Proof that Gauss- Seidel iteration method converges for any initial x if the matrix is self-adjoint and...
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I am trying to create a direct proof that if the matrix A is self-adjoint and positive-definite, then the Gauss-Seidel iteration converges for any initial ${bf x}_{0}$
I think I need to prove $rho({bf T}) < 1$ so
I begin by letting my ${bf A} = {bf D} - {bf L} - {bf L}^*$ and ${bf T}_{G}= ({bf D}-{bf L})^{-1}{bf L}^*$ where
I can use the fact that I can multiply ${bf D}^{1/2}$ to both sides of ${bf T}_{G}$ since Eigenvalues are insensitive giving me the following:
${bf T}_1 = {bf D}^{1/2}{bf T}_{G}{bf D}^{-1/2} = ({bf I} -
{bf L}_{1})^{-1}{bf L}_{1}^*$, where ${bf L}_{1} =
{bf D}^{-1/2}{bf L}{bf D}^{-1/2}$.
I'm stuck as to what to do next. Any help is appreciated!
linear-algebra numerical-methods numerical-linear-algebra positive-definite
asked Nov 29 at 4:38
Jennifer Laundey
84
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I am trying to create a direct proof that if the matrix A is self-adjoint and positive-definite, then the Gauss-Seidel iteration converges for any initial ${bf x}_{0}$
I think I need to prove $rho({bf T}) < 1$ so
I begin by letting my ${bf A} = {bf D} - {bf L} - {bf L}^*$ and ${bf T}_{G}= ({bf D}-{bf L})^{-1}{bf L}^*$ where
I can use the fact that I can multiply ${bf D}^{1/2}$ to both sides of ${bf T}_{G}$ since Eigenvalues are insensitive giving me the following:
${bf T}_1 = {bf D}^{1/2}{bf T}_{G}{bf D}^{-1/2} = ({bf I} -
{bf L}_{1})^{-1}{bf L}_{1}^*$, where ${bf L}_{1} =
{bf D}^{-1/2}{bf L}{bf D}^{-1/2}$.
I'm stuck as to what to do next. Any help is appreciated!
linear-algebra numerical-methods numerical-linear-algebra positive-definite
asked Nov 29 at 4:38
Jennifer Laundey
84
1
Check, e.g., Theorem 4.5 here
– Algebraic Pavel
Nov 29 at 9:59
add a comment |
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0
down vote
favorite
up vote
0
down vote
favorite
I am trying to create a direct proof that if the matrix A is self-adjoint and positive-definite, then the Gauss-Seidel iteration converges for any initial ${bf x}_{0}$
I think I need to prove $rho({bf T}) < 1$ so
I begin by letting my ${bf A} = {bf D} - {bf L} - {bf L}^*$ and ${bf T}_{G}= ({bf D}-{bf L})^{-1}{bf L}^*$ where
I can use the fact that I can multiply ${bf D}^{1/2}$ to both sides of ${bf T}_{G}$ since Eigenvalues are insensitive giving me the following:
${bf T}_1 = {bf D}^{1/2}{bf T}_{G}{bf D}^{-1/2} = ({bf I} -
{bf L}_{1})^{-1}{bf L}_{1}^*$, where ${bf L}_{1} =
{bf D}^{-1/2}{bf L}{bf D}^{-1/2}$.
I'm stuck as to what to do next. Any help is appreciated!
linear-algebra numerical-methods numerical-linear-algebra positive-definite
asked Nov 29 at 4:38
Jennifer Laundey
84
I am trying to create a direct proof that if the matrix A is self-adjoint and positive-definite, then the Gauss-Seidel iteration converges for any initial ${bf x}_{0}$
I think I need to prove $rho({bf T}) < 1$ so
I begin by letting my ${bf A} = {bf D} - {bf L} - {bf L}^*$ and ${bf T}_{G}= ({bf D}-{bf L})^{-1}{bf L}^*$ where
I can use the fact that I can multiply ${bf D}^{1/2}$ to both sides of ${bf T}_{G}$ since Eigenvalues are insensitive giving me the following:
${bf T}_1 = {bf D}^{1/2}{bf T}_{G}{bf D}^{-1/2} = ({bf I} -
{bf L}_{1})^{-1}{bf L}_{1}^*$, where ${bf L}_{1} =
{bf D}^{-1/2}{bf L}{bf D}^{-1/2}$.
I'm stuck as to what to do next. Any help is appreciated!
linear-algebra numerical-methods numerical-linear-algebra positive-definite
linear-algebra numerical-methods numerical-linear-algebra positive-definite
asked Nov 29 at 4:38
Jennifer Laundey
84
asked Nov 29 at 4:38
Jennifer Laundey
84
edited Nov 30 at 3:44
asked Nov 29 at 4:38
Jennifer Laundey
84
asked Nov 29 at 4:38
Jennifer Laundey
84
asked Nov 29 at 4:38
Jennifer Laundey
84
84
1
Check, e.g., Theorem 4.5 here
– Algebraic Pavel
Nov 29 at 9:59
add a comment |
1
Check, e.g., Theorem 4.5 here
– Algebraic Pavel
Nov 29 at 9:59
1
1
Check, e.g., Theorem 4.5 here
– Algebraic Pavel
Nov 29 at 9:59
Check, e.g., Theorem 4.5 here
– Algebraic Pavel
Nov 29 at 9:59
add a comment |
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Check, e.g., Theorem 4.5 here
– Algebraic Pavel
Nov 29 at 9:59