theoretical results on the conditioning of numerical eigenvalue problems
$begingroup$
The question I am working on is as follows. A is an $ntimes n$ matrix. A is diagonalizable by a similarity transformation s.t. $T^{-1}AT=B ={rm diag}(lambda_1,..,lambda_n). $ $tilde{A}=A+{G}. $
If $tilde{lambda}$ is an eigenvalue of $tilde{A},$ show
$$ |(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_pgeq 1. $$
My strategy was to use algebra to solve for the left side matrix and then calculate all of the possible norms for it. $tilde{lambda}$ and $lambda$ happened to cancel out nicely, but this does not incorporate using the opposite to show $tilde{A}-tilde{lambda}I$ is invertible.
The hint is generally confusing and am not sure to do what with it.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 14 '18 at 21:02
Jennifer LaundeyJennifer Laundey
84
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add a comment |
$begingroup$
The question I am working on is as follows. A is an $ntimes n$ matrix. A is diagonalizable by a similarity transformation s.t. $T^{-1}AT=B ={rm diag}(lambda_1,..,lambda_n). $ $tilde{A}=A+{G}. $
If $tilde{lambda}$ is an eigenvalue of $tilde{A},$ show
$$ |(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_pgeq 1. $$
My strategy was to use algebra to solve for the left side matrix and then calculate all of the possible norms for it. $tilde{lambda}$ and $lambda$ happened to cancel out nicely, but this does not incorporate using the opposite to show $tilde{A}-tilde{lambda}I$ is invertible.
The hint is generally confusing and am not sure to do what with it.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 14 '18 at 21:02
Jennifer LaundeyJennifer Laundey
84
$endgroup$
add a comment |
$begingroup$
The question I am working on is as follows. A is an $ntimes n$ matrix. A is diagonalizable by a similarity transformation s.t. $T^{-1}AT=B ={rm diag}(lambda_1,..,lambda_n). $ $tilde{A}=A+{G}. $
If $tilde{lambda}$ is an eigenvalue of $tilde{A},$ show
$$ |(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_pgeq 1. $$
My strategy was to use algebra to solve for the left side matrix and then calculate all of the possible norms for it. $tilde{lambda}$ and $lambda$ happened to cancel out nicely, but this does not incorporate using the opposite to show $tilde{A}-tilde{lambda}I$ is invertible.
The hint is generally confusing and am not sure to do what with it.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 14 '18 at 21:02
Jennifer LaundeyJennifer Laundey
84
$endgroup$
The question I am working on is as follows. A is an $ntimes n$ matrix. A is diagonalizable by a similarity transformation s.t. $T^{-1}AT=B ={rm diag}(lambda_1,..,lambda_n). $ $tilde{A}=A+{G}. $
If $tilde{lambda}$ is an eigenvalue of $tilde{A},$ show
$$ |(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_pgeq 1. $$
My strategy was to use algebra to solve for the left side matrix and then calculate all of the possible norms for it. $tilde{lambda}$ and $lambda$ happened to cancel out nicely, but this does not incorporate using the opposite to show $tilde{A}-tilde{lambda}I$ is invertible.
The hint is generally confusing and am not sure to do what with it.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 14 '18 at 21:02
Jennifer LaundeyJennifer Laundey
84
asked Dec 14 '18 at 21:02
Jennifer LaundeyJennifer Laundey
84
edited Dec 15 '18 at 5:54
Jennifer Laundey
asked Dec 14 '18 at 21:02
Jennifer LaundeyJennifer Laundey
84
asked Dec 14 '18 at 21:02
Jennifer LaundeyJennifer Laundey
84
asked Dec 14 '18 at 21:02
Jennifer LaundeyJennifer Laundey
84
84
add a comment |
add a comment |
1 Answer
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$begingroup$
To incorporate your professor's hint: if the opposite is true, then there is a value of $p$ for which we have
$$
|(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_p < 1
$$
However, note that $(tilde A - tilde lambda I)$ is invertible if and only if $T^{-1}(tilde A - tilde lambda I)T$ is invertible, and
$$
T^{-1}(tilde A - tilde lambda I)T = \
T^{-1}((A + G) - tilde lambda I)T = \
(T^{-1}AT^{-1} - tilde lambda I) + T^{-1}GT = \
(B - tilde lambda I) + T^{-1}GT
$$
This matrix will be invertible if and only if the matrix
$$
(B - tilde lambda I)^{-1}[(B - tilde lambda I) + T^{-1}GT] = \
I + (B - tilde lambda I)^{-1}T^{-1}GT
$$
is invertible. Now, use the fact that $I + M$ will be invertible whenever $|M| < 1$ for some norm $|cdot|$.
answered Dec 14 '18 at 21:22
OmnomnomnomOmnomnomnom
128k791185
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1 Answer
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$begingroup$
To incorporate your professor's hint: if the opposite is true, then there is a value of $p$ for which we have
$$
|(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_p < 1
$$
However, note that $(tilde A - tilde lambda I)$ is invertible if and only if $T^{-1}(tilde A - tilde lambda I)T$ is invertible, and
$$
T^{-1}(tilde A - tilde lambda I)T = \
T^{-1}((A + G) - tilde lambda I)T = \
(T^{-1}AT^{-1} - tilde lambda I) + T^{-1}GT = \
(B - tilde lambda I) + T^{-1}GT
$$
This matrix will be invertible if and only if the matrix
$$
(B - tilde lambda I)^{-1}[(B - tilde lambda I) + T^{-1}GT] = \
I + (B - tilde lambda I)^{-1}T^{-1}GT
$$
is invertible. Now, use the fact that $I + M$ will be invertible whenever $|M| < 1$ for some norm $|cdot|$.
answered Dec 14 '18 at 21:22
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
add a comment |
$begingroup$
To incorporate your professor's hint: if the opposite is true, then there is a value of $p$ for which we have
$$
|(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_p < 1
$$
However, note that $(tilde A - tilde lambda I)$ is invertible if and only if $T^{-1}(tilde A - tilde lambda I)T$ is invertible, and
$$
T^{-1}(tilde A - tilde lambda I)T = \
T^{-1}((A + G) - tilde lambda I)T = \
(T^{-1}AT^{-1} - tilde lambda I) + T^{-1}GT = \
(B - tilde lambda I) + T^{-1}GT
$$
This matrix will be invertible if and only if the matrix
$$
(B - tilde lambda I)^{-1}[(B - tilde lambda I) + T^{-1}GT] = \
I + (B - tilde lambda I)^{-1}T^{-1}GT
$$
is invertible. Now, use the fact that $I + M$ will be invertible whenever $|M| < 1$ for some norm $|cdot|$.
answered Dec 14 '18 at 21:22
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
add a comment |
$begingroup$
To incorporate your professor's hint: if the opposite is true, then there is a value of $p$ for which we have
$$
|(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_p < 1
$$
However, note that $(tilde A - tilde lambda I)$ is invertible if and only if $T^{-1}(tilde A - tilde lambda I)T$ is invertible, and
$$
T^{-1}(tilde A - tilde lambda I)T = \
T^{-1}((A + G) - tilde lambda I)T = \
(T^{-1}AT^{-1} - tilde lambda I) + T^{-1}GT = \
(B - tilde lambda I) + T^{-1}GT
$$
This matrix will be invertible if and only if the matrix
$$
(B - tilde lambda I)^{-1}[(B - tilde lambda I) + T^{-1}GT] = \
I + (B - tilde lambda I)^{-1}T^{-1}GT
$$
is invertible. Now, use the fact that $I + M$ will be invertible whenever $|M| < 1$ for some norm $|cdot|$.
answered Dec 14 '18 at 21:22
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
To incorporate your professor's hint: if the opposite is true, then there is a value of $p$ for which we have
$$
|(B-tilde{lambda}I)^{-1}T^{-1}{bf G}T|_p < 1
$$
However, note that $(tilde A - tilde lambda I)$ is invertible if and only if $T^{-1}(tilde A - tilde lambda I)T$ is invertible, and
$$
T^{-1}(tilde A - tilde lambda I)T = \
T^{-1}((A + G) - tilde lambda I)T = \
(T^{-1}AT^{-1} - tilde lambda I) + T^{-1}GT = \
(B - tilde lambda I) + T^{-1}GT
$$
This matrix will be invertible if and only if the matrix
$$
(B - tilde lambda I)^{-1}[(B - tilde lambda I) + T^{-1}GT] = \
I + (B - tilde lambda I)^{-1}T^{-1}GT
$$
is invertible. Now, use the fact that $I + M$ will be invertible whenever $|M| < 1$ for some norm $|cdot|$.
answered Dec 14 '18 at 21:22
OmnomnomnomOmnomnomnom
128k791185
answered Dec 14 '18 at 21:22
OmnomnomnomOmnomnomnom
128k791185
answered Dec 14 '18 at 21:22
OmnomnomnomOmnomnomnom
128k791185
answered Dec 14 '18 at 21:22
OmnomnomnomOmnomnomnom
128k791185
128k791185
add a comment |
add a comment |
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