Why don't similar matrices have same eigenvectors and eigenvalues?
up vote
2
down vote
favorite
What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:
$R=STS^{-1}$
Therefore:
$Rv = STS^{-1}v$
$2v = STS^{-1}v$
$2S^{-1}v = TS^{-1}v$
$2S^{-1}Sv = Tv$
$2Iv = Tv$
$Tv = 2v$
Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?
linear-algebra matrices eigenvalues-eigenvectors inverse
asked 5 hours ago
Justin Sanders
111
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Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
2
down vote
favorite
What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:
$R=STS^{-1}$
Therefore:
$Rv = STS^{-1}v$
$2v = STS^{-1}v$
$2S^{-1}v = TS^{-1}v$
$2S^{-1}Sv = Tv$
$2Iv = Tv$
$Tv = 2v$
Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?
linear-algebra matrices eigenvalues-eigenvectors inverse
asked 5 hours ago
Justin Sanders
111
New contributor
Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
– Theo Bendit
5 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:
$R=STS^{-1}$
Therefore:
$Rv = STS^{-1}v$
$2v = STS^{-1}v$
$2S^{-1}v = TS^{-1}v$
$2S^{-1}Sv = Tv$
$2Iv = Tv$
$Tv = 2v$
Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?
linear-algebra matrices eigenvalues-eigenvectors inverse
asked 5 hours ago
Justin Sanders
111
New contributor
Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:
$R=STS^{-1}$
Therefore:
$Rv = STS^{-1}v$
$2v = STS^{-1}v$
$2S^{-1}v = TS^{-1}v$
$2S^{-1}Sv = Tv$
$2Iv = Tv$
$Tv = 2v$
Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?
linear-algebra matrices eigenvalues-eigenvectors inverse
linear-algebra matrices eigenvalues-eigenvectors inverse
asked 5 hours ago
Justin Sanders
111
New contributor
Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Justin Sanders
111
New contributor
Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Justin Sanders
111
New contributor
Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Justin Sanders
111
asked 5 hours ago
Justin Sanders
111
111
New contributor
Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Justin Sanders is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
– Theo Bendit
5 hours ago
add a comment |
Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
– Theo Bendit
5 hours ago
Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
– Theo Bendit
5 hours ago
Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
– Theo Bendit
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
When you went from
$$
2S^{-1}v=TS^{-1}v
$$
to
$$
2S^{-1}Sv=Tv
$$
you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
$$
2S^{-1}vS=TS^{-1}vS
$$
Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.
answered 5 hours ago
GenericMathMan
311
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GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.
answered 5 hours ago
José Carlos Santos
147k22117218
add a comment |
up vote
2
down vote
Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.
Similar matrices do have the same eigenvalues, to wit:
if
$B = PAP^{-1}, tag 1$
then
$B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$
whence
$det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$
since $det(P^{-1}) = (det(P))^{-1}; tag 4$
thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.
However, similar matrices do not in general share eigenvectors; if
$B vec v = lambda vec v, tag 5$
then
$PAP^{-1} vec v = lambda vec v, tag 6$
or
$AP^{-1} vec v = lambda P^{-1} vec v, tag 7$
that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.
So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.
answered 5 hours ago
Robert Lewis
42.9k22863
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
When you went from
$$
2S^{-1}v=TS^{-1}v
$$
to
$$
2S^{-1}Sv=Tv
$$
you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
$$
2S^{-1}vS=TS^{-1}vS
$$
Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.
answered 5 hours ago
GenericMathMan
311
New contributor
GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
When you went from
$$
2S^{-1}v=TS^{-1}v
$$
to
$$
2S^{-1}Sv=Tv
$$
you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
$$
2S^{-1}vS=TS^{-1}vS
$$
Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.
answered 5 hours ago
GenericMathMan
311
New contributor
GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
up vote
2
down vote
When you went from
$$
2S^{-1}v=TS^{-1}v
$$
to
$$
2S^{-1}Sv=Tv
$$
you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
$$
2S^{-1}vS=TS^{-1}vS
$$
Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.
answered 5 hours ago
GenericMathMan
311
New contributor
GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
When you went from
$$
2S^{-1}v=TS^{-1}v
$$
to
$$
2S^{-1}Sv=Tv
$$
you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
$$
2S^{-1}vS=TS^{-1}vS
$$
Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.
answered 5 hours ago
GenericMathMan
311
New contributor
GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
GenericMathMan
311
New contributor
GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
GenericMathMan
311
answered 5 hours ago
GenericMathMan
311
311
New contributor
GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
GenericMathMan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
up vote
2
down vote
It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.
answered 5 hours ago
José Carlos Santos
147k22117218
add a comment |
up vote
2
down vote
It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.
answered 5 hours ago
José Carlos Santos
147k22117218
add a comment |
up vote
2
down vote
up vote
2
down vote
It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.
answered 5 hours ago
José Carlos Santos
147k22117218
It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.
answered 5 hours ago
José Carlos Santos
147k22117218
answered 5 hours ago
José Carlos Santos
147k22117218
answered 5 hours ago
José Carlos Santos
147k22117218
answered 5 hours ago
José Carlos Santos
147k22117218
147k22117218
add a comment |
add a comment |
up vote
2
down vote
Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.
Similar matrices do have the same eigenvalues, to wit:
if
$B = PAP^{-1}, tag 1$
then
$B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$
whence
$det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$
since $det(P^{-1}) = (det(P))^{-1}; tag 4$
thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.
However, similar matrices do not in general share eigenvectors; if
$B vec v = lambda vec v, tag 5$
then
$PAP^{-1} vec v = lambda vec v, tag 6$
or
$AP^{-1} vec v = lambda P^{-1} vec v, tag 7$
that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.
So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.
answered 5 hours ago
Robert Lewis
42.9k22863
add a comment |
up vote
2
down vote
Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.
Similar matrices do have the same eigenvalues, to wit:
if
$B = PAP^{-1}, tag 1$
then
$B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$
whence
$det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$
since $det(P^{-1}) = (det(P))^{-1}; tag 4$
thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.
However, similar matrices do not in general share eigenvectors; if
$B vec v = lambda vec v, tag 5$
then
$PAP^{-1} vec v = lambda vec v, tag 6$
or
$AP^{-1} vec v = lambda P^{-1} vec v, tag 7$
that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.
So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.
answered 5 hours ago
Robert Lewis
42.9k22863
add a comment |
up vote
2
down vote
up vote
2
down vote
Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.
Similar matrices do have the same eigenvalues, to wit:
if
$B = PAP^{-1}, tag 1$
then
$B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$
whence
$det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$
since $det(P^{-1}) = (det(P))^{-1}; tag 4$
thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.
However, similar matrices do not in general share eigenvectors; if
$B vec v = lambda vec v, tag 5$
then
$PAP^{-1} vec v = lambda vec v, tag 6$
or
$AP^{-1} vec v = lambda P^{-1} vec v, tag 7$
that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.
So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.
answered 5 hours ago
Robert Lewis
42.9k22863
Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.
Similar matrices do have the same eigenvalues, to wit:
if
$B = PAP^{-1}, tag 1$
then
$B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$
whence
$det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$
since $det(P^{-1}) = (det(P))^{-1}; tag 4$
thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.
However, similar matrices do not in general share eigenvectors; if
$B vec v = lambda vec v, tag 5$
then
$PAP^{-1} vec v = lambda vec v, tag 6$
or
$AP^{-1} vec v = lambda P^{-1} vec v, tag 7$
that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.
So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.
answered 5 hours ago
Robert Lewis
42.9k22863
answered 5 hours ago
Robert Lewis
42.9k22863
answered 5 hours ago
Robert Lewis
42.9k22863
answered 5 hours ago
Robert Lewis
42.9k22863
42.9k22863
add a comment |
add a comment |
Justin Sanders is a new contributor. Be nice, and check out our Code of Conduct.
Justin Sanders is a new contributor. Be nice, and check out our Code of Conduct.
Justin Sanders is a new contributor. Be nice, and check out our Code of Conduct.
Justin Sanders is a new contributor. Be nice, and check out our Code of Conduct.
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Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
– Theo Bendit
5 hours ago