Finding the Upper Bound of the difference between the Inverse of the 2 matrix
$begingroup$
Given that $ K = A^{-1} - B^{-1} + A^{-1} - A^{-1}BA^{-1}$, we need to find the upper bound of $K$ where matrix $A = C + Irho$ and $ B = C_{x} + Irho $ has dimension $ntimes n$, $ C = RDR^{T}$ where $D$ is a diagonal matrix whose eigenvalues are in descending order and $C$ is a matrix of full rank and $C_{x}$ is an approximated matrix and $rho$ is a constant. If we partition $C$ as
begin{equation}
C =
left({begin{array}{cc} R_{m} & R_{(n-m)} end{array}}right)
left(begin{array}{cc} D_{mm} & 0\ 0 & D_{(n-m)(n-m)} end{array}right)
left(begin{array}{c} R'_m \ R'_{(n-m)} end{array}right)
end{equation}
Where by Eckart - Young theorem $ C_{x} = R_{m}D_{m}R_{m}^T$.
Is it right to proceed this way?
First, break $K$ into 2 parts, thus $ A^{-1} - B^{-1} $ and $ A^{-1} - A^{-1}BA^{-1} $ and using the fact that
$$ parallel A^{-1} - B^{-1} + A^{-1} - A^{-1}BA^{-1} parallel_F ;leq ;parallel A^{-1} - B^{-1}parallel_F + parallel A^{-1} - A^{-1}BA^{-1} parallel_{F} $$
And considering first
begin{eqnarray}
parallel A^{-1} - B^{-1}parallel_F ^2 & = & parallel ( C + rho)^{-1} - (C_x + rho)^{-1}parallel_F ^2 \
& = & parallel rho^{-2}R_{n-m}(D_{n-m} + rho^{-1})^{-1}R_{n-m}^Tparallel_F^2 quad text { That is by substitution }\
& = & rho^{-2}{tr bigg ( R_{n-m}(D_{n-m}^{-2} +rho^{-1})^{-2}R_{n-m}^{T} bigg )}\
& = & rho^{-2} { sum_{m+1}^{n} bigg(frac{1}{lambda_{i}^2} + frac{1}{rho}bigg)^2}\
end{eqnarray}
Thus the above was achieved by using the Woodbury identity and using the fact that $R_{n-m}^{T}R_{n-m} =I $.
Before I proceed, is this idea correct ? If not can anyone guide me on how to start? And what about the second part?
matrices matrix-calculus matrix-decomposition matrix-rank
asked Dec 8 '18 at 10:02
KsmithKsmith
213
$endgroup$
add a comment |
$begingroup$
Given that $ K = A^{-1} - B^{-1} + A^{-1} - A^{-1}BA^{-1}$, we need to find the upper bound of $K$ where matrix $A = C + Irho$ and $ B = C_{x} + Irho $ has dimension $ntimes n$, $ C = RDR^{T}$ where $D$ is a diagonal matrix whose eigenvalues are in descending order and $C$ is a matrix of full rank and $C_{x}$ is an approximated matrix and $rho$ is a constant. If we partition $C$ as
begin{equation}
C =
left({begin{array}{cc} R_{m} & R_{(n-m)} end{array}}right)
left(begin{array}{cc} D_{mm} & 0\ 0 & D_{(n-m)(n-m)} end{array}right)
left(begin{array}{c} R'_m \ R'_{(n-m)} end{array}right)
end{equation}
Where by Eckart - Young theorem $ C_{x} = R_{m}D_{m}R_{m}^T$.
Is it right to proceed this way?
First, break $K$ into 2 parts, thus $ A^{-1} - B^{-1} $ and $ A^{-1} - A^{-1}BA^{-1} $ and using the fact that
$$ parallel A^{-1} - B^{-1} + A^{-1} - A^{-1}BA^{-1} parallel_F ;leq ;parallel A^{-1} - B^{-1}parallel_F + parallel A^{-1} - A^{-1}BA^{-1} parallel_{F} $$
And considering first
begin{eqnarray}
parallel A^{-1} - B^{-1}parallel_F ^2 & = & parallel ( C + rho)^{-1} - (C_x + rho)^{-1}parallel_F ^2 \
& = & parallel rho^{-2}R_{n-m}(D_{n-m} + rho^{-1})^{-1}R_{n-m}^Tparallel_F^2 quad text { That is by substitution }\
& = & rho^{-2}{tr bigg ( R_{n-m}(D_{n-m}^{-2} +rho^{-1})^{-2}R_{n-m}^{T} bigg )}\
& = & rho^{-2} { sum_{m+1}^{n} bigg(frac{1}{lambda_{i}^2} + frac{1}{rho}bigg)^2}\
end{eqnarray}
Thus the above was achieved by using the Woodbury identity and using the fact that $R_{n-m}^{T}R_{n-m} =I $.
Before I proceed, is this idea correct ? If not can anyone guide me on how to start? And what about the second part?
matrices matrix-calculus matrix-decomposition matrix-rank
asked Dec 8 '18 at 10:02
KsmithKsmith
213
$endgroup$
add a comment |
$begingroup$
Given that $ K = A^{-1} - B^{-1} + A^{-1} - A^{-1}BA^{-1}$, we need to find the upper bound of $K$ where matrix $A = C + Irho$ and $ B = C_{x} + Irho $ has dimension $ntimes n$, $ C = RDR^{T}$ where $D$ is a diagonal matrix whose eigenvalues are in descending order and $C$ is a matrix of full rank and $C_{x}$ is an approximated matrix and $rho$ is a constant. If we partition $C$ as
begin{equation}
C =
left({begin{array}{cc} R_{m} & R_{(n-m)} end{array}}right)
left(begin{array}{cc} D_{mm} & 0\ 0 & D_{(n-m)(n-m)} end{array}right)
left(begin{array}{c} R'_m \ R'_{(n-m)} end{array}right)
end{equation}
Where by Eckart - Young theorem $ C_{x} = R_{m}D_{m}R_{m}^T$.
Is it right to proceed this way?
First, break $K$ into 2 parts, thus $ A^{-1} - B^{-1} $ and $ A^{-1} - A^{-1}BA^{-1} $ and using the fact that
$$ parallel A^{-1} - B^{-1} + A^{-1} - A^{-1}BA^{-1} parallel_F ;leq ;parallel A^{-1} - B^{-1}parallel_F + parallel A^{-1} - A^{-1}BA^{-1} parallel_{F} $$
And considering first
begin{eqnarray}
parallel A^{-1} - B^{-1}parallel_F ^2 & = & parallel ( C + rho)^{-1} - (C_x + rho)^{-1}parallel_F ^2 \
& = & parallel rho^{-2}R_{n-m}(D_{n-m} + rho^{-1})^{-1}R_{n-m}^Tparallel_F^2 quad text { That is by substitution }\
& = & rho^{-2}{tr bigg ( R_{n-m}(D_{n-m}^{-2} +rho^{-1})^{-2}R_{n-m}^{T} bigg )}\
& = & rho^{-2} { sum_{m+1}^{n} bigg(frac{1}{lambda_{i}^2} + frac{1}{rho}bigg)^2}\
end{eqnarray}
Thus the above was achieved by using the Woodbury identity and using the fact that $R_{n-m}^{T}R_{n-m} =I $.
Before I proceed, is this idea correct ? If not can anyone guide me on how to start? And what about the second part?
matrices matrix-calculus matrix-decomposition matrix-rank
asked Dec 8 '18 at 10:02
KsmithKsmith
213
$endgroup$
Given that $ K = A^{-1} - B^{-1} + A^{-1} - A^{-1}BA^{-1}$, we need to find the upper bound of $K$ where matrix $A = C + Irho$ and $ B = C_{x} + Irho $ has dimension $ntimes n$, $ C = RDR^{T}$ where $D$ is a diagonal matrix whose eigenvalues are in descending order and $C$ is a matrix of full rank and $C_{x}$ is an approximated matrix and $rho$ is a constant. If we partition $C$ as
begin{equation}
C =
left({begin{array}{cc} R_{m} & R_{(n-m)} end{array}}right)
left(begin{array}{cc} D_{mm} & 0\ 0 & D_{(n-m)(n-m)} end{array}right)
left(begin{array}{c} R'_m \ R'_{(n-m)} end{array}right)
end{equation}
Where by Eckart - Young theorem $ C_{x} = R_{m}D_{m}R_{m}^T$.
Is it right to proceed this way?
First, break $K$ into 2 parts, thus $ A^{-1} - B^{-1} $ and $ A^{-1} - A^{-1}BA^{-1} $ and using the fact that
$$ parallel A^{-1} - B^{-1} + A^{-1} - A^{-1}BA^{-1} parallel_F ;leq ;parallel A^{-1} - B^{-1}parallel_F + parallel A^{-1} - A^{-1}BA^{-1} parallel_{F} $$
And considering first
begin{eqnarray}
parallel A^{-1} - B^{-1}parallel_F ^2 & = & parallel ( C + rho)^{-1} - (C_x + rho)^{-1}parallel_F ^2 \
& = & parallel rho^{-2}R_{n-m}(D_{n-m} + rho^{-1})^{-1}R_{n-m}^Tparallel_F^2 quad text { That is by substitution }\
& = & rho^{-2}{tr bigg ( R_{n-m}(D_{n-m}^{-2} +rho^{-1})^{-2}R_{n-m}^{T} bigg )}\
& = & rho^{-2} { sum_{m+1}^{n} bigg(frac{1}{lambda_{i}^2} + frac{1}{rho}bigg)^2}\
end{eqnarray}
Thus the above was achieved by using the Woodbury identity and using the fact that $R_{n-m}^{T}R_{n-m} =I $.
Before I proceed, is this idea correct ? If not can anyone guide me on how to start? And what about the second part?
matrices matrix-calculus matrix-decomposition matrix-rank
matrices matrix-calculus matrix-decomposition matrix-rank
asked Dec 8 '18 at 10:02
KsmithKsmith
213
asked Dec 8 '18 at 10:02
KsmithKsmith
213
asked Dec 8 '18 at 10:02
KsmithKsmith
213
asked Dec 8 '18 at 10:02
KsmithKsmith
213
asked Dec 8 '18 at 10:02
KsmithKsmith
213
213
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