $T$ with corresponding eigenvalues of absolute value 1 if and only if $T$ is both self adjoint and orthogonal.

2) Theorem:

Let $T$ be a linear operator on a finite-dimensional complex inner product space $V$. Then $V$ has an orthonormal basis of eigenvectors with absolute value 1 if and only if $T$ is unitary.

So my question is why doesnt a unitary operator have to be self-adjoint to have orthonormal basis of eigenvectors with corresponding eigenvalues of absolute value 1 but orthogonal operators have to?

linear-algebra

share|cite|improve this question

edited Mar 14 at 20:41

Alan Muniz

2,61311030

asked Mar 14 at 20:30

StudentStudent

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Orthogonal operators could still be missing eigenvectors since they may have complex eigenvalues, for example a 90 degree rotation of a plane.
  $endgroup$
  – Joppy
  Mar 14 at 21:13
  

  
  
  
  

add a comment | 

$begingroup$

1) Theorem:

Let $T$ be a linear operator on a finite-dimensional real inner product space $V$. Then $V$ has an orthonormal basis of eigenvectors of $T$ with corresponding eigenvalues of absolute value 1 if and only if $T$ is both self adjoint and orthogonal.

2) Theorem:

Let $T$ be a linear operator on a finite-dimensional complex inner product space $V$. Then $V$ has an orthonormal basis of eigenvectors with absolute value 1 if and only if $T$ is unitary.

So my question is why doesnt a unitary operator have to be self-adjoint to have orthonormal basis of eigenvectors with corresponding eigenvalues of absolute value 1 but orthogonal operators have to?

linear-algebra

share|cite|improve this question

edited Mar 14 at 20:41

Alan Muniz

2,61311030

asked Mar 14 at 20:30

StudentStudent

1

$endgroup$

share|cite|improve this question

edited Mar 14 at 20:41

Alan Muniz

2,61311030

asked Mar 14 at 20:30

StudentStudent

1

share|cite|improve this question

edited Mar 14 at 20:41

Alan Muniz

2,61311030

asked Mar 14 at 20:30

StudentStudent

1

edited Mar 14 at 20:41

Alan Muniz

2,61311030

edited Mar 14 at 20:41

Alan Muniz

2,61311030

asked Mar 14 at 20:30

StudentStudent

1

asked Mar 14 at 20:30

StudentStudent

1