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On the relation between traces and characteristic polynomials
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I know that the trace of an $n times n$ matrix is invariant and results from adding the diagonal entries. I also know that the characteristic polynomial of that very same matrix is also invariant, and somehow that invariance seems more primitive than that of the trace.
What is the relation between those two invariances?
Does the trace derive from a primordial invariant represented by the underlying polynomial?
How exactly can that relation be characterized, both logically and numerically? I heard sometime that the trace merely multiplies the polynomial by some factor ($times (n-1)$) or something along those lines, I really do not recall.
Could you please elaborate on that? Thanks in advance.
matrices polynomials terminology matrix-calculus trace
edited Jul 19 at 15:01
Rodrigo de Azevedo
12.6k41751
asked Jul 19 at 9:20
Javier Arias
876617
add a comment |Â
up vote
0
down vote
favorite
I know that the trace of an $n times n$ matrix is invariant and results from adding the diagonal entries. I also know that the characteristic polynomial of that very same matrix is also invariant, and somehow that invariance seems more primitive than that of the trace.
What is the relation between those two invariances?
Does the trace derive from a primordial invariant represented by the underlying polynomial?
How exactly can that relation be characterized, both logically and numerically? I heard sometime that the trace merely multiplies the polynomial by some factor ($times (n-1)$) or something along those lines, I really do not recall.
Could you please elaborate on that? Thanks in advance.
matrices polynomials terminology matrix-calculus trace
edited Jul 19 at 15:01
Rodrigo de Azevedo
12.6k41751
asked Jul 19 at 9:20
Javier Arias
876617
The trace is, up to sign, a coefficient of the characteristic polynomial. The coefficients of the characteristic polynomial, up to sign, are the traces of the action of the matrix on the exterior powers of the underlying vector space.
– Lord Shark the Unknown
Jul 19 at 9:23
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that the trace of an $n times n$ matrix is invariant and results from adding the diagonal entries. I also know that the characteristic polynomial of that very same matrix is also invariant, and somehow that invariance seems more primitive than that of the trace.
What is the relation between those two invariances?
Does the trace derive from a primordial invariant represented by the underlying polynomial?
How exactly can that relation be characterized, both logically and numerically? I heard sometime that the trace merely multiplies the polynomial by some factor ($times (n-1)$) or something along those lines, I really do not recall.
Could you please elaborate on that? Thanks in advance.
matrices polynomials terminology matrix-calculus trace
edited Jul 19 at 15:01
Rodrigo de Azevedo
12.6k41751
asked Jul 19 at 9:20
Javier Arias
876617
I know that the trace of an $n times n$ matrix is invariant and results from adding the diagonal entries. I also know that the characteristic polynomial of that very same matrix is also invariant, and somehow that invariance seems more primitive than that of the trace.
What is the relation between those two invariances?
Does the trace derive from a primordial invariant represented by the underlying polynomial?
How exactly can that relation be characterized, both logically and numerically? I heard sometime that the trace merely multiplies the polynomial by some factor ($times (n-1)$) or something along those lines, I really do not recall.
Could you please elaborate on that? Thanks in advance.
matrices polynomials terminology matrix-calculus trace
edited Jul 19 at 15:01
Rodrigo de Azevedo
12.6k41751
asked Jul 19 at 9:20
Javier Arias
876617
edited Jul 19 at 15:01
Rodrigo de Azevedo
12.6k41751
edited Jul 19 at 15:01
Rodrigo de Azevedo
12.6k41751
edited Jul 19 at 15:01
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 19 at 9:20
Javier Arias
876617
asked Jul 19 at 9:20
Javier Arias
876617
asked Jul 19 at 9:20
Javier Arias
876617
876617
The trace is, up to sign, a coefficient of the characteristic polynomial. The coefficients of the characteristic polynomial, up to sign, are the traces of the action of the matrix on the exterior powers of the underlying vector space.
– Lord Shark the Unknown
Jul 19 at 9:23
add a comment |Â
The trace is, up to sign, a coefficient of the characteristic polynomial. The coefficients of the characteristic polynomial, up to sign, are the traces of the action of the matrix on the exterior powers of the underlying vector space.
– Lord Shark the Unknown
Jul 19 at 9:23
The trace is, up to sign, a coefficient of the characteristic polynomial. The coefficients of the characteristic polynomial, up to sign, are the traces of the action of the matrix on the exterior powers of the underlying vector space.
– Lord Shark the Unknown
Jul 19 at 9:23
The trace is, up to sign, a coefficient of the characteristic polynomial. The coefficients of the characteristic polynomial, up to sign, are the traces of the action of the matrix on the exterior powers of the underlying vector space.
– Lord Shark the Unknown
Jul 19 at 9:23
add a comment |Â
1 Answer
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0
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Assuming that you define the characteristic polynomial $P(lambda)$ of the matrix $A$ as $P(lambda)=det(A-lambdaoperatornameId_n)$, then $operatornametrA$ is the coefficient of $lambda^n-1$ in $P(lambda)$ times $(-1)^n-1$. Therefore, the invariance of $P(lambda)$ implies the invariance of $operatornametrA$.
answered Jul 19 at 9:28
José Carlos Santos
114k1698177
Does that mean that the primary invariance, so to speak, is that of the polynomial, and that of the trace is derived from it?
– Javier Arias
Jul 19 at 9:52
I wouldn't say that, since the invariance of the trace can be proved without even knowing what the characteristic polynomial is or what a determinant is. It follows from the fact that $operatornametr(AB)=operatornametr(BA)$.
– José Carlos Santos
Jul 19 at 9:55
so those two invariances are orthogonal?? It does not seem to me. I mean, the invariance of the trace may be proved (logically) without knowing the characteristic polynomial, but, as far as I understand it, it cannot be numerically determined without the characteristic being known? Correct? If that is the case, it seems like there is one primary and one secondary invariance there, so to speak.
– Javier Arias
Jul 19 at 9:59
I cannot answer your question, since I don't understand it. What do you by “characteristicâ€�
– José Carlos Santos
Jul 19 at 10:01
i mean the characteristic polynomial
– Javier Arias
Jul 19 at 10:03
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assuming that you define the characteristic polynomial $P(lambda)$ of the matrix $A$ as $P(lambda)=det(A-lambdaoperatornameId_n)$, then $operatornametrA$ is the coefficient of $lambda^n-1$ in $P(lambda)$ times $(-1)^n-1$. Therefore, the invariance of $P(lambda)$ implies the invariance of $operatornametrA$.
answered Jul 19 at 9:28
José Carlos Santos
114k1698177
Does that mean that the primary invariance, so to speak, is that of the polynomial, and that of the trace is derived from it?
– Javier Arias
Jul 19 at 9:52
I wouldn't say that, since the invariance of the trace can be proved without even knowing what the characteristic polynomial is or what a determinant is. It follows from the fact that $operatornametr(AB)=operatornametr(BA)$.
– José Carlos Santos
Jul 19 at 9:55
so those two invariances are orthogonal?? It does not seem to me. I mean, the invariance of the trace may be proved (logically) without knowing the characteristic polynomial, but, as far as I understand it, it cannot be numerically determined without the characteristic being known? Correct? If that is the case, it seems like there is one primary and one secondary invariance there, so to speak.
– Javier Arias
Jul 19 at 9:59
I cannot answer your question, since I don't understand it. What do you by “characteristicâ€�
– José Carlos Santos
Jul 19 at 10:01
i mean the characteristic polynomial
– Javier Arias
Jul 19 at 10:03
 |Â
show 6 more comments
up vote
0
down vote
Assuming that you define the characteristic polynomial $P(lambda)$ of the matrix $A$ as $P(lambda)=det(A-lambdaoperatornameId_n)$, then $operatornametrA$ is the coefficient of $lambda^n-1$ in $P(lambda)$ times $(-1)^n-1$. Therefore, the invariance of $P(lambda)$ implies the invariance of $operatornametrA$.
answered Jul 19 at 9:28
José Carlos Santos
114k1698177
Does that mean that the primary invariance, so to speak, is that of the polynomial, and that of the trace is derived from it?
– Javier Arias
Jul 19 at 9:52
I wouldn't say that, since the invariance of the trace can be proved without even knowing what the characteristic polynomial is or what a determinant is. It follows from the fact that $operatornametr(AB)=operatornametr(BA)$.
– José Carlos Santos
Jul 19 at 9:55
so those two invariances are orthogonal?? It does not seem to me. I mean, the invariance of the trace may be proved (logically) without knowing the characteristic polynomial, but, as far as I understand it, it cannot be numerically determined without the characteristic being known? Correct? If that is the case, it seems like there is one primary and one secondary invariance there, so to speak.
– Javier Arias
Jul 19 at 9:59
I cannot answer your question, since I don't understand it. What do you by “characteristicâ€�
– José Carlos Santos
Jul 19 at 10:01
i mean the characteristic polynomial
– Javier Arias
Jul 19 at 10:03
 |Â
show 6 more comments
up vote
0
down vote
up vote
0
down vote
Assuming that you define the characteristic polynomial $P(lambda)$ of the matrix $A$ as $P(lambda)=det(A-lambdaoperatornameId_n)$, then $operatornametrA$ is the coefficient of $lambda^n-1$ in $P(lambda)$ times $(-1)^n-1$. Therefore, the invariance of $P(lambda)$ implies the invariance of $operatornametrA$.
answered Jul 19 at 9:28
José Carlos Santos
114k1698177
Assuming that you define the characteristic polynomial $P(lambda)$ of the matrix $A$ as $P(lambda)=det(A-lambdaoperatornameId_n)$, then $operatornametrA$ is the coefficient of $lambda^n-1$ in $P(lambda)$ times $(-1)^n-1$. Therefore, the invariance of $P(lambda)$ implies the invariance of $operatornametrA$.
answered Jul 19 at 9:28
José Carlos Santos
114k1698177
edited Jul 19 at 9:34
answered Jul 19 at 9:28
José Carlos Santos
114k1698177
answered Jul 19 at 9:28
José Carlos Santos
114k1698177
answered Jul 19 at 9:28
José Carlos Santos
114k1698177
114k1698177
Does that mean that the primary invariance, so to speak, is that of the polynomial, and that of the trace is derived from it?
– Javier Arias
Jul 19 at 9:52
I wouldn't say that, since the invariance of the trace can be proved without even knowing what the characteristic polynomial is or what a determinant is. It follows from the fact that $operatornametr(AB)=operatornametr(BA)$.
– José Carlos Santos
Jul 19 at 9:55
so those two invariances are orthogonal?? It does not seem to me. I mean, the invariance of the trace may be proved (logically) without knowing the characteristic polynomial, but, as far as I understand it, it cannot be numerically determined without the characteristic being known? Correct? If that is the case, it seems like there is one primary and one secondary invariance there, so to speak.
– Javier Arias
Jul 19 at 9:59
I cannot answer your question, since I don't understand it. What do you by “characteristicâ€�
– José Carlos Santos
Jul 19 at 10:01
i mean the characteristic polynomial
– Javier Arias
Jul 19 at 10:03
 |Â
show 6 more comments
Does that mean that the primary invariance, so to speak, is that of the polynomial, and that of the trace is derived from it?
– Javier Arias
Jul 19 at 9:52
I wouldn't say that, since the invariance of the trace can be proved without even knowing what the characteristic polynomial is or what a determinant is. It follows from the fact that $operatornametr(AB)=operatornametr(BA)$.
– José Carlos Santos
Jul 19 at 9:55
so those two invariances are orthogonal?? It does not seem to me. I mean, the invariance of the trace may be proved (logically) without knowing the characteristic polynomial, but, as far as I understand it, it cannot be numerically determined without the characteristic being known? Correct? If that is the case, it seems like there is one primary and one secondary invariance there, so to speak.
– Javier Arias
Jul 19 at 9:59
I cannot answer your question, since I don't understand it. What do you by “characteristicâ€�
– José Carlos Santos
Jul 19 at 10:01
i mean the characteristic polynomial
– Javier Arias
Jul 19 at 10:03
Does that mean that the primary invariance, so to speak, is that of the polynomial, and that of the trace is derived from it?
– Javier Arias
Jul 19 at 9:52
Does that mean that the primary invariance, so to speak, is that of the polynomial, and that of the trace is derived from it?
– Javier Arias
Jul 19 at 9:52
I wouldn't say that, since the invariance of the trace can be proved without even knowing what the characteristic polynomial is or what a determinant is. It follows from the fact that $operatornametr(AB)=operatornametr(BA)$.
– José Carlos Santos
Jul 19 at 9:55
I wouldn't say that, since the invariance of the trace can be proved without even knowing what the characteristic polynomial is or what a determinant is. It follows from the fact that $operatornametr(AB)=operatornametr(BA)$.
– José Carlos Santos
Jul 19 at 9:55
so those two invariances are orthogonal?? It does not seem to me. I mean, the invariance of the trace may be proved (logically) without knowing the characteristic polynomial, but, as far as I understand it, it cannot be numerically determined without the characteristic being known? Correct? If that is the case, it seems like there is one primary and one secondary invariance there, so to speak.
– Javier Arias
Jul 19 at 9:59
so those two invariances are orthogonal?? It does not seem to me. I mean, the invariance of the trace may be proved (logically) without knowing the characteristic polynomial, but, as far as I understand it, it cannot be numerically determined without the characteristic being known? Correct? If that is the case, it seems like there is one primary and one secondary invariance there, so to speak.
– Javier Arias
Jul 19 at 9:59
I cannot answer your question, since I don't understand it. What do you by “characteristicâ€�
– José Carlos Santos
Jul 19 at 10:01
I cannot answer your question, since I don't understand it. What do you by “characteristicâ€�
– José Carlos Santos
Jul 19 at 10:01
i mean the characteristic polynomial
– Javier Arias
Jul 19 at 10:03
i mean the characteristic polynomial
– Javier Arias
Jul 19 at 10:03
 |Â
show 6 more comments
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The trace is, up to sign, a coefficient of the characteristic polynomial. The coefficients of the characteristic polynomial, up to sign, are the traces of the action of the matrix on the exterior powers of the underlying vector space.
– Lord Shark the Unknown
Jul 19 at 9:23