complex and real spectral theorem for matrices
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I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $mathbbR$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices.
Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $mathbbC$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real?
Thanks in advance.
matrices eigenvalues-eigenvectors spectral-theory
asked Aug 9 at 22:41
Charles
533420
add a comment |Â
up vote
0
down vote
favorite
I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $mathbbR$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices.
Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $mathbbC$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real?
Thanks in advance.
matrices eigenvalues-eigenvectors spectral-theory
asked Aug 9 at 22:41
Charles
533420
$lambda$ is an eigenvalue iff $ker (A-lambda I)$ is non trivial. If$A-lambda I$ is real, then the kernel contains a purely real element.
– copper.hat
Aug 9 at 23:00
"IfA−λI is real, then the kernel contains a purely real element." this is precisely what I want to prove.
– Charles
Aug 9 at 23:11
If $B$ is real and $B (u+iv) = 0$ where $u,v$ are real, then $Bu = 0 $ **and** $Bv=0$, so you always have a real eigenvector (assuming that $U+iv =neq 0$, of course).
– copper.hat
Aug 9 at 23:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $mathbbR$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices.
Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $mathbbC$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real?
Thanks in advance.
matrices eigenvalues-eigenvectors spectral-theory
asked Aug 9 at 22:41
Charles
533420
I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $mathbbR$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices.
Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $mathbbC$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real?
Thanks in advance.
matrices eigenvalues-eigenvectors spectral-theory
asked Aug 9 at 22:41
Charles
533420
asked Aug 9 at 22:41
Charles
533420
asked Aug 9 at 22:41
Charles
533420
asked Aug 9 at 22:41
Charles
533420
533420
$lambda$ is an eigenvalue iff $ker (A-lambda I)$ is non trivial. If$A-lambda I$ is real, then the kernel contains a purely real element.
– copper.hat
Aug 9 at 23:00
"IfA−λI is real, then the kernel contains a purely real element." this is precisely what I want to prove.
– Charles
Aug 9 at 23:11
If $B$ is real and $B (u+iv) = 0$ where $u,v$ are real, then $Bu = 0 $ **and** $Bv=0$, so you always have a real eigenvector (assuming that $U+iv =neq 0$, of course).
– copper.hat
Aug 9 at 23:12
add a comment |Â
$lambda$ is an eigenvalue iff $ker (A-lambda I)$ is non trivial. If$A-lambda I$ is real, then the kernel contains a purely real element.
– copper.hat
Aug 9 at 23:00
"IfA−λI is real, then the kernel contains a purely real element." this is precisely what I want to prove.
– Charles
Aug 9 at 23:11
If $B$ is real and $B (u+iv) = 0$ where $u,v$ are real, then $Bu = 0 $ **and** $Bv=0$, so you always have a real eigenvector (assuming that $U+iv =neq 0$, of course).
– copper.hat
Aug 9 at 23:12
$lambda$ is an eigenvalue iff $ker (A-lambda I)$ is non trivial. If$A-lambda I$ is real, then the kernel contains a purely real element.
– copper.hat
Aug 9 at 23:00
$lambda$ is an eigenvalue iff $ker (A-lambda I)$ is non trivial. If$A-lambda I$ is real, then the kernel contains a purely real element.
– copper.hat
Aug 9 at 23:00
"IfA−λI is real, then the kernel contains a purely real element." this is precisely what I want to prove.
– Charles
Aug 9 at 23:11
"IfA−λI is real, then the kernel contains a purely real element." this is precisely what I want to prove.
– Charles
Aug 9 at 23:11
If $B$ is real and $B (u+iv) = 0$ where $u,v$ are real, then $Bu = 0 $ **and** $Bv=0$, so you always have a real eigenvector (assuming that $U+iv =neq 0$, of course).
– copper.hat
Aug 9 at 23:12
If $B$ is real and $B (u+iv) = 0$ where $u,v$ are real, then $Bu = 0 $ **and** $Bv=0$, so you always have a real eigenvector (assuming that $U+iv =neq 0$, of course).
– copper.hat
Aug 9 at 23:12
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Suppose $v$ is a complex eigenvector of $A$, a real symmetric matrix, with corresponding real eigenvalue $lambda$. Note that
beginalign
lambda overlinev=overlinelambda v = overlineAv= Aoverlinev
endalign
then $overlinev$ is also an eigenvector. Hence $v+overlinev$ is a real eigenvector. So pick this one.
answered Aug 9 at 22:48
Jacky Chong
16.1k2925
great! Can I also see from your proof that these new real eigenvectors are orthogonal with each other?
– Charles
Aug 9 at 23:06
From the complex spectral theorem, you can find $v_1, v_2, ldots, v_n$ linearly independent complex eigenvectors. Take the real component of each $v_i$.
– Jacky Chong
Aug 9 at 23:08
add a comment |Â
up vote
0
down vote
Because if $lambda$ is an eigenvalue of $A$ and $lambdainmathbb R$, then $A$ has an eigenvector in $mathbbR^n$ whose corresponding eigenvalue is $lambda$. In fact, asserting that $A$ has a real eigenvector corresponding to $lambda$ is equivalent to asserting that $lambda$ is a real root of the characteristic polynomial of $A$.
answered Aug 9 at 22:50
José Carlos Santos
115k1699177
I can't see why your first claim is true here. Can you explain it?
– Charles
Aug 9 at 23:07
@Charles THat's what I explain in the second sentence. If $f$ a linear map from $mathbbR^n$ into itself, how can you determine whether it has a real eigenvector? You can do it by proving that the characteristic polynomial $P(x)$ of $f$ has a real root $lambda$, which will be then an eigenvalue of $f$. Then (and only in that case) $f$ will have a real eigenvector corresponding to $lambda$.
– José Carlos Santos
Aug 9 at 23:12
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Suppose $v$ is a complex eigenvector of $A$, a real symmetric matrix, with corresponding real eigenvalue $lambda$. Note that
beginalign
lambda overlinev=overlinelambda v = overlineAv= Aoverlinev
endalign
then $overlinev$ is also an eigenvector. Hence $v+overlinev$ is a real eigenvector. So pick this one.
answered Aug 9 at 22:48
Jacky Chong
16.1k2925
great! Can I also see from your proof that these new real eigenvectors are orthogonal with each other?
– Charles
Aug 9 at 23:06
From the complex spectral theorem, you can find $v_1, v_2, ldots, v_n$ linearly independent complex eigenvectors. Take the real component of each $v_i$.
– Jacky Chong
Aug 9 at 23:08
add a comment |Â
up vote
2
down vote
accepted
Suppose $v$ is a complex eigenvector of $A$, a real symmetric matrix, with corresponding real eigenvalue $lambda$. Note that
beginalign
lambda overlinev=overlinelambda v = overlineAv= Aoverlinev
endalign
then $overlinev$ is also an eigenvector. Hence $v+overlinev$ is a real eigenvector. So pick this one.
answered Aug 9 at 22:48
Jacky Chong
16.1k2925
great! Can I also see from your proof that these new real eigenvectors are orthogonal with each other?
– Charles
Aug 9 at 23:06
From the complex spectral theorem, you can find $v_1, v_2, ldots, v_n$ linearly independent complex eigenvectors. Take the real component of each $v_i$.
– Jacky Chong
Aug 9 at 23:08
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose $v$ is a complex eigenvector of $A$, a real symmetric matrix, with corresponding real eigenvalue $lambda$. Note that
beginalign
lambda overlinev=overlinelambda v = overlineAv= Aoverlinev
endalign
then $overlinev$ is also an eigenvector. Hence $v+overlinev$ is a real eigenvector. So pick this one.
answered Aug 9 at 22:48
Jacky Chong
16.1k2925
Suppose $v$ is a complex eigenvector of $A$, a real symmetric matrix, with corresponding real eigenvalue $lambda$. Note that
beginalign
lambda overlinev=overlinelambda v = overlineAv= Aoverlinev
endalign
then $overlinev$ is also an eigenvector. Hence $v+overlinev$ is a real eigenvector. So pick this one.
answered Aug 9 at 22:48
Jacky Chong
16.1k2925
answered Aug 9 at 22:48
Jacky Chong
16.1k2925
answered Aug 9 at 22:48
Jacky Chong
16.1k2925
answered Aug 9 at 22:48
Jacky Chong
16.1k2925
16.1k2925
great! Can I also see from your proof that these new real eigenvectors are orthogonal with each other?
– Charles
Aug 9 at 23:06
From the complex spectral theorem, you can find $v_1, v_2, ldots, v_n$ linearly independent complex eigenvectors. Take the real component of each $v_i$.
– Jacky Chong
Aug 9 at 23:08
add a comment |Â
great! Can I also see from your proof that these new real eigenvectors are orthogonal with each other?
– Charles
Aug 9 at 23:06
From the complex spectral theorem, you can find $v_1, v_2, ldots, v_n$ linearly independent complex eigenvectors. Take the real component of each $v_i$.
– Jacky Chong
Aug 9 at 23:08
great! Can I also see from your proof that these new real eigenvectors are orthogonal with each other?
– Charles
Aug 9 at 23:06
great! Can I also see from your proof that these new real eigenvectors are orthogonal with each other?
– Charles
Aug 9 at 23:06
From the complex spectral theorem, you can find $v_1, v_2, ldots, v_n$ linearly independent complex eigenvectors. Take the real component of each $v_i$.
– Jacky Chong
Aug 9 at 23:08
From the complex spectral theorem, you can find $v_1, v_2, ldots, v_n$ linearly independent complex eigenvectors. Take the real component of each $v_i$.
– Jacky Chong
Aug 9 at 23:08
add a comment |Â
up vote
0
down vote
Because if $lambda$ is an eigenvalue of $A$ and $lambdainmathbb R$, then $A$ has an eigenvector in $mathbbR^n$ whose corresponding eigenvalue is $lambda$. In fact, asserting that $A$ has a real eigenvector corresponding to $lambda$ is equivalent to asserting that $lambda$ is a real root of the characteristic polynomial of $A$.
answered Aug 9 at 22:50
José Carlos Santos
115k1699177
I can't see why your first claim is true here. Can you explain it?
– Charles
Aug 9 at 23:07
@Charles THat's what I explain in the second sentence. If $f$ a linear map from $mathbbR^n$ into itself, how can you determine whether it has a real eigenvector? You can do it by proving that the characteristic polynomial $P(x)$ of $f$ has a real root $lambda$, which will be then an eigenvalue of $f$. Then (and only in that case) $f$ will have a real eigenvector corresponding to $lambda$.
– José Carlos Santos
Aug 9 at 23:12
add a comment |Â
up vote
0
down vote
Because if $lambda$ is an eigenvalue of $A$ and $lambdainmathbb R$, then $A$ has an eigenvector in $mathbbR^n$ whose corresponding eigenvalue is $lambda$. In fact, asserting that $A$ has a real eigenvector corresponding to $lambda$ is equivalent to asserting that $lambda$ is a real root of the characteristic polynomial of $A$.
answered Aug 9 at 22:50
José Carlos Santos
115k1699177
I can't see why your first claim is true here. Can you explain it?
– Charles
Aug 9 at 23:07
@Charles THat's what I explain in the second sentence. If $f$ a linear map from $mathbbR^n$ into itself, how can you determine whether it has a real eigenvector? You can do it by proving that the characteristic polynomial $P(x)$ of $f$ has a real root $lambda$, which will be then an eigenvalue of $f$. Then (and only in that case) $f$ will have a real eigenvector corresponding to $lambda$.
– José Carlos Santos
Aug 9 at 23:12
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Because if $lambda$ is an eigenvalue of $A$ and $lambdainmathbb R$, then $A$ has an eigenvector in $mathbbR^n$ whose corresponding eigenvalue is $lambda$. In fact, asserting that $A$ has a real eigenvector corresponding to $lambda$ is equivalent to asserting that $lambda$ is a real root of the characteristic polynomial of $A$.
answered Aug 9 at 22:50
José Carlos Santos
115k1699177
Because if $lambda$ is an eigenvalue of $A$ and $lambdainmathbb R$, then $A$ has an eigenvector in $mathbbR^n$ whose corresponding eigenvalue is $lambda$. In fact, asserting that $A$ has a real eigenvector corresponding to $lambda$ is equivalent to asserting that $lambda$ is a real root of the characteristic polynomial of $A$.
answered Aug 9 at 22:50
José Carlos Santos
115k1699177
answered Aug 9 at 22:50
José Carlos Santos
115k1699177
answered Aug 9 at 22:50
José Carlos Santos
115k1699177
answered Aug 9 at 22:50
José Carlos Santos
115k1699177
115k1699177
I can't see why your first claim is true here. Can you explain it?
– Charles
Aug 9 at 23:07
@Charles THat's what I explain in the second sentence. If $f$ a linear map from $mathbbR^n$ into itself, how can you determine whether it has a real eigenvector? You can do it by proving that the characteristic polynomial $P(x)$ of $f$ has a real root $lambda$, which will be then an eigenvalue of $f$. Then (and only in that case) $f$ will have a real eigenvector corresponding to $lambda$.
– José Carlos Santos
Aug 9 at 23:12
add a comment |Â
I can't see why your first claim is true here. Can you explain it?
– Charles
Aug 9 at 23:07
@Charles THat's what I explain in the second sentence. If $f$ a linear map from $mathbbR^n$ into itself, how can you determine whether it has a real eigenvector? You can do it by proving that the characteristic polynomial $P(x)$ of $f$ has a real root $lambda$, which will be then an eigenvalue of $f$. Then (and only in that case) $f$ will have a real eigenvector corresponding to $lambda$.
– José Carlos Santos
Aug 9 at 23:12
I can't see why your first claim is true here. Can you explain it?
– Charles
Aug 9 at 23:07
I can't see why your first claim is true here. Can you explain it?
– Charles
Aug 9 at 23:07
@Charles THat's what I explain in the second sentence. If $f$ a linear map from $mathbbR^n$ into itself, how can you determine whether it has a real eigenvector? You can do it by proving that the characteristic polynomial $P(x)$ of $f$ has a real root $lambda$, which will be then an eigenvalue of $f$. Then (and only in that case) $f$ will have a real eigenvector corresponding to $lambda$.
– José Carlos Santos
Aug 9 at 23:12
@Charles THat's what I explain in the second sentence. If $f$ a linear map from $mathbbR^n$ into itself, how can you determine whether it has a real eigenvector? You can do it by proving that the characteristic polynomial $P(x)$ of $f$ has a real root $lambda$, which will be then an eigenvalue of $f$. Then (and only in that case) $f$ will have a real eigenvector corresponding to $lambda$.
– José Carlos Santos
Aug 9 at 23:12
add a comment |Â
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$lambda$ is an eigenvalue iff $ker (A-lambda I)$ is non trivial. If$A-lambda I$ is real, then the kernel contains a purely real element.
– copper.hat
Aug 9 at 23:00
"IfA−λI is real, then the kernel contains a purely real element." this is precisely what I want to prove.
– Charles
Aug 9 at 23:11
If $B$ is real and $B (u+iv) = 0$ where $u,v$ are real, then $Bu = 0 $ **and** $Bv=0$, so you always have a real eigenvector (assuming that $U+iv =neq 0$, of course).
– copper.hat
Aug 9 at 23:12