Spectral theorem over $mathbb {R}$
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It is clear to me that a self-adjoint linear map on a real/complex inner product space has real eigenvalues, and its eigenspaces for distinct eigenvalues are orthogonal.
In order to prove it is diagonalisable, we also need that the geometric multiplicity of the eigenspaces equals the algebraic multiplicity. How is this done?
In particular, if the arguments differ, how is this demonstrated over $mathbb{R}$? I have seen proofs over $mathbb {C}$, and they use algebraic closure, so I imagine the real case is more involved, and would like to see a proof that works here.
linear-algebra diagonalization
asked Dec 5 at 18:21
Joshua Tilley
539313
add a comment |
up vote
2
down vote
favorite
It is clear to me that a self-adjoint linear map on a real/complex inner product space has real eigenvalues, and its eigenspaces for distinct eigenvalues are orthogonal.
In order to prove it is diagonalisable, we also need that the geometric multiplicity of the eigenspaces equals the algebraic multiplicity. How is this done?
In particular, if the arguments differ, how is this demonstrated over $mathbb{R}$? I have seen proofs over $mathbb {C}$, and they use algebraic closure, so I imagine the real case is more involved, and would like to see a proof that works here.
linear-algebra diagonalization
asked Dec 5 at 18:21
Joshua Tilley
539313
I can see that you would be able to pass to the complexification of your space and use spectral theorem from $mathbb {C}$. Follow ups would then be, prove the eigenvectors are (can be taken as) real. Furthermore, if there is a way to do this natively (without algebraic closure), and it isn't too painful, I'd like to see that.
– Joshua Tilley
Dec 5 at 18:34
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
It is clear to me that a self-adjoint linear map on a real/complex inner product space has real eigenvalues, and its eigenspaces for distinct eigenvalues are orthogonal.
In order to prove it is diagonalisable, we also need that the geometric multiplicity of the eigenspaces equals the algebraic multiplicity. How is this done?
In particular, if the arguments differ, how is this demonstrated over $mathbb{R}$? I have seen proofs over $mathbb {C}$, and they use algebraic closure, so I imagine the real case is more involved, and would like to see a proof that works here.
linear-algebra diagonalization
asked Dec 5 at 18:21
Joshua Tilley
539313
It is clear to me that a self-adjoint linear map on a real/complex inner product space has real eigenvalues, and its eigenspaces for distinct eigenvalues are orthogonal.
In order to prove it is diagonalisable, we also need that the geometric multiplicity of the eigenspaces equals the algebraic multiplicity. How is this done?
In particular, if the arguments differ, how is this demonstrated over $mathbb{R}$? I have seen proofs over $mathbb {C}$, and they use algebraic closure, so I imagine the real case is more involved, and would like to see a proof that works here.
linear-algebra diagonalization
linear-algebra diagonalization
asked Dec 5 at 18:21
Joshua Tilley
539313
asked Dec 5 at 18:21
Joshua Tilley
539313
edited Dec 5 at 18:27
asked Dec 5 at 18:21
Joshua Tilley
539313
asked Dec 5 at 18:21
Joshua Tilley
539313
asked Dec 5 at 18:21
Joshua Tilley
539313
539313
I can see that you would be able to pass to the complexification of your space and use spectral theorem from $mathbb {C}$. Follow ups would then be, prove the eigenvectors are (can be taken as) real. Furthermore, if there is a way to do this natively (without algebraic closure), and it isn't too painful, I'd like to see that.
– Joshua Tilley
Dec 5 at 18:34
add a comment |
I can see that you would be able to pass to the complexification of your space and use spectral theorem from $mathbb {C}$. Follow ups would then be, prove the eigenvectors are (can be taken as) real. Furthermore, if there is a way to do this natively (without algebraic closure), and it isn't too painful, I'd like to see that.
– Joshua Tilley
Dec 5 at 18:34
I can see that you would be able to pass to the complexification of your space and use spectral theorem from $mathbb {C}$. Follow ups would then be, prove the eigenvectors are (can be taken as) real. Furthermore, if there is a way to do this natively (without algebraic closure), and it isn't too painful, I'd like to see that.
– Joshua Tilley
Dec 5 at 18:34
I can see that you would be able to pass to the complexification of your space and use spectral theorem from $mathbb {C}$. Follow ups would then be, prove the eigenvectors are (can be taken as) real. Furthermore, if there is a way to do this natively (without algebraic closure), and it isn't too painful, I'd like to see that.
– Joshua Tilley
Dec 5 at 18:34
add a comment |
1 Answer
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By induction. Let $T$ be a real self-adjoint endomorphism of a real IPS, $V$. Let $lambda$ be an eigenvalue of $T$ since $T$ is self-adjoint, $T$ has a real eigenvalue. Then by definition, there is a $lambda$-eigenvector $v$ for $T$. Let $V'={v}^perp$. Then since $T$ is self-adjoint, if $win V'$, then $langle Tw,vrangle = langle w,Tvrangle=langle w,lambda vrangle= 0$. Thus $V'$ is $T$-invariant. Hence $V$ splits as a direct sum $Bbb{R}voplus V'$ and $T$ splits with it as $T=lambda oplus T|_{V'}$. Then by induction $T|_{V'}$ has an orthonormal basis of eigenvectors, and together with $v$, these give an orthonormal basis for $V$ consisting of eigenvectors of $T$.
This argument doesn't depend on the base field being $Bbb{R}$ or $Bbb{C}$, but I assumed it was $Bbb{R}$ to make the writing simpler.
answered Dec 5 at 18:33
jgon
11.9k21840
Nice. As a note, the claim that $T$ has a real eigenvalue follows from passing to the complexification of $V$, using algebraic closure of $mathbb{C}$, and then using the fact that there its eigenvalues are real, and that complexifying doesn't change the eigenvalues, yes?
– Joshua Tilley
Dec 5 at 18:37
@JoshuaTilley Oh yes, sorry.
– jgon
Dec 5 at 18:40
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
By induction. Let $T$ be a real self-adjoint endomorphism of a real IPS, $V$. Let $lambda$ be an eigenvalue of $T$ since $T$ is self-adjoint, $T$ has a real eigenvalue. Then by definition, there is a $lambda$-eigenvector $v$ for $T$. Let $V'={v}^perp$. Then since $T$ is self-adjoint, if $win V'$, then $langle Tw,vrangle = langle w,Tvrangle=langle w,lambda vrangle= 0$. Thus $V'$ is $T$-invariant. Hence $V$ splits as a direct sum $Bbb{R}voplus V'$ and $T$ splits with it as $T=lambda oplus T|_{V'}$. Then by induction $T|_{V'}$ has an orthonormal basis of eigenvectors, and together with $v$, these give an orthonormal basis for $V$ consisting of eigenvectors of $T$.
This argument doesn't depend on the base field being $Bbb{R}$ or $Bbb{C}$, but I assumed it was $Bbb{R}$ to make the writing simpler.
answered Dec 5 at 18:33
jgon
11.9k21840
Nice. As a note, the claim that $T$ has a real eigenvalue follows from passing to the complexification of $V$, using algebraic closure of $mathbb{C}$, and then using the fact that there its eigenvalues are real, and that complexifying doesn't change the eigenvalues, yes?
– Joshua Tilley
Dec 5 at 18:37
@JoshuaTilley Oh yes, sorry.
– jgon
Dec 5 at 18:40
add a comment |
up vote
2
down vote
accepted
By induction. Let $T$ be a real self-adjoint endomorphism of a real IPS, $V$. Let $lambda$ be an eigenvalue of $T$ since $T$ is self-adjoint, $T$ has a real eigenvalue. Then by definition, there is a $lambda$-eigenvector $v$ for $T$. Let $V'={v}^perp$. Then since $T$ is self-adjoint, if $win V'$, then $langle Tw,vrangle = langle w,Tvrangle=langle w,lambda vrangle= 0$. Thus $V'$ is $T$-invariant. Hence $V$ splits as a direct sum $Bbb{R}voplus V'$ and $T$ splits with it as $T=lambda oplus T|_{V'}$. Then by induction $T|_{V'}$ has an orthonormal basis of eigenvectors, and together with $v$, these give an orthonormal basis for $V$ consisting of eigenvectors of $T$.
This argument doesn't depend on the base field being $Bbb{R}$ or $Bbb{C}$, but I assumed it was $Bbb{R}$ to make the writing simpler.
answered Dec 5 at 18:33
jgon
11.9k21840
Nice. As a note, the claim that $T$ has a real eigenvalue follows from passing to the complexification of $V$, using algebraic closure of $mathbb{C}$, and then using the fact that there its eigenvalues are real, and that complexifying doesn't change the eigenvalues, yes?
– Joshua Tilley
Dec 5 at 18:37
@JoshuaTilley Oh yes, sorry.
– jgon
Dec 5 at 18:40
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
By induction. Let $T$ be a real self-adjoint endomorphism of a real IPS, $V$. Let $lambda$ be an eigenvalue of $T$ since $T$ is self-adjoint, $T$ has a real eigenvalue. Then by definition, there is a $lambda$-eigenvector $v$ for $T$. Let $V'={v}^perp$. Then since $T$ is self-adjoint, if $win V'$, then $langle Tw,vrangle = langle w,Tvrangle=langle w,lambda vrangle= 0$. Thus $V'$ is $T$-invariant. Hence $V$ splits as a direct sum $Bbb{R}voplus V'$ and $T$ splits with it as $T=lambda oplus T|_{V'}$. Then by induction $T|_{V'}$ has an orthonormal basis of eigenvectors, and together with $v$, these give an orthonormal basis for $V$ consisting of eigenvectors of $T$.
This argument doesn't depend on the base field being $Bbb{R}$ or $Bbb{C}$, but I assumed it was $Bbb{R}$ to make the writing simpler.
answered Dec 5 at 18:33
jgon
11.9k21840
By induction. Let $T$ be a real self-adjoint endomorphism of a real IPS, $V$. Let $lambda$ be an eigenvalue of $T$ since $T$ is self-adjoint, $T$ has a real eigenvalue. Then by definition, there is a $lambda$-eigenvector $v$ for $T$. Let $V'={v}^perp$. Then since $T$ is self-adjoint, if $win V'$, then $langle Tw,vrangle = langle w,Tvrangle=langle w,lambda vrangle= 0$. Thus $V'$ is $T$-invariant. Hence $V$ splits as a direct sum $Bbb{R}voplus V'$ and $T$ splits with it as $T=lambda oplus T|_{V'}$. Then by induction $T|_{V'}$ has an orthonormal basis of eigenvectors, and together with $v$, these give an orthonormal basis for $V$ consisting of eigenvectors of $T$.
This argument doesn't depend on the base field being $Bbb{R}$ or $Bbb{C}$, but I assumed it was $Bbb{R}$ to make the writing simpler.
answered Dec 5 at 18:33
jgon
11.9k21840
answered Dec 5 at 18:33
jgon
11.9k21840
answered Dec 5 at 18:33
jgon
11.9k21840
answered Dec 5 at 18:33
jgon
11.9k21840
11.9k21840
Nice. As a note, the claim that $T$ has a real eigenvalue follows from passing to the complexification of $V$, using algebraic closure of $mathbb{C}$, and then using the fact that there its eigenvalues are real, and that complexifying doesn't change the eigenvalues, yes?
– Joshua Tilley
Dec 5 at 18:37
@JoshuaTilley Oh yes, sorry.
– jgon
Dec 5 at 18:40
add a comment |
Nice. As a note, the claim that $T$ has a real eigenvalue follows from passing to the complexification of $V$, using algebraic closure of $mathbb{C}$, and then using the fact that there its eigenvalues are real, and that complexifying doesn't change the eigenvalues, yes?
– Joshua Tilley
Dec 5 at 18:37
@JoshuaTilley Oh yes, sorry.
– jgon
Dec 5 at 18:40
Nice. As a note, the claim that $T$ has a real eigenvalue follows from passing to the complexification of $V$, using algebraic closure of $mathbb{C}$, and then using the fact that there its eigenvalues are real, and that complexifying doesn't change the eigenvalues, yes?
– Joshua Tilley
Dec 5 at 18:37
Nice. As a note, the claim that $T$ has a real eigenvalue follows from passing to the complexification of $V$, using algebraic closure of $mathbb{C}$, and then using the fact that there its eigenvalues are real, and that complexifying doesn't change the eigenvalues, yes?
– Joshua Tilley
Dec 5 at 18:37
@JoshuaTilley Oh yes, sorry.
– jgon
Dec 5 at 18:40
@JoshuaTilley Oh yes, sorry.
– jgon
Dec 5 at 18:40
add a comment |
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I can see that you would be able to pass to the complexification of your space and use spectral theorem from $mathbb {C}$. Follow ups would then be, prove the eigenvectors are (can be taken as) real. Furthermore, if there is a way to do this natively (without algebraic closure), and it isn't too painful, I'd like to see that.
– Joshua Tilley
Dec 5 at 18:34