How to prove $det(e^A) = e^{operatorname{tr}(A)}$?












19












$begingroup$


Prove $$det(e^A) = e^{operatorname{tr}(A)}$$ for all matrices $A in mathbb{C}_{n×n}$.






linear-algebra matrices determinant trace







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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


















19












$begingroup$


Prove $$det(e^A) = e^{operatorname{tr}(A)}$$ for all matrices $A in mathbb{C}_{n×n}$.






linear-algebra matrices determinant trace







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















19












19








19


9



$begingroup$


Prove $$det(e^A) = e^{operatorname{tr}(A)}$$ for all matrices $A in mathbb{C}_{n×n}$.






linear-algebra matrices determinant trace







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 4 '13 at 12:24









Julien

38.6k358129




38.6k358129










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




















  • $begingroup$
    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












4 Answers
4






active

oldest

votes


















22












$begingroup$

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 $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















19












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















12












$begingroup$

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 $t=1$, $det(e^{A})= e^{text{tr}(A)}$.






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$endgroup$













  • $begingroup$
    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



















2












$begingroup$

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 $e^A = e^De^N.$






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$endgroup$













  • $begingroup$
    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













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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









22












$begingroup$

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 $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















22












$begingroup$

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 $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














22












22








22





$begingroup$

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 $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$






share|cite|improve this answer











$endgroup$



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 $$P^{-1}e^TP=e^Aqquad Rightarrowqquad det (e^T)=det (P^{-1}e^TP)=det(e^A).$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 18:56







user337254

















answered Mar 6 '13 at 15:29









JulienJulien

38.6k358129




38.6k358129












  • $begingroup$
    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










  • $begingroup$
    @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











19












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















19












$begingroup$

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*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














19












19








19





$begingroup$

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*}






share|cite|improve this answer











$endgroup$



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*}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 6 '13 at 16:14

























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


















  • $begingroup$
    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












$begingroup$
Posted a more explizit proof
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:40




$begingroup$
Posted a more explizit proof
$endgroup$
– Dominic Michaelis
Mar 6 '13 at 15:40











12












$begingroup$

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 $t=1$, $det(e^{A})= e^{text{tr}(A)}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















12












$begingroup$

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 $t=1$, $det(e^{A})= e^{text{tr}(A)}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














12












12








12





$begingroup$

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 $t=1$, $det(e^{A})= e^{text{tr}(A)}$.






share|cite|improve this answer











$endgroup$



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 $t=1$, $det(e^{A})= e^{text{tr}(A)}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 6 '13 at 16:52

























answered Mar 6 '13 at 16:05









SeiriosSeirios

24k34699




24k34699












  • $begingroup$
    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












  • $begingroup$
    @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











2












$begingroup$

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 $e^A = e^De^N.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















2












$begingroup$

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 $e^A = e^De^N.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












2








2





$begingroup$

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 $e^A = e^De^N.$






share|cite|improve this answer











$endgroup$



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 $e^A = e^De^N.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 6 '13 at 15:52

























answered Mar 6 '13 at 15:33









ThomasThomas

35.4k1056116




35.4k1056116












  • $begingroup$
    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










  • $begingroup$
    @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




















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