trace and operator norm of $exp^A$
up vote
3
down vote
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If $A$ is an n by n complex matrix,
1.How to compute $tr(exp^A)$ .Can we use the Taylor expansion as following:
$exp^A=sum_{k=0}^{infty}frac{A^k}{k!},$then $tr(exp^A)=sum_{k=0}^{infty}tr(frac{A^k}{k!})$.
2.How to compute the operator norm of $exp^A$?
linear-algebra complex-analysis operator-theory operator-algebras c-star-algebras
asked Nov 17 at 17:26
mathrookie
724512
add a comment |
up vote
3
down vote
favorite
If $A$ is an n by n complex matrix,
1.How to compute $tr(exp^A)$ .Can we use the Taylor expansion as following:
$exp^A=sum_{k=0}^{infty}frac{A^k}{k!},$then $tr(exp^A)=sum_{k=0}^{infty}tr(frac{A^k}{k!})$.
2.How to compute the operator norm of $exp^A$?
linear-algebra complex-analysis operator-theory operator-algebras c-star-algebras
asked Nov 17 at 17:26
mathrookie
724512
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If $A$ is an n by n complex matrix,
1.How to compute $tr(exp^A)$ .Can we use the Taylor expansion as following:
$exp^A=sum_{k=0}^{infty}frac{A^k}{k!},$then $tr(exp^A)=sum_{k=0}^{infty}tr(frac{A^k}{k!})$.
2.How to compute the operator norm of $exp^A$?
linear-algebra complex-analysis operator-theory operator-algebras c-star-algebras
asked Nov 17 at 17:26
mathrookie
724512
If $A$ is an n by n complex matrix,
1.How to compute $tr(exp^A)$ .Can we use the Taylor expansion as following:
$exp^A=sum_{k=0}^{infty}frac{A^k}{k!},$then $tr(exp^A)=sum_{k=0}^{infty}tr(frac{A^k}{k!})$.
2.How to compute the operator norm of $exp^A$?
linear-algebra complex-analysis operator-theory operator-algebras c-star-algebras
linear-algebra complex-analysis operator-theory operator-algebras c-star-algebras
asked Nov 17 at 17:26
mathrookie
724512
asked Nov 17 at 17:26
mathrookie
724512
asked Nov 17 at 17:26
mathrookie
724512
asked Nov 17 at 17:26
mathrookie
724512
asked Nov 17 at 17:26
mathrookie
724512
724512
add a comment |
add a comment |
1 Answer
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up vote
2
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The trace of an finite-dimensional operator is defined the sum of its diagonal elements, which in turn equals the sum of its eigenvalues. So let ${lambda_j in mathbb{C} | jin{1,ldots,n}}$ the set of eigenvalues of $A$ then we have
$$mathrm{tr}{A}=sum_{j=1}^n A_{jj} = sum_{j=1}^n lambda_{j}$$. Because the trace is invariant w.r.t. similarity transforms of $A$, we have $$mathrm{tr}{(exp{A})} = sum_{j=1}^n exp{lambda_{j}} .$$
answered Nov 17 at 17:52
M1183
943
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The trace of an finite-dimensional operator is defined the sum of its diagonal elements, which in turn equals the sum of its eigenvalues. So let ${lambda_j in mathbb{C} | jin{1,ldots,n}}$ the set of eigenvalues of $A$ then we have
$$mathrm{tr}{A}=sum_{j=1}^n A_{jj} = sum_{j=1}^n lambda_{j}$$. Because the trace is invariant w.r.t. similarity transforms of $A$, we have $$mathrm{tr}{(exp{A})} = sum_{j=1}^n exp{lambda_{j}} .$$
answered Nov 17 at 17:52
M1183
943
add a comment |
up vote
2
down vote
The trace of an finite-dimensional operator is defined the sum of its diagonal elements, which in turn equals the sum of its eigenvalues. So let ${lambda_j in mathbb{C} | jin{1,ldots,n}}$ the set of eigenvalues of $A$ then we have
$$mathrm{tr}{A}=sum_{j=1}^n A_{jj} = sum_{j=1}^n lambda_{j}$$. Because the trace is invariant w.r.t. similarity transforms of $A$, we have $$mathrm{tr}{(exp{A})} = sum_{j=1}^n exp{lambda_{j}} .$$
answered Nov 17 at 17:52
M1183
943
add a comment |
up vote
2
down vote
up vote
2
down vote
The trace of an finite-dimensional operator is defined the sum of its diagonal elements, which in turn equals the sum of its eigenvalues. So let ${lambda_j in mathbb{C} | jin{1,ldots,n}}$ the set of eigenvalues of $A$ then we have
$$mathrm{tr}{A}=sum_{j=1}^n A_{jj} = sum_{j=1}^n lambda_{j}$$. Because the trace is invariant w.r.t. similarity transforms of $A$, we have $$mathrm{tr}{(exp{A})} = sum_{j=1}^n exp{lambda_{j}} .$$
answered Nov 17 at 17:52
M1183
943
The trace of an finite-dimensional operator is defined the sum of its diagonal elements, which in turn equals the sum of its eigenvalues. So let ${lambda_j in mathbb{C} | jin{1,ldots,n}}$ the set of eigenvalues of $A$ then we have
$$mathrm{tr}{A}=sum_{j=1}^n A_{jj} = sum_{j=1}^n lambda_{j}$$. Because the trace is invariant w.r.t. similarity transforms of $A$, we have $$mathrm{tr}{(exp{A})} = sum_{j=1}^n exp{lambda_{j}} .$$
answered Nov 17 at 17:52
M1183
943
answered Nov 17 at 17:52
M1183
943
answered Nov 17 at 17:52
M1183
943
answered Nov 17 at 17:52
M1183
943
943
add a comment |
add a comment |
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