Symmetric matrix with n rows has n linearly independent eigenvectors?
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I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?
EDIT:
The comments led me to a nice answer from another user which I think I understand.
Essentially the argument is:
Take any eigenvector of the symmetric matrix $A$, call it $x$. Then $A$ maps the set ${v : langle v, x rangle=0}=:V$ to itself since take $v in V$ then $(Av)^Tx=v^TA^Tx=v^TAx=lambda v^T x=0$ but this space V has 1 dimension less than $V cup text{ span} (x)$. Then because all endomorphisms which are linear functions have eigenvalues and $A$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $V$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Jan 6 at 11:08
DanDan
998
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add a comment |
$begingroup$
I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?
EDIT:
The comments led me to a nice answer from another user which I think I understand.
Essentially the argument is:
Take any eigenvector of the symmetric matrix $A$, call it $x$. Then $A$ maps the set ${v : langle v, x rangle=0}=:V$ to itself since take $v in V$ then $(Av)^Tx=v^TA^Tx=v^TAx=lambda v^T x=0$ but this space V has 1 dimension less than $V cup text{ span} (x)$. Then because all endomorphisms which are linear functions have eigenvalues and $A$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $V$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Jan 6 at 11:08
DanDan
998
$endgroup$
$begingroup$
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
$endgroup$
– A.Γ.
Jan 6 at 11:44
$begingroup$
Possible duplicate math.stackexchange.com/questions/2777494/…
$endgroup$
– A.Γ.
Jan 6 at 12:00
add a comment |
$begingroup$
I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?
EDIT:
The comments led me to a nice answer from another user which I think I understand.
Essentially the argument is:
Take any eigenvector of the symmetric matrix $A$, call it $x$. Then $A$ maps the set ${v : langle v, x rangle=0}=:V$ to itself since take $v in V$ then $(Av)^Tx=v^TA^Tx=v^TAx=lambda v^T x=0$ but this space V has 1 dimension less than $V cup text{ span} (x)$. Then because all endomorphisms which are linear functions have eigenvalues and $A$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $V$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Jan 6 at 11:08
DanDan
998
$endgroup$
I'm trying to better understand SPD matrices and have started with symmetric matrices. It's apparent to me that symmetric matrices have real eigenvalues, but how to show that the algebraic multiplicity of each eigenvalue is no greater than it's geometric multiplicity?
EDIT:
The comments led me to a nice answer from another user which I think I understand.
Essentially the argument is:
Take any eigenvector of the symmetric matrix $A$, call it $x$. Then $A$ maps the set ${v : langle v, x rangle=0}=:V$ to itself since take $v in V$ then $(Av)^Tx=v^TA^Tx=v^TAx=lambda v^T x=0$ but this space V has 1 dimension less than $V cup text{ span} (x)$. Then because all endomorphisms which are linear functions have eigenvalues and $A$ restricted to V is an endomorphism, A has an eigenvalue/eigenvector pair on $V$. You can repeat the process and therefore all the eigenvectors are distinct and orthonormal.
linear-algebra eigenvalues-eigenvectors diagonalization
linear-algebra eigenvalues-eigenvectors diagonalization
asked Jan 6 at 11:08
DanDan
998
asked Jan 6 at 11:08
DanDan
998
edited Jan 6 at 13:16
Dan
asked Jan 6 at 11:08
DanDan
998
asked Jan 6 at 11:08
DanDan
998
asked Jan 6 at 11:08
DanDan
998
998
$begingroup$
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
$endgroup$
– A.Γ.
Jan 6 at 11:44
$begingroup$
Possible duplicate math.stackexchange.com/questions/2777494/…
$endgroup$
– A.Γ.
Jan 6 at 12:00
add a comment |
$begingroup$
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
$endgroup$
– A.Γ.
Jan 6 at 11:44
$begingroup$
Possible duplicate math.stackexchange.com/questions/2777494/…
$endgroup$
– A.Γ.
Jan 6 at 12:00
$begingroup$
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
$endgroup$
– A.Γ.
Jan 6 at 11:44
$begingroup$
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
$endgroup$
– A.Γ.
Jan 6 at 11:44
$begingroup$
Possible duplicate math.stackexchange.com/questions/2777494/…
$endgroup$
– A.Γ.
Jan 6 at 12:00
$begingroup$
Possible duplicate math.stackexchange.com/questions/2777494/…
$endgroup$
– A.Γ.
Jan 6 at 12:00
add a comment |
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$begingroup$
"$n$ distinct eigenvectors" - you must mean $n$ linearly independent eigenvectors?
$endgroup$
– A.Γ.
Jan 6 at 11:44
$begingroup$
Possible duplicate math.stackexchange.com/questions/2777494/…
$endgroup$
– A.Γ.
Jan 6 at 12:00