What is the real characteristic equation? [duplicate]
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This question already has an answer here:
$|A-lambda I|$ and $|lambda I-A|$
1 answer
My book says that you form the character equation as
$begin{vmatrix}
lambda I - A
end{vmatrix}$. However I see occasionally on stack exchange and other resources that define the character equation as $begin{vmatrix}A - lambda Iend{vmatrix}$. Why does this appear to be such a trivial matter? Here are some examples below where I have found the latter being used.
Geometric multiplicity of repeated Eigenvalues
https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors (under "Eigenvalues and the characteristic polynomial"
linear-algebra eigenvalues-eigenvectors
asked Dec 11 '18 at 1:15
Evan KimEvan Kim
3658
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Dec 11 '18 at 1:21
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$begingroup$
This question already has an answer here:
$|A-lambda I|$ and $|lambda I-A|$
1 answer
My book says that you form the character equation as
$begin{vmatrix}
lambda I - A
end{vmatrix}$. However I see occasionally on stack exchange and other resources that define the character equation as $begin{vmatrix}A - lambda Iend{vmatrix}$. Why does this appear to be such a trivial matter? Here are some examples below where I have found the latter being used.
Geometric multiplicity of repeated Eigenvalues
https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors (under "Eigenvalues and the characteristic polynomial"
linear-algebra eigenvalues-eigenvectors
asked Dec 11 '18 at 1:15
Evan KimEvan Kim
3658
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marked as duplicate by amWhy linear-algebra
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Dec 11 '18 at 1:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
$|A-lambda I|$ and $|lambda I-A|$
1 answer
My book says that you form the character equation as
$begin{vmatrix}
lambda I - A
end{vmatrix}$. However I see occasionally on stack exchange and other resources that define the character equation as $begin{vmatrix}A - lambda Iend{vmatrix}$. Why does this appear to be such a trivial matter? Here are some examples below where I have found the latter being used.
Geometric multiplicity of repeated Eigenvalues
https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors (under "Eigenvalues and the characteristic polynomial"
linear-algebra eigenvalues-eigenvectors
asked Dec 11 '18 at 1:15
Evan KimEvan Kim
3658
$endgroup$
This question already has an answer here:
$|A-lambda I|$ and $|lambda I-A|$
1 answer
My book says that you form the character equation as
$begin{vmatrix}
lambda I - A
end{vmatrix}$. However I see occasionally on stack exchange and other resources that define the character equation as $begin{vmatrix}A - lambda Iend{vmatrix}$. Why does this appear to be such a trivial matter? Here are some examples below where I have found the latter being used.
Geometric multiplicity of repeated Eigenvalues
https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors (under "Eigenvalues and the characteristic polynomial"
This question already has an answer here:
$|A-lambda I|$ and $|lambda I-A|$
1 answer
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Dec 11 '18 at 1:15
Evan KimEvan Kim
3658
asked Dec 11 '18 at 1:15
Evan KimEvan Kim
3658
asked Dec 11 '18 at 1:15
Evan KimEvan Kim
3658
asked Dec 11 '18 at 1:15
Evan KimEvan Kim
3658
asked Dec 11 '18 at 1:15
Evan KimEvan Kim
3658
3658
marked as duplicate by amWhy linear-algebra
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1 Answer
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The two only differ by a factor of $(-1)^n$, so it's not a big deal. In my opinion, the better convention is $det (lambda I - A)$, because that guarantees that the characteristic polynomial is always monic (has leading coefficient $1$). But it is a convention; both conventions are basically fine.
answered Dec 11 '18 at 1:18
Qiaochu YuanQiaochu Yuan
279k32590935
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1 Answer
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1 Answer
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$begingroup$
The two only differ by a factor of $(-1)^n$, so it's not a big deal. In my opinion, the better convention is $det (lambda I - A)$, because that guarantees that the characteristic polynomial is always monic (has leading coefficient $1$). But it is a convention; both conventions are basically fine.
answered Dec 11 '18 at 1:18
Qiaochu YuanQiaochu Yuan
279k32590935
$endgroup$
add a comment |
$begingroup$
The two only differ by a factor of $(-1)^n$, so it's not a big deal. In my opinion, the better convention is $det (lambda I - A)$, because that guarantees that the characteristic polynomial is always monic (has leading coefficient $1$). But it is a convention; both conventions are basically fine.
answered Dec 11 '18 at 1:18
Qiaochu YuanQiaochu Yuan
279k32590935
$endgroup$
add a comment |
$begingroup$
The two only differ by a factor of $(-1)^n$, so it's not a big deal. In my opinion, the better convention is $det (lambda I - A)$, because that guarantees that the characteristic polynomial is always monic (has leading coefficient $1$). But it is a convention; both conventions are basically fine.
answered Dec 11 '18 at 1:18
Qiaochu YuanQiaochu Yuan
279k32590935
$endgroup$
The two only differ by a factor of $(-1)^n$, so it's not a big deal. In my opinion, the better convention is $det (lambda I - A)$, because that guarantees that the characteristic polynomial is always monic (has leading coefficient $1$). But it is a convention; both conventions are basically fine.
answered Dec 11 '18 at 1:18
Qiaochu YuanQiaochu Yuan
279k32590935
answered Dec 11 '18 at 1:18
Qiaochu YuanQiaochu Yuan
279k32590935
answered Dec 11 '18 at 1:18
Qiaochu YuanQiaochu Yuan
279k32590935
answered Dec 11 '18 at 1:18
Qiaochu YuanQiaochu Yuan
279k32590935
279k32590935
add a comment |
add a comment |
