Is the characteristic polynomial of a matrix $det(lambda I-A)$ or$ det(A-lambda I)$?
up vote
9
down vote
favorite
I haven't been able to get a very clear answer on this. In the exercise that I performed to find the Characteristic Polynomial of a given Matrix, I used the determinant of $(lambda I-A)$ to find the answer.
I don't actually attend any courses or do anything that requires me to solve these problems, or even presents them to me regularly. My solving this problem is a result of me asking my friend who is in a college math course for his homework, because I'm personally interested in learning more about math. As such, I have to study the problems he gives me on my own, and can only use the internet, for the most part, in order to get the knowledge I need to solve them. While looking for the definition of 'Characteristic Polynomial', this website defines a Characteristic Polynomial as "If $A$ is an $ntimes n$ matrix, then the Characteristic Polynomial of $A$ is the function $f(lambda )=det(lambda I−A)$". A few other resources I found also say this, while others say that it is the determinant of $(A-lambda I)$.
I solved the problem, which, yes, did use an $ntimes n$ Matrix ($3times 3$ to be specific), using the former equation of $(lambda I-A)$, and I presented the solution to my friend. He concurred, and we decided that was our answer, however when he entered it into whatever website assigns him his homework, it said that we were incorrect. We thought about it for a while, and even looked to see if the issue was that we hadn't simplified the problem properly, but everything checked out (I initially read $[lambda I-A]$ as the correct determinant, so in my head, I didn't realize that some other websites used the inverse, and didn't think to try that). Eventually, we both gave up, and I used an online calculator to solve it, and was presented with an answer achieved by using $(A-lambda I)$. We put that in to the website, and sure enough it was correct.
This causes some concern with me. Is $(A-lambda I)$ always used to find the Characteristic Polynomial? How am I to remember that it is this way, and not the other? Why do we choose to define the Characteristic Polynomial as one determinant over the other? Is the website that I cited outright incorrect, and thus should be considered disreputed, or is there something unique that I fail to understand regarding certain Matrices and their polynomials?
matrices
edited Nov 18 at 21:43
Xander Henderson
14k103552
asked Nov 17 at 23:14
Ace Otero
461
add a comment |
up vote
9
down vote
favorite
I haven't been able to get a very clear answer on this. In the exercise that I performed to find the Characteristic Polynomial of a given Matrix, I used the determinant of $(lambda I-A)$ to find the answer.
I don't actually attend any courses or do anything that requires me to solve these problems, or even presents them to me regularly. My solving this problem is a result of me asking my friend who is in a college math course for his homework, because I'm personally interested in learning more about math. As such, I have to study the problems he gives me on my own, and can only use the internet, for the most part, in order to get the knowledge I need to solve them. While looking for the definition of 'Characteristic Polynomial', this website defines a Characteristic Polynomial as "If $A$ is an $ntimes n$ matrix, then the Characteristic Polynomial of $A$ is the function $f(lambda )=det(lambda I−A)$". A few other resources I found also say this, while others say that it is the determinant of $(A-lambda I)$.
I solved the problem, which, yes, did use an $ntimes n$ Matrix ($3times 3$ to be specific), using the former equation of $(lambda I-A)$, and I presented the solution to my friend. He concurred, and we decided that was our answer, however when he entered it into whatever website assigns him his homework, it said that we were incorrect. We thought about it for a while, and even looked to see if the issue was that we hadn't simplified the problem properly, but everything checked out (I initially read $[lambda I-A]$ as the correct determinant, so in my head, I didn't realize that some other websites used the inverse, and didn't think to try that). Eventually, we both gave up, and I used an online calculator to solve it, and was presented with an answer achieved by using $(A-lambda I)$. We put that in to the website, and sure enough it was correct.
This causes some concern with me. Is $(A-lambda I)$ always used to find the Characteristic Polynomial? How am I to remember that it is this way, and not the other? Why do we choose to define the Characteristic Polynomial as one determinant over the other? Is the website that I cited outright incorrect, and thus should be considered disreputed, or is there something unique that I fail to understand regarding certain Matrices and their polynomials?
matrices
edited Nov 18 at 21:43
Xander Henderson
14k103552
asked Nov 17 at 23:14
Ace Otero
461
12
Purely a matter of convention. Since $det(A-lambda I)=(-1)^{n} det(lambda I-A)$ the two polynomials have the same roots and that's what matters.
– Kavi Rama Murthy
Nov 17 at 23:16
add a comment |
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I haven't been able to get a very clear answer on this. In the exercise that I performed to find the Characteristic Polynomial of a given Matrix, I used the determinant of $(lambda I-A)$ to find the answer.
I don't actually attend any courses or do anything that requires me to solve these problems, or even presents them to me regularly. My solving this problem is a result of me asking my friend who is in a college math course for his homework, because I'm personally interested in learning more about math. As such, I have to study the problems he gives me on my own, and can only use the internet, for the most part, in order to get the knowledge I need to solve them. While looking for the definition of 'Characteristic Polynomial', this website defines a Characteristic Polynomial as "If $A$ is an $ntimes n$ matrix, then the Characteristic Polynomial of $A$ is the function $f(lambda )=det(lambda I−A)$". A few other resources I found also say this, while others say that it is the determinant of $(A-lambda I)$.
I solved the problem, which, yes, did use an $ntimes n$ Matrix ($3times 3$ to be specific), using the former equation of $(lambda I-A)$, and I presented the solution to my friend. He concurred, and we decided that was our answer, however when he entered it into whatever website assigns him his homework, it said that we were incorrect. We thought about it for a while, and even looked to see if the issue was that we hadn't simplified the problem properly, but everything checked out (I initially read $[lambda I-A]$ as the correct determinant, so in my head, I didn't realize that some other websites used the inverse, and didn't think to try that). Eventually, we both gave up, and I used an online calculator to solve it, and was presented with an answer achieved by using $(A-lambda I)$. We put that in to the website, and sure enough it was correct.
This causes some concern with me. Is $(A-lambda I)$ always used to find the Characteristic Polynomial? How am I to remember that it is this way, and not the other? Why do we choose to define the Characteristic Polynomial as one determinant over the other? Is the website that I cited outright incorrect, and thus should be considered disreputed, or is there something unique that I fail to understand regarding certain Matrices and their polynomials?
matrices
edited Nov 18 at 21:43
Xander Henderson
14k103552
asked Nov 17 at 23:14
Ace Otero
461
I haven't been able to get a very clear answer on this. In the exercise that I performed to find the Characteristic Polynomial of a given Matrix, I used the determinant of $(lambda I-A)$ to find the answer.
I don't actually attend any courses or do anything that requires me to solve these problems, or even presents them to me regularly. My solving this problem is a result of me asking my friend who is in a college math course for his homework, because I'm personally interested in learning more about math. As such, I have to study the problems he gives me on my own, and can only use the internet, for the most part, in order to get the knowledge I need to solve them. While looking for the definition of 'Characteristic Polynomial', this website defines a Characteristic Polynomial as "If $A$ is an $ntimes n$ matrix, then the Characteristic Polynomial of $A$ is the function $f(lambda )=det(lambda I−A)$". A few other resources I found also say this, while others say that it is the determinant of $(A-lambda I)$.
I solved the problem, which, yes, did use an $ntimes n$ Matrix ($3times 3$ to be specific), using the former equation of $(lambda I-A)$, and I presented the solution to my friend. He concurred, and we decided that was our answer, however when he entered it into whatever website assigns him his homework, it said that we were incorrect. We thought about it for a while, and even looked to see if the issue was that we hadn't simplified the problem properly, but everything checked out (I initially read $[lambda I-A]$ as the correct determinant, so in my head, I didn't realize that some other websites used the inverse, and didn't think to try that). Eventually, we both gave up, and I used an online calculator to solve it, and was presented with an answer achieved by using $(A-lambda I)$. We put that in to the website, and sure enough it was correct.
This causes some concern with me. Is $(A-lambda I)$ always used to find the Characteristic Polynomial? How am I to remember that it is this way, and not the other? Why do we choose to define the Characteristic Polynomial as one determinant over the other? Is the website that I cited outright incorrect, and thus should be considered disreputed, or is there something unique that I fail to understand regarding certain Matrices and their polynomials?
matrices
matrices
edited Nov 18 at 21:43
Xander Henderson
14k103552
asked Nov 17 at 23:14
Ace Otero
461
edited Nov 18 at 21:43
Xander Henderson
14k103552
asked Nov 17 at 23:14
Ace Otero
461
edited Nov 18 at 21:43
Xander Henderson
14k103552
edited Nov 18 at 21:43
Xander Henderson
14k103552
edited Nov 18 at 21:43
Xander Henderson
14k103552
14k103552
asked Nov 17 at 23:14
Ace Otero
461
asked Nov 17 at 23:14
Ace Otero
461
asked Nov 17 at 23:14
Ace Otero
461
461
12
Purely a matter of convention. Since $det(A-lambda I)=(-1)^{n} det(lambda I-A)$ the two polynomials have the same roots and that's what matters.
– Kavi Rama Murthy
Nov 17 at 23:16
add a comment |
12
Purely a matter of convention. Since $det(A-lambda I)=(-1)^{n} det(lambda I-A)$ the two polynomials have the same roots and that's what matters.
– Kavi Rama Murthy
Nov 17 at 23:16
12
12
Purely a matter of convention. Since $det(A-lambda I)=(-1)^{n} det(lambda I-A)$ the two polynomials have the same roots and that's what matters.
– Kavi Rama Murthy
Nov 17 at 23:16
Purely a matter of convention. Since $det(A-lambda I)=(-1)^{n} det(lambda I-A)$ the two polynomials have the same roots and that's what matters.
– Kavi Rama Murthy
Nov 17 at 23:16
add a comment |
3 Answers
3
active
oldest
votes
up vote
20
down vote
Both definitions are used by some people. The $lambda I - A$ choice is more common, since it yields a monic polynomial.
Finally, note that $det(B) =(-1)^n det(-B)$ If $B$ is an $ntimes n$ Matrix. Thus, since one is mostly interested in the zeros of the characteristic polynomial, it doesn't matter too much which definition you take.
answered Nov 17 at 23:19
PhoemueX
27.1k22455
1
That definition is actually less common computationally because it leads to more errors if you have to reverse the sign on the entries instead of on the variable.
– Matt Samuel
Nov 18 at 11:58
add a comment |
up vote
12
down vote
Both definitions are common; if it matters you need to make sure which one is in place.
An advantage of the definition $det(lambda I - A)$ is that the leading coefficient of the characteristic polynomial is $1$.
However, the other definition $det(A - lambda I)$ arises somewhat more smoothly and is more convenient for calculations in that on has the $A$ "as given" and just has to put the $-lambda$ on the diagonal, as opposed to changing all the signs.
By it arising more smoothly I mean that on usually starts from the idea of finding an eigenvalue, so a $lambda$ such that
$$Av = lambda v $$
for some non-zero $v$,
or
$$Av- lambda v = 0 $$
that is
$$(A-lambda I )v = 0 $$
Of course one could also do this the other way around but to me it feels more intuitive this way.
answered Nov 17 at 23:26
quid♦
36.7k95093
So then, what would have caused us to find two different answers using the different formulas? If they're interchangeable, should we not have reached, more or less, the same answer with either one?
– Ace Otero
Nov 18 at 0:15
The polynomials you get with the different definitions differ by a factor of $(-1)^n$ where $n$ is the dimension of the matrix. In you case $n=3$ this is $-1$ and so all the signs of the coefficients are flipped. As said in another answer this does not affect the roots of the polynomial (that is the eigenvalues of $A$), which is what we mostly care about.
– quid♦
Nov 18 at 0:18
1
Good answer. Note that there is a sort of symmetry in a projective sense: the definition as $p(lambda) = det(lambda I - A)$ has a "nicer" value at $lambda=infty$ (leading term), the definition as $p(lambda)=det(A-lambda I)$ has a "nicer" value at $lambda=0$, i.e., $p(0)=det(A)$. It depends on the application which one is more convenient.
– Federico Poloni
Nov 18 at 12:14
add a comment |
up vote
2
down vote
Finding $A-lambda I$ is more straight forward. The eigen values in both cases are the same. You pick the one that you like better.
answered Nov 17 at 23:22
Mohammad Riazi-Kermani
40.3k41958
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
20
down vote
Both definitions are used by some people. The $lambda I - A$ choice is more common, since it yields a monic polynomial.
Finally, note that $det(B) =(-1)^n det(-B)$ If $B$ is an $ntimes n$ Matrix. Thus, since one is mostly interested in the zeros of the characteristic polynomial, it doesn't matter too much which definition you take.
answered Nov 17 at 23:19
PhoemueX
27.1k22455
1
That definition is actually less common computationally because it leads to more errors if you have to reverse the sign on the entries instead of on the variable.
– Matt Samuel
Nov 18 at 11:58
add a comment |
up vote
20
down vote
Both definitions are used by some people. The $lambda I - A$ choice is more common, since it yields a monic polynomial.
Finally, note that $det(B) =(-1)^n det(-B)$ If $B$ is an $ntimes n$ Matrix. Thus, since one is mostly interested in the zeros of the characteristic polynomial, it doesn't matter too much which definition you take.
answered Nov 17 at 23:19
PhoemueX
27.1k22455
1
That definition is actually less common computationally because it leads to more errors if you have to reverse the sign on the entries instead of on the variable.
– Matt Samuel
Nov 18 at 11:58
add a comment |
up vote
20
down vote
up vote
20
down vote
Both definitions are used by some people. The $lambda I - A$ choice is more common, since it yields a monic polynomial.
Finally, note that $det(B) =(-1)^n det(-B)$ If $B$ is an $ntimes n$ Matrix. Thus, since one is mostly interested in the zeros of the characteristic polynomial, it doesn't matter too much which definition you take.
answered Nov 17 at 23:19
PhoemueX
27.1k22455
Both definitions are used by some people. The $lambda I - A$ choice is more common, since it yields a monic polynomial.
Finally, note that $det(B) =(-1)^n det(-B)$ If $B$ is an $ntimes n$ Matrix. Thus, since one is mostly interested in the zeros of the characteristic polynomial, it doesn't matter too much which definition you take.
answered Nov 17 at 23:19
PhoemueX
27.1k22455
answered Nov 17 at 23:19
PhoemueX
27.1k22455
answered Nov 17 at 23:19
PhoemueX
27.1k22455
answered Nov 17 at 23:19
PhoemueX
27.1k22455
27.1k22455
1
That definition is actually less common computationally because it leads to more errors if you have to reverse the sign on the entries instead of on the variable.
– Matt Samuel
Nov 18 at 11:58
add a comment |
1
That definition is actually less common computationally because it leads to more errors if you have to reverse the sign on the entries instead of on the variable.
– Matt Samuel
Nov 18 at 11:58
1
1
That definition is actually less common computationally because it leads to more errors if you have to reverse the sign on the entries instead of on the variable.
– Matt Samuel
Nov 18 at 11:58
That definition is actually less common computationally because it leads to more errors if you have to reverse the sign on the entries instead of on the variable.
– Matt Samuel
Nov 18 at 11:58
add a comment |
up vote
12
down vote
Both definitions are common; if it matters you need to make sure which one is in place.
An advantage of the definition $det(lambda I - A)$ is that the leading coefficient of the characteristic polynomial is $1$.
However, the other definition $det(A - lambda I)$ arises somewhat more smoothly and is more convenient for calculations in that on has the $A$ "as given" and just has to put the $-lambda$ on the diagonal, as opposed to changing all the signs.
By it arising more smoothly I mean that on usually starts from the idea of finding an eigenvalue, so a $lambda$ such that
$$Av = lambda v $$
for some non-zero $v$,
or
$$Av- lambda v = 0 $$
that is
$$(A-lambda I )v = 0 $$
Of course one could also do this the other way around but to me it feels more intuitive this way.
answered Nov 17 at 23:26
quid♦
36.7k95093
So then, what would have caused us to find two different answers using the different formulas? If they're interchangeable, should we not have reached, more or less, the same answer with either one?
– Ace Otero
Nov 18 at 0:15
The polynomials you get with the different definitions differ by a factor of $(-1)^n$ where $n$ is the dimension of the matrix. In you case $n=3$ this is $-1$ and so all the signs of the coefficients are flipped. As said in another answer this does not affect the roots of the polynomial (that is the eigenvalues of $A$), which is what we mostly care about.
– quid♦
Nov 18 at 0:18
1
Good answer. Note that there is a sort of symmetry in a projective sense: the definition as $p(lambda) = det(lambda I - A)$ has a "nicer" value at $lambda=infty$ (leading term), the definition as $p(lambda)=det(A-lambda I)$ has a "nicer" value at $lambda=0$, i.e., $p(0)=det(A)$. It depends on the application which one is more convenient.
– Federico Poloni
Nov 18 at 12:14
add a comment |
up vote
12
down vote
Both definitions are common; if it matters you need to make sure which one is in place.
An advantage of the definition $det(lambda I - A)$ is that the leading coefficient of the characteristic polynomial is $1$.
However, the other definition $det(A - lambda I)$ arises somewhat more smoothly and is more convenient for calculations in that on has the $A$ "as given" and just has to put the $-lambda$ on the diagonal, as opposed to changing all the signs.
By it arising more smoothly I mean that on usually starts from the idea of finding an eigenvalue, so a $lambda$ such that
$$Av = lambda v $$
for some non-zero $v$,
or
$$Av- lambda v = 0 $$
that is
$$(A-lambda I )v = 0 $$
Of course one could also do this the other way around but to me it feels more intuitive this way.
answered Nov 17 at 23:26
quid♦
36.7k95093
So then, what would have caused us to find two different answers using the different formulas? If they're interchangeable, should we not have reached, more or less, the same answer with either one?
– Ace Otero
Nov 18 at 0:15
The polynomials you get with the different definitions differ by a factor of $(-1)^n$ where $n$ is the dimension of the matrix. In you case $n=3$ this is $-1$ and so all the signs of the coefficients are flipped. As said in another answer this does not affect the roots of the polynomial (that is the eigenvalues of $A$), which is what we mostly care about.
– quid♦
Nov 18 at 0:18
1
Good answer. Note that there is a sort of symmetry in a projective sense: the definition as $p(lambda) = det(lambda I - A)$ has a "nicer" value at $lambda=infty$ (leading term), the definition as $p(lambda)=det(A-lambda I)$ has a "nicer" value at $lambda=0$, i.e., $p(0)=det(A)$. It depends on the application which one is more convenient.
– Federico Poloni
Nov 18 at 12:14
add a comment |
up vote
12
down vote
up vote
12
down vote
Both definitions are common; if it matters you need to make sure which one is in place.
An advantage of the definition $det(lambda I - A)$ is that the leading coefficient of the characteristic polynomial is $1$.
However, the other definition $det(A - lambda I)$ arises somewhat more smoothly and is more convenient for calculations in that on has the $A$ "as given" and just has to put the $-lambda$ on the diagonal, as opposed to changing all the signs.
By it arising more smoothly I mean that on usually starts from the idea of finding an eigenvalue, so a $lambda$ such that
$$Av = lambda v $$
for some non-zero $v$,
or
$$Av- lambda v = 0 $$
that is
$$(A-lambda I )v = 0 $$
Of course one could also do this the other way around but to me it feels more intuitive this way.
answered Nov 17 at 23:26
quid♦
36.7k95093
Both definitions are common; if it matters you need to make sure which one is in place.
An advantage of the definition $det(lambda I - A)$ is that the leading coefficient of the characteristic polynomial is $1$.
However, the other definition $det(A - lambda I)$ arises somewhat more smoothly and is more convenient for calculations in that on has the $A$ "as given" and just has to put the $-lambda$ on the diagonal, as opposed to changing all the signs.
By it arising more smoothly I mean that on usually starts from the idea of finding an eigenvalue, so a $lambda$ such that
$$Av = lambda v $$
for some non-zero $v$,
or
$$Av- lambda v = 0 $$
that is
$$(A-lambda I )v = 0 $$
Of course one could also do this the other way around but to me it feels more intuitive this way.
answered Nov 17 at 23:26
quid♦
36.7k95093
answered Nov 17 at 23:26
quid♦
36.7k95093
answered Nov 17 at 23:26
quid♦
36.7k95093
answered Nov 17 at 23:26
quid♦
36.7k95093
36.7k95093
So then, what would have caused us to find two different answers using the different formulas? If they're interchangeable, should we not have reached, more or less, the same answer with either one?
– Ace Otero
Nov 18 at 0:15
The polynomials you get with the different definitions differ by a factor of $(-1)^n$ where $n$ is the dimension of the matrix. In you case $n=3$ this is $-1$ and so all the signs of the coefficients are flipped. As said in another answer this does not affect the roots of the polynomial (that is the eigenvalues of $A$), which is what we mostly care about.
– quid♦
Nov 18 at 0:18
1
Good answer. Note that there is a sort of symmetry in a projective sense: the definition as $p(lambda) = det(lambda I - A)$ has a "nicer" value at $lambda=infty$ (leading term), the definition as $p(lambda)=det(A-lambda I)$ has a "nicer" value at $lambda=0$, i.e., $p(0)=det(A)$. It depends on the application which one is more convenient.
– Federico Poloni
Nov 18 at 12:14
add a comment |
So then, what would have caused us to find two different answers using the different formulas? If they're interchangeable, should we not have reached, more or less, the same answer with either one?
– Ace Otero
Nov 18 at 0:15
The polynomials you get with the different definitions differ by a factor of $(-1)^n$ where $n$ is the dimension of the matrix. In you case $n=3$ this is $-1$ and so all the signs of the coefficients are flipped. As said in another answer this does not affect the roots of the polynomial (that is the eigenvalues of $A$), which is what we mostly care about.
– quid♦
Nov 18 at 0:18
1
Good answer. Note that there is a sort of symmetry in a projective sense: the definition as $p(lambda) = det(lambda I - A)$ has a "nicer" value at $lambda=infty$ (leading term), the definition as $p(lambda)=det(A-lambda I)$ has a "nicer" value at $lambda=0$, i.e., $p(0)=det(A)$. It depends on the application which one is more convenient.
– Federico Poloni
Nov 18 at 12:14
So then, what would have caused us to find two different answers using the different formulas? If they're interchangeable, should we not have reached, more or less, the same answer with either one?
– Ace Otero
Nov 18 at 0:15
So then, what would have caused us to find two different answers using the different formulas? If they're interchangeable, should we not have reached, more or less, the same answer with either one?
– Ace Otero
Nov 18 at 0:15
The polynomials you get with the different definitions differ by a factor of $(-1)^n$ where $n$ is the dimension of the matrix. In you case $n=3$ this is $-1$ and so all the signs of the coefficients are flipped. As said in another answer this does not affect the roots of the polynomial (that is the eigenvalues of $A$), which is what we mostly care about.
– quid♦
Nov 18 at 0:18
The polynomials you get with the different definitions differ by a factor of $(-1)^n$ where $n$ is the dimension of the matrix. In you case $n=3$ this is $-1$ and so all the signs of the coefficients are flipped. As said in another answer this does not affect the roots of the polynomial (that is the eigenvalues of $A$), which is what we mostly care about.
– quid♦
Nov 18 at 0:18
1
1
Good answer. Note that there is a sort of symmetry in a projective sense: the definition as $p(lambda) = det(lambda I - A)$ has a "nicer" value at $lambda=infty$ (leading term), the definition as $p(lambda)=det(A-lambda I)$ has a "nicer" value at $lambda=0$, i.e., $p(0)=det(A)$. It depends on the application which one is more convenient.
– Federico Poloni
Nov 18 at 12:14
Good answer. Note that there is a sort of symmetry in a projective sense: the definition as $p(lambda) = det(lambda I - A)$ has a "nicer" value at $lambda=infty$ (leading term), the definition as $p(lambda)=det(A-lambda I)$ has a "nicer" value at $lambda=0$, i.e., $p(0)=det(A)$. It depends on the application which one is more convenient.
– Federico Poloni
Nov 18 at 12:14
add a comment |
up vote
2
down vote
Finding $A-lambda I$ is more straight forward. The eigen values in both cases are the same. You pick the one that you like better.
answered Nov 17 at 23:22
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
2
down vote
Finding $A-lambda I$ is more straight forward. The eigen values in both cases are the same. You pick the one that you like better.
answered Nov 17 at 23:22
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
2
down vote
up vote
2
down vote
Finding $A-lambda I$ is more straight forward. The eigen values in both cases are the same. You pick the one that you like better.
answered Nov 17 at 23:22
Mohammad Riazi-Kermani
40.3k41958
Finding $A-lambda I$ is more straight forward. The eigen values in both cases are the same. You pick the one that you like better.
answered Nov 17 at 23:22
Mohammad Riazi-Kermani
40.3k41958
answered Nov 17 at 23:22
Mohammad Riazi-Kermani
40.3k41958
answered Nov 17 at 23:22
Mohammad Riazi-Kermani
40.3k41958
answered Nov 17 at 23:22
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002931%2fis-the-characteristic-polynomial-of-a-matrix-det-lambda-i-a-or-deta-lam%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
12
Purely a matter of convention. Since $det(A-lambda I)=(-1)^{n} det(lambda I-A)$ the two polynomials have the same roots and that's what matters.
– Kavi Rama Murthy
Nov 17 at 23:16