Mixing
How to prove $det(e^A) = e^{operatorname{tr}(A)}$?
$begingroup$
Prove $$det(e^A) = e^{operatorname{tr}(A)}$$ for all matrices $A in mathbb{C}_{n×n}$.
linear-algebra matrices determinant trace
edited Apr 4 '13 at 12:24
Julien
38.7k358129
asked Mar 6 '13 at 15:23
JohnJohn
96115
$endgroup$
add a comment |
$begingroup$
Prove $$det(e^A) = e^{operatorname{tr}(A)}$$ for all matrices $A in mathbb{C}_{n×n}$.
linear-algebra matrices determinant trace
edited Apr 4 '13 at 12:24
Julien
38.7k358129
asked Mar 6 '13 at 15:23
JohnJohn
96115
$endgroup$
$begingroup$
All the answers so far use a triangularized form at some point. If you know that every complex square matrix is triangularizable, it brings the problem back to triangular matrices.
$endgroup$
– Julien
Mar 6 '13 at 16:04
add a comment |
$begingroup$
Prove $$det(e^A) = e^{operatorname{tr}(A)}$$ for all matrices $A in mathbb{C}_{n×n}$.
linear-algebra matrices determinant trace
edited Apr 4 '13 at 12:24
Julien
38.7k358129
asked Mar 6 '13 at 15:23
JohnJohn
96115
$endgroup$
Prove $$det(e^A) = e^{operatorname{tr}(A)}$$ for all matrices $A in mathbb{C}_{n×n}$.
linear-algebra matrices determinant trace
linear-algebra matrices determinant trace
edited Apr 4 '13 at 12:24
Julien
38.7k358129
asked Mar 6 '13 at 15:23
JohnJohn
96115
edited Apr 4 '13 at 12:24
Julien
38.7k358129
asked Mar 6 '13 at 15:23
JohnJohn
96115
edited Apr 4 '13 at 12:24
Julien
38.7k358129
edited Apr 4 '13 at 12:24
Julien
38.7k358129
edited Apr 4 '13 at 12:24
Julien
38.7k358129
38.7k358129
asked Mar 6 '13 at 15:23
JohnJohn
96115
asked Mar 6 '13 at 15:23
JohnJohn
96115
asked Mar 6 '13 at 15:23
JohnJohn
96115
96115
$begingroup$
All the answers so far use a triangularized form at some point. If you know that every complex square matrix is triangularizable, it brings the problem back to triangular matrices.
$endgroup$
– Julien
Mar 6 '13 at 16:04
add a comment |
$begingroup$
All the answers so far use a triangularized form at some point. If you know that every complex square matrix is triangularizable, it brings the problem back to triangular matrices.
$endgroup$
– Julien
Mar 6 '13 at 16:04
$begingroup$
All the answers so far use a triangularized form at some point. If you know that every complex square matrix is triangularizable, it brings the problem back to triangular matrices.
$endgroup$
– Julien
Mar 6 '13 at 16:04
$begingroup$
All the answers so far use a triangularized form at some point. If you know that every complex square matrix is triangularizable, it brings the problem back to triangular matrices.
$endgroup$
– Julien
Mar 6 '13 at 16:04
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in $M_n(mathbb{C})$.
But actually, it suffices to triangularize
$$
A=P^{-1}TP
$$
with $P$ invertible and $T$ upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in $mathbb{C}$.
Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
Observe that each $T^k$ is upper-triangular with $lambda_1^k,ldots,lambda_n^k$ on the diagonal. It follows that $e^T$ is upper triangular with $e^{lambda_1},ldots,e^{lambda_n}$ on the diagonal. So
$$
det e^T=e^{lambda_1}cdots e^{lambda_n}=e^{lambda_1+ldots+lambda_n}=e^{mbox{tr};T}
$$
Finally, observe that $mbox{tr} ;A=mbox{tr};T$, and that $P^{-1}T^kP=A^k$ for all $k$, so $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$
answered Mar 6 '13 at 15:29
JulienJulien
38.7k358129
$endgroup$
$begingroup$
Nice answer (+1)
$endgroup$
– Thomas
Mar 6 '13 at 16:09
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@1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$.
$endgroup$
– Alexander Cska
Nov 18 '17 at 15:21
add a comment |
$begingroup$
Hint:
Use that every complex matrix has a jordan normal form and that the determinant of a triangular matrix is the product of the diagonal.
use that $exp(A)=exp(S^{-1} J S ) = S^{-1} exp(J) S $
And that the trace doesn't change under transformations.
begin{align*}
det(exp(A))&=det(exp(S J S^{-1}))\
&=det(S exp(J) S^{-1})\
&=det(S) det(exp(J)) det (S^{-1})\
&=det(exp (J))\
&=prod_{i=1}^n exp(j_{ii})\
&=exp(sum_{i=1}^n{j_{ii}})\
&=exp(text{tr}J)
end{align*}
answered Mar 6 '13 at 15:26
Dominic MichaelisDominic Michaelis
17.7k43570
$endgroup$
$begingroup$
Can i have another hint. its a hard question
$endgroup$
– John
Mar 6 '13 at 15:29
$begingroup$
posted antoher hint
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
is ta tthe jordan normal form
$endgroup$
– John
Mar 6 '13 at 15:33
$begingroup$
$A$ is the normal matrix and $D$ is the jordan normal form of $A$
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
Posted a more explizit proof
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:40
|
show 13 more comments
$begingroup$
Let $f(t)= det(e^{tA})$. Then $f'(t)=D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$. But $A$ and $e^{tA}$ commute, and $^ttext{com}(e^{tA})e^{tA}=det(e^{tA}) operatorname{I}_n$. Therefore, $f'(t)=text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{text{tr}(A)t}$. For $t=1$, $det(e^{A})= e^{text{tr}(A)}$.
answered Mar 6 '13 at 16:05
SeiriosSeirios
24.1k34699
$endgroup$
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Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=det(e^{sA})I_n$.
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– Julien
Mar 6 '13 at 16:18
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@julien: Thank you, I edited my answer.
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– Seirios
Mar 6 '13 at 16:52
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Why $D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$?
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– math.n00b
Aug 16 '14 at 14:44
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The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula
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– Seirios
Aug 16 '14 at 15:17
add a comment |
$begingroup$
You can do it in these steps (still requires some work):
$quad bf (1)$ $A$ is diagonalizable
$quad bf (2)$ $A$ is nilpotent
$quad bf (3)$ $A$ is arbitrary
$bf (1)$ This shouldn't be too hard. Start with assuming that $A = CDC^{-1}$ for $D$ a diagonal matrix.
$bf (2)$ Use that every nilpotent matrix is similar to a upper triangular matrix $D$ with $0$s on the diagonal. So $A = CDC^{-1}$.
$bf (3)$ Use that every matrix can be written as the sum $A = D + N$ of a nilpotent matrix $N$ and a diagonalizable matrix $D$ and $D$ and $N$ commute. So
$$
det(e^{A}) = det(e^De^N) =det(e^{D})det(e^{N}) = e^{text{Tr}(D)}e^{text{Tr}(N)} = e^{text{Tr}(D) + text{Tr}(N)} = e^{text{Tr}(A)}.
$$
We have used here that $D$ and $N$ commute so that $e^A = e^De^N.$
answered Mar 6 '13 at 15:33
ThomasThomas
35.5k1056116
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so is this basically all i need to write out
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– John
Mar 6 '13 at 15:38
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@John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know.
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– Thomas
Mar 6 '13 at 15:39
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can you help me a bit more please regarding those details
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– John
Mar 6 '13 at 15:40
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@John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map.
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– Thomas
Mar 6 '13 at 15:42
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You need to add that $D$ and $N$ commute. Also, since you triangularize $D$, why don't you triangularize $A$ directly (which is what I did)?
$endgroup$
– Julien
Mar 6 '13 at 15:50
|
show 5 more comments
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in $M_n(mathbb{C})$.
But actually, it suffices to triangularize
$$
A=P^{-1}TP
$$
with $P$ invertible and $T$ upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in $mathbb{C}$.
Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
Observe that each $T^k$ is upper-triangular with $lambda_1^k,ldots,lambda_n^k$ on the diagonal. It follows that $e^T$ is upper triangular with $e^{lambda_1},ldots,e^{lambda_n}$ on the diagonal. So
$$
det e^T=e^{lambda_1}cdots e^{lambda_n}=e^{lambda_1+ldots+lambda_n}=e^{mbox{tr};T}
$$
Finally, observe that $mbox{tr} ;A=mbox{tr};T$, and that $P^{-1}T^kP=A^k$ for all $k$, so $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$
answered Mar 6 '13 at 15:29
JulienJulien
38.7k358129
$endgroup$
$begingroup$
Nice answer (+1)
$endgroup$
– Thomas
Mar 6 '13 at 16:09
$begingroup$
@1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$.
$endgroup$
– Alexander Cska
Nov 18 '17 at 15:21
add a comment |
$begingroup$
Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in $M_n(mathbb{C})$.
But actually, it suffices to triangularize
$$
A=P^{-1}TP
$$
with $P$ invertible and $T$ upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in $mathbb{C}$.
Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
Observe that each $T^k$ is upper-triangular with $lambda_1^k,ldots,lambda_n^k$ on the diagonal. It follows that $e^T$ is upper triangular with $e^{lambda_1},ldots,e^{lambda_n}$ on the diagonal. So
$$
det e^T=e^{lambda_1}cdots e^{lambda_n}=e^{lambda_1+ldots+lambda_n}=e^{mbox{tr};T}
$$
Finally, observe that $mbox{tr} ;A=mbox{tr};T$, and that $P^{-1}T^kP=A^k$ for all $k$, so $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$
answered Mar 6 '13 at 15:29
JulienJulien
38.7k358129
$endgroup$
$begingroup$
Nice answer (+1)
$endgroup$
– Thomas
Mar 6 '13 at 16:09
$begingroup$
@1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$.
$endgroup$
– Alexander Cska
Nov 18 '17 at 15:21
add a comment |
$begingroup$
Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in $M_n(mathbb{C})$.
But actually, it suffices to triangularize
$$
A=P^{-1}TP
$$
with $P$ invertible and $T$ upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in $mathbb{C}$.
Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
Observe that each $T^k$ is upper-triangular with $lambda_1^k,ldots,lambda_n^k$ on the diagonal. It follows that $e^T$ is upper triangular with $e^{lambda_1},ldots,e^{lambda_n}$ on the diagonal. So
$$
det e^T=e^{lambda_1}cdots e^{lambda_n}=e^{lambda_1+ldots+lambda_n}=e^{mbox{tr};T}
$$
Finally, observe that $mbox{tr} ;A=mbox{tr};T$, and that $P^{-1}T^kP=A^k$ for all $k$, so $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$
answered Mar 6 '13 at 15:29
JulienJulien
38.7k358129
$endgroup$
Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in $M_n(mathbb{C})$.
But actually, it suffices to triangularize
$$
A=P^{-1}TP
$$
with $P$ invertible and $T$ upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in $mathbb{C}$.
Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
Observe that each $T^k$ is upper-triangular with $lambda_1^k,ldots,lambda_n^k$ on the diagonal. It follows that $e^T$ is upper triangular with $e^{lambda_1},ldots,e^{lambda_n}$ on the diagonal. So
$$
det e^T=e^{lambda_1}cdots e^{lambda_n}=e^{lambda_1+ldots+lambda_n}=e^{mbox{tr};T}
$$
Finally, observe that $mbox{tr} ;A=mbox{tr};T$, and that $P^{-1}T^kP=A^k$ for all $k$, so $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$
answered Mar 6 '13 at 15:29
JulienJulien
38.7k358129
edited Jan 11 at 18:56
user337254
answered Mar 6 '13 at 15:29
JulienJulien
38.7k358129
answered Mar 6 '13 at 15:29
JulienJulien
38.7k358129
answered Mar 6 '13 at 15:29
JulienJulien
38.7k358129
38.7k358129
$begingroup$
Nice answer (+1)
$endgroup$
– Thomas
Mar 6 '13 at 16:09
$begingroup$
@1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$.
$endgroup$
– Alexander Cska
Nov 18 '17 at 15:21
add a comment |
$begingroup$
Nice answer (+1)
$endgroup$
– Thomas
Mar 6 '13 at 16:09
$begingroup$
@1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$.
$endgroup$
– Alexander Cska
Nov 18 '17 at 15:21
$begingroup$
Nice answer (+1)
$endgroup$
– Thomas
Mar 6 '13 at 16:09
$begingroup$
Nice answer (+1)
$endgroup$
– Thomas
Mar 6 '13 at 16:09
$begingroup$
@1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$.
$endgroup$
– Alexander Cska
Nov 18 '17 at 15:21
$begingroup$
@1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$.
$endgroup$
– Alexander Cska
Nov 18 '17 at 15:21
add a comment |
$begingroup$
Hint:
Use that every complex matrix has a jordan normal form and that the determinant of a triangular matrix is the product of the diagonal.
use that $exp(A)=exp(S^{-1} J S ) = S^{-1} exp(J) S $
And that the trace doesn't change under transformations.
begin{align*}
det(exp(A))&=det(exp(S J S^{-1}))\
&=det(S exp(J) S^{-1})\
&=det(S) det(exp(J)) det (S^{-1})\
&=det(exp (J))\
&=prod_{i=1}^n exp(j_{ii})\
&=exp(sum_{i=1}^n{j_{ii}})\
&=exp(text{tr}J)
end{align*}
answered Mar 6 '13 at 15:26
Dominic MichaelisDominic Michaelis
17.7k43570
$endgroup$
$begingroup$
Can i have another hint. its a hard question
$endgroup$
– John
Mar 6 '13 at 15:29
$begingroup$
posted antoher hint
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
is ta tthe jordan normal form
$endgroup$
– John
Mar 6 '13 at 15:33
$begingroup$
$A$ is the normal matrix and $D$ is the jordan normal form of $A$
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
Posted a more explizit proof
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:40
|
show 13 more comments
$begingroup$
Hint:
Use that every complex matrix has a jordan normal form and that the determinant of a triangular matrix is the product of the diagonal.
use that $exp(A)=exp(S^{-1} J S ) = S^{-1} exp(J) S $
And that the trace doesn't change under transformations.
begin{align*}
det(exp(A))&=det(exp(S J S^{-1}))\
&=det(S exp(J) S^{-1})\
&=det(S) det(exp(J)) det (S^{-1})\
&=det(exp (J))\
&=prod_{i=1}^n exp(j_{ii})\
&=exp(sum_{i=1}^n{j_{ii}})\
&=exp(text{tr}J)
end{align*}
answered Mar 6 '13 at 15:26
Dominic MichaelisDominic Michaelis
17.7k43570
$endgroup$
$begingroup$
Can i have another hint. its a hard question
$endgroup$
– John
Mar 6 '13 at 15:29
$begingroup$
posted antoher hint
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
is ta tthe jordan normal form
$endgroup$
– John
Mar 6 '13 at 15:33
$begingroup$
$A$ is the normal matrix and $D$ is the jordan normal form of $A$
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
Posted a more explizit proof
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:40
|
show 13 more comments
$begingroup$
Hint:
Use that every complex matrix has a jordan normal form and that the determinant of a triangular matrix is the product of the diagonal.
use that $exp(A)=exp(S^{-1} J S ) = S^{-1} exp(J) S $
And that the trace doesn't change under transformations.
begin{align*}
det(exp(A))&=det(exp(S J S^{-1}))\
&=det(S exp(J) S^{-1})\
&=det(S) det(exp(J)) det (S^{-1})\
&=det(exp (J))\
&=prod_{i=1}^n exp(j_{ii})\
&=exp(sum_{i=1}^n{j_{ii}})\
&=exp(text{tr}J)
end{align*}
answered Mar 6 '13 at 15:26
Dominic MichaelisDominic Michaelis
17.7k43570
$endgroup$
Hint:
Use that every complex matrix has a jordan normal form and that the determinant of a triangular matrix is the product of the diagonal.
use that $exp(A)=exp(S^{-1} J S ) = S^{-1} exp(J) S $
And that the trace doesn't change under transformations.
begin{align*}
det(exp(A))&=det(exp(S J S^{-1}))\
&=det(S exp(J) S^{-1})\
&=det(S) det(exp(J)) det (S^{-1})\
&=det(exp (J))\
&=prod_{i=1}^n exp(j_{ii})\
&=exp(sum_{i=1}^n{j_{ii}})\
&=exp(text{tr}J)
end{align*}
answered Mar 6 '13 at 15:26
Dominic MichaelisDominic Michaelis
17.7k43570
edited Mar 6 '13 at 16:14
answered Mar 6 '13 at 15:26
Dominic MichaelisDominic Michaelis
17.7k43570
answered Mar 6 '13 at 15:26
Dominic MichaelisDominic Michaelis
17.7k43570
answered Mar 6 '13 at 15:26
Dominic MichaelisDominic Michaelis
17.7k43570
17.7k43570
$begingroup$
Can i have another hint. its a hard question
$endgroup$
– John
Mar 6 '13 at 15:29
$begingroup$
posted antoher hint
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
is ta tthe jordan normal form
$endgroup$
– John
Mar 6 '13 at 15:33
$begingroup$
$A$ is the normal matrix and $D$ is the jordan normal form of $A$
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
Posted a more explizit proof
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:40
|
show 13 more comments
$begingroup$
Can i have another hint. its a hard question
$endgroup$
– John
Mar 6 '13 at 15:29
$begingroup$
posted antoher hint
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
is ta tthe jordan normal form
$endgroup$
– John
Mar 6 '13 at 15:33
$begingroup$
$A$ is the normal matrix and $D$ is the jordan normal form of $A$
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
Posted a more explizit proof
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:40
$begingroup$
Can i have another hint. its a hard question
$endgroup$
– John
Mar 6 '13 at 15:29
$begingroup$
Can i have another hint. its a hard question
$endgroup$
– John
Mar 6 '13 at 15:29
$begingroup$
posted antoher hint
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
posted antoher hint
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
is ta tthe jordan normal form
$endgroup$
– John
Mar 6 '13 at 15:33
$begingroup$
is ta tthe jordan normal form
$endgroup$
– John
Mar 6 '13 at 15:33
$begingroup$
$A$ is the normal matrix and $D$ is the jordan normal form of $A$
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
$begingroup$
$A$ is the normal matrix and $D$ is the jordan normal form of $A$
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:33
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Posted a more explizit proof
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– Dominic Michaelis
Mar 6 '13 at 15:40
$begingroup$
Posted a more explizit proof
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:40
|
show 13 more comments
$begingroup$
Let $f(t)= det(e^{tA})$. Then $f'(t)=D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$. But $A$ and $e^{tA}$ commute, and $^ttext{com}(e^{tA})e^{tA}=det(e^{tA}) operatorname{I}_n$. Therefore, $f'(t)=text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{text{tr}(A)t}$. For $t=1$, $det(e^{A})= e^{text{tr}(A)}$.
answered Mar 6 '13 at 16:05
SeiriosSeirios
24.1k34699
$endgroup$
$begingroup$
Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=det(e^{sA})I_n$.
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– Julien
Mar 6 '13 at 16:18
$begingroup$
@julien: Thank you, I edited my answer.
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– Seirios
Mar 6 '13 at 16:52
$begingroup$
Why $D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$?
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– math.n00b
Aug 16 '14 at 14:44
$begingroup$
The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula
$endgroup$
– Seirios
Aug 16 '14 at 15:17
add a comment |
$begingroup$
Let $f(t)= det(e^{tA})$. Then $f'(t)=D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$. But $A$ and $e^{tA}$ commute, and $^ttext{com}(e^{tA})e^{tA}=det(e^{tA}) operatorname{I}_n$. Therefore, $f'(t)=text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{text{tr}(A)t}$. For $t=1$, $det(e^{A})= e^{text{tr}(A)}$.
answered Mar 6 '13 at 16:05
SeiriosSeirios
24.1k34699
$endgroup$
$begingroup$
Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=det(e^{sA})I_n$.
$endgroup$
– Julien
Mar 6 '13 at 16:18
$begingroup$
@julien: Thank you, I edited my answer.
$endgroup$
– Seirios
Mar 6 '13 at 16:52
$begingroup$
Why $D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$?
$endgroup$
– math.n00b
Aug 16 '14 at 14:44
$begingroup$
The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula
$endgroup$
– Seirios
Aug 16 '14 at 15:17
add a comment |
$begingroup$
Let $f(t)= det(e^{tA})$. Then $f'(t)=D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$. But $A$ and $e^{tA}$ commute, and $^ttext{com}(e^{tA})e^{tA}=det(e^{tA}) operatorname{I}_n$. Therefore, $f'(t)=text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{text{tr}(A)t}$. For $t=1$, $det(e^{A})= e^{text{tr}(A)}$.
answered Mar 6 '13 at 16:05
SeiriosSeirios
24.1k34699
$endgroup$
Let $f(t)= det(e^{tA})$. Then $f'(t)=D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$. But $A$ and $e^{tA}$ commute, and $^ttext{com}(e^{tA})e^{tA}=det(e^{tA}) operatorname{I}_n$. Therefore, $f'(t)=text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{text{tr}(A)t}$. For $t=1$, $det(e^{A})= e^{text{tr}(A)}$.
answered Mar 6 '13 at 16:05
SeiriosSeirios
24.1k34699
edited Mar 6 '13 at 16:52
answered Mar 6 '13 at 16:05
SeiriosSeirios
24.1k34699
answered Mar 6 '13 at 16:05
SeiriosSeirios
24.1k34699
answered Mar 6 '13 at 16:05
SeiriosSeirios
24.1k34699
24.1k34699
$begingroup$
Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=det(e^{sA})I_n$.
$endgroup$
– Julien
Mar 6 '13 at 16:18
$begingroup$
@julien: Thank you, I edited my answer.
$endgroup$
– Seirios
Mar 6 '13 at 16:52
$begingroup$
Why $D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$?
$endgroup$
– math.n00b
Aug 16 '14 at 14:44
$begingroup$
The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula
$endgroup$
– Seirios
Aug 16 '14 at 15:17
add a comment |
$begingroup$
Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=det(e^{sA})I_n$.
$endgroup$
– Julien
Mar 6 '13 at 16:18
$begingroup$
@julien: Thank you, I edited my answer.
$endgroup$
– Seirios
Mar 6 '13 at 16:52
$begingroup$
Why $D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$?
$endgroup$
– math.n00b
Aug 16 '14 at 14:44
$begingroup$
The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula
$endgroup$
– Seirios
Aug 16 '14 at 15:17
$begingroup$
Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=det(e^{sA})I_n$.
$endgroup$
– Julien
Mar 6 '13 at 16:18
$begingroup$
Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=det(e^{sA})I_n$.
$endgroup$
– Julien
Mar 6 '13 at 16:18
$begingroup$
@julien: Thank you, I edited my answer.
$endgroup$
– Seirios
Mar 6 '13 at 16:52
$begingroup$
@julien: Thank you, I edited my answer.
$endgroup$
– Seirios
Mar 6 '13 at 16:52
$begingroup$
Why $D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$?
$endgroup$
– math.n00b
Aug 16 '14 at 14:44
$begingroup$
Why $D det(e^{tA}) cdot Ae^{tA}=text{tr} left(^t text{com}(e^{tA})Ae^{tA} right)$?
$endgroup$
– math.n00b
Aug 16 '14 at 14:44
$begingroup$
The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula
$endgroup$
– Seirios
Aug 16 '14 at 15:17
$begingroup$
The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula
$endgroup$
– Seirios
Aug 16 '14 at 15:17
add a comment |
$begingroup$
You can do it in these steps (still requires some work):
$quad bf (1)$ $A$ is diagonalizable
$quad bf (2)$ $A$ is nilpotent
$quad bf (3)$ $A$ is arbitrary
$bf (1)$ This shouldn't be too hard. Start with assuming that $A = CDC^{-1}$ for $D$ a diagonal matrix.
$bf (2)$ Use that every nilpotent matrix is similar to a upper triangular matrix $D$ with $0$s on the diagonal. So $A = CDC^{-1}$.
$bf (3)$ Use that every matrix can be written as the sum $A = D + N$ of a nilpotent matrix $N$ and a diagonalizable matrix $D$ and $D$ and $N$ commute. So
$$
det(e^{A}) = det(e^De^N) =det(e^{D})det(e^{N}) = e^{text{Tr}(D)}e^{text{Tr}(N)} = e^{text{Tr}(D) + text{Tr}(N)} = e^{text{Tr}(A)}.
$$
We have used here that $D$ and $N$ commute so that $e^A = e^De^N.$
answered Mar 6 '13 at 15:33
ThomasThomas
35.5k1056116
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$begingroup$
so is this basically all i need to write out
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– John
Mar 6 '13 at 15:38
$begingroup$
@John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know.
$endgroup$
– Thomas
Mar 6 '13 at 15:39
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can you help me a bit more please regarding those details
$endgroup$
– John
Mar 6 '13 at 15:40
$begingroup$
@John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map.
$endgroup$
– Thomas
Mar 6 '13 at 15:42
$begingroup$
You need to add that $D$ and $N$ commute. Also, since you triangularize $D$, why don't you triangularize $A$ directly (which is what I did)?
$endgroup$
– Julien
Mar 6 '13 at 15:50
|
show 5 more comments
$begingroup$
You can do it in these steps (still requires some work):
$quad bf (1)$ $A$ is diagonalizable
$quad bf (2)$ $A$ is nilpotent
$quad bf (3)$ $A$ is arbitrary
$bf (1)$ This shouldn't be too hard. Start with assuming that $A = CDC^{-1}$ for $D$ a diagonal matrix.
$bf (2)$ Use that every nilpotent matrix is similar to a upper triangular matrix $D$ with $0$s on the diagonal. So $A = CDC^{-1}$.
$bf (3)$ Use that every matrix can be written as the sum $A = D + N$ of a nilpotent matrix $N$ and a diagonalizable matrix $D$ and $D$ and $N$ commute. So
$$
det(e^{A}) = det(e^De^N) =det(e^{D})det(e^{N}) = e^{text{Tr}(D)}e^{text{Tr}(N)} = e^{text{Tr}(D) + text{Tr}(N)} = e^{text{Tr}(A)}.
$$
We have used here that $D$ and $N$ commute so that $e^A = e^De^N.$
answered Mar 6 '13 at 15:33
ThomasThomas
35.5k1056116
$endgroup$
$begingroup$
so is this basically all i need to write out
$endgroup$
– John
Mar 6 '13 at 15:38
$begingroup$
@John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know.
$endgroup$
– Thomas
Mar 6 '13 at 15:39
$begingroup$
can you help me a bit more please regarding those details
$endgroup$
– John
Mar 6 '13 at 15:40
$begingroup$
@John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map.
$endgroup$
– Thomas
Mar 6 '13 at 15:42
$begingroup$
You need to add that $D$ and $N$ commute. Also, since you triangularize $D$, why don't you triangularize $A$ directly (which is what I did)?
$endgroup$
– Julien
Mar 6 '13 at 15:50
|
show 5 more comments
$begingroup$
You can do it in these steps (still requires some work):
$quad bf (1)$ $A$ is diagonalizable
$quad bf (2)$ $A$ is nilpotent
$quad bf (3)$ $A$ is arbitrary
$bf (1)$ This shouldn't be too hard. Start with assuming that $A = CDC^{-1}$ for $D$ a diagonal matrix.
$bf (2)$ Use that every nilpotent matrix is similar to a upper triangular matrix $D$ with $0$s on the diagonal. So $A = CDC^{-1}$.
$bf (3)$ Use that every matrix can be written as the sum $A = D + N$ of a nilpotent matrix $N$ and a diagonalizable matrix $D$ and $D$ and $N$ commute. So
$$
det(e^{A}) = det(e^De^N) =det(e^{D})det(e^{N}) = e^{text{Tr}(D)}e^{text{Tr}(N)} = e^{text{Tr}(D) + text{Tr}(N)} = e^{text{Tr}(A)}.
$$
We have used here that $D$ and $N$ commute so that $e^A = e^De^N.$
answered Mar 6 '13 at 15:33
ThomasThomas
35.5k1056116
$endgroup$
You can do it in these steps (still requires some work):
$quad bf (1)$ $A$ is diagonalizable
$quad bf (2)$ $A$ is nilpotent
$quad bf (3)$ $A$ is arbitrary
$bf (1)$ This shouldn't be too hard. Start with assuming that $A = CDC^{-1}$ for $D$ a diagonal matrix.
$bf (2)$ Use that every nilpotent matrix is similar to a upper triangular matrix $D$ with $0$s on the diagonal. So $A = CDC^{-1}$.
$bf (3)$ Use that every matrix can be written as the sum $A = D + N$ of a nilpotent matrix $N$ and a diagonalizable matrix $D$ and $D$ and $N$ commute. So
$$
det(e^{A}) = det(e^De^N) =det(e^{D})det(e^{N}) = e^{text{Tr}(D)}e^{text{Tr}(N)} = e^{text{Tr}(D) + text{Tr}(N)} = e^{text{Tr}(A)}.
$$
We have used here that $D$ and $N$ commute so that $e^A = e^De^N.$
answered Mar 6 '13 at 15:33
ThomasThomas
35.5k1056116
edited Mar 6 '13 at 15:52
answered Mar 6 '13 at 15:33
ThomasThomas
35.5k1056116
answered Mar 6 '13 at 15:33
ThomasThomas
35.5k1056116
answered Mar 6 '13 at 15:33
ThomasThomas
35.5k1056116
35.5k1056116
$begingroup$
so is this basically all i need to write out
$endgroup$
– John
Mar 6 '13 at 15:38
$begingroup$
@John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know.
$endgroup$
– Thomas
Mar 6 '13 at 15:39
$begingroup$
can you help me a bit more please regarding those details
$endgroup$
– John
Mar 6 '13 at 15:40
$begingroup$
@John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map.
$endgroup$
– Thomas
Mar 6 '13 at 15:42
$begingroup$
You need to add that $D$ and $N$ commute. Also, since you triangularize $D$, why don't you triangularize $A$ directly (which is what I did)?
$endgroup$
– Julien
Mar 6 '13 at 15:50
|
show 5 more comments
$begingroup$
so is this basically all i need to write out
$endgroup$
– John
Mar 6 '13 at 15:38
$begingroup$
@John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know.
$endgroup$
– Thomas
Mar 6 '13 at 15:39
$begingroup$
can you help me a bit more please regarding those details
$endgroup$
– John
Mar 6 '13 at 15:40
$begingroup$
@John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map.
$endgroup$
– Thomas
Mar 6 '13 at 15:42
$begingroup$
You need to add that $D$ and $N$ commute. Also, since you triangularize $D$, why don't you triangularize $A$ directly (which is what I did)?
$endgroup$
– Julien
Mar 6 '13 at 15:50
$begingroup$
so is this basically all i need to write out
$endgroup$
– John
Mar 6 '13 at 15:38
$begingroup$
so is this basically all i need to write out
$endgroup$
– John
Mar 6 '13 at 15:38
$begingroup$
@John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know.
$endgroup$
– Thomas
Mar 6 '13 at 15:39
$begingroup$
@John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know.
$endgroup$
– Thomas
Mar 6 '13 at 15:39
$begingroup$
can you help me a bit more please regarding those details
$endgroup$
– John
Mar 6 '13 at 15:40
$begingroup$
can you help me a bit more please regarding those details
$endgroup$
– John
Mar 6 '13 at 15:40
$begingroup$
@John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map.
$endgroup$
– Thomas
Mar 6 '13 at 15:42
$begingroup$
@John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map.
$endgroup$
– Thomas
Mar 6 '13 at 15:42
$begingroup$
You need to add that $D$ and $N$ commute. Also, since you triangularize $D$, why don't you triangularize $A$ directly (which is what I did)?
$endgroup$
– Julien
Mar 6 '13 at 15:50
$begingroup$
You need to add that $D$ and $N$ commute. Also, since you triangularize $D$, why don't you triangularize $A$ directly (which is what I did)?
$endgroup$
– Julien
Mar 6 '13 at 15:50
|
show 5 more comments
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$begingroup$
All the answers so far use a triangularized form at some point. If you know that every complex square matrix is triangularizable, it brings the problem back to triangular matrices.
$endgroup$
– Julien
Mar 6 '13 at 16:04