Mixing
Limit of matrix inverse: $lim_{lambda to infty} (A + lambda I)^{-1} = mathbf{0}$?
$begingroup$
Let matrix $A in mathbb{R}^{ntimes n}$ be positive semidefinite.
Is it then true to that
$$
(A + lambda I)^{-1} to mathbf{0} quad (lambda to infty) quad ?
$$If so, is the fact that $A$ is positive definite irrelevant here?
My thoughts so far:
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1}
$$
I think that $lim_{lambda to infty} Big( frac{1}{lambda}A + I Big)^{-1} = I^{-1} = I$, but I don't know if I can just pass the $lim$ through the inverse $(cdot)^{-1}$ like that. If this is the case, then
$$
lim_{lambda to infty} (A + lambda I)^{-1} = lim_{lambda to infty} (1/lambda) lim_{lambda to infty} (A/lambda + I)^{-1} = 0 cdot I = mathbf{0}
$$
as I'd like to show.
Where this comes from:
I'm trying to justify a claim made in an econometrics lecture. Namely,
$$
textrm{Var}(hat{beta}^{textrm{ridge}}) = sigma^2 (X^{T}X + lambda I)^{-1} X^T X [(X^T X + lambda I)^{-1}]^T to mathbf{0}
$$
where $hat{beta}^textrm{ridge}$ is the ridge estimator in a linear model, $X in mathbb{R}^{n times p}$ is the design matrix, and the equality is known. The limit, however, wasn't justified.
linear-algebra matrices limits
asked Jan 24 at 17:22
zxmknzxmkn
340213
$endgroup$
add a comment |
$begingroup$
Let matrix $A in mathbb{R}^{ntimes n}$ be positive semidefinite.
Is it then true to that
$$
(A + lambda I)^{-1} to mathbf{0} quad (lambda to infty) quad ?
$$If so, is the fact that $A$ is positive definite irrelevant here?
My thoughts so far:
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1}
$$
I think that $lim_{lambda to infty} Big( frac{1}{lambda}A + I Big)^{-1} = I^{-1} = I$, but I don't know if I can just pass the $lim$ through the inverse $(cdot)^{-1}$ like that. If this is the case, then
$$
lim_{lambda to infty} (A + lambda I)^{-1} = lim_{lambda to infty} (1/lambda) lim_{lambda to infty} (A/lambda + I)^{-1} = 0 cdot I = mathbf{0}
$$
as I'd like to show.
Where this comes from:
I'm trying to justify a claim made in an econometrics lecture. Namely,
$$
textrm{Var}(hat{beta}^{textrm{ridge}}) = sigma^2 (X^{T}X + lambda I)^{-1} X^T X [(X^T X + lambda I)^{-1}]^T to mathbf{0}
$$
where $hat{beta}^textrm{ridge}$ is the ridge estimator in a linear model, $X in mathbb{R}^{n times p}$ is the design matrix, and the equality is known. The limit, however, wasn't justified.
linear-algebra matrices limits
asked Jan 24 at 17:22
zxmknzxmkn
340213
$endgroup$
4
$begingroup$
$A$ can be any matrix above. The point is, the inverse of a matrix is a continuous function in a neighbourhood of the identity, therefore since $A - lambda I$ is going to eventually be invertible, we may pass the limit inside the inverse by continuity, giving the desired result by the continuity of scalar multiplication.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 17:27
3
$begingroup$
If $|cdot|$ is a matrix norm, then the Neumann series guarantees that $A+lambda I$ is invertible with $$(A+lambda I)^{-1} = sum_{n=0}^{infty} frac{(-1)^n}{lambda^{n+1}}A^n, $$ which converges uniformly on the region $|lambda| geq |A|+delta$ for any given $delta > 0$. By the Weierstrass M-test, the limit as $lambdatoinfty$ can be evaluated term-wise, proving the desired claim.
$endgroup$
– Sangchul Lee
Jan 24 at 17:34
$begingroup$
@астонвіллаолофмэллбэрг Great! That completes my line of reasoning. For others looking on, here's why there is a neighborhood of $I$ in $M_n(mathbb{R})$ in which $(cdot)^{-1}$ is continuous: $(cdot)^{-1} : GL_n(mathbb{R}) to GL_n(mathbb{R})$ is continuous and $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$ (see: math.stackexchange.com/a/810675/369800). [To understand the proof just linked: determinant continuous (see: math.stackexchange.com/a/121834/369800) and adjoint continuous (see: math.stackexchange.com/a/2031642/369800)]
$endgroup$
– zxmkn
Jan 24 at 19:07
$begingroup$
Recall that the inverse matrix is the adjugate matrix divided by the discriminant. Thus a "singularity" of the inversion only happens when the discriminant vanishes.
$endgroup$
– Alexey
Jan 26 at 20:45
add a comment |
$begingroup$
Let matrix $A in mathbb{R}^{ntimes n}$ be positive semidefinite.
Is it then true to that
$$
(A + lambda I)^{-1} to mathbf{0} quad (lambda to infty) quad ?
$$If so, is the fact that $A$ is positive definite irrelevant here?
My thoughts so far:
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1}
$$
I think that $lim_{lambda to infty} Big( frac{1}{lambda}A + I Big)^{-1} = I^{-1} = I$, but I don't know if I can just pass the $lim$ through the inverse $(cdot)^{-1}$ like that. If this is the case, then
$$
lim_{lambda to infty} (A + lambda I)^{-1} = lim_{lambda to infty} (1/lambda) lim_{lambda to infty} (A/lambda + I)^{-1} = 0 cdot I = mathbf{0}
$$
as I'd like to show.
Where this comes from:
I'm trying to justify a claim made in an econometrics lecture. Namely,
$$
textrm{Var}(hat{beta}^{textrm{ridge}}) = sigma^2 (X^{T}X + lambda I)^{-1} X^T X [(X^T X + lambda I)^{-1}]^T to mathbf{0}
$$
where $hat{beta}^textrm{ridge}$ is the ridge estimator in a linear model, $X in mathbb{R}^{n times p}$ is the design matrix, and the equality is known. The limit, however, wasn't justified.
linear-algebra matrices limits
asked Jan 24 at 17:22
zxmknzxmkn
340213
$endgroup$
Let matrix $A in mathbb{R}^{ntimes n}$ be positive semidefinite.
Is it then true to that
$$
(A + lambda I)^{-1} to mathbf{0} quad (lambda to infty) quad ?
$$If so, is the fact that $A$ is positive definite irrelevant here?
My thoughts so far:
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1}
$$
I think that $lim_{lambda to infty} Big( frac{1}{lambda}A + I Big)^{-1} = I^{-1} = I$, but I don't know if I can just pass the $lim$ through the inverse $(cdot)^{-1}$ like that. If this is the case, then
$$
lim_{lambda to infty} (A + lambda I)^{-1} = lim_{lambda to infty} (1/lambda) lim_{lambda to infty} (A/lambda + I)^{-1} = 0 cdot I = mathbf{0}
$$
as I'd like to show.
Where this comes from:
I'm trying to justify a claim made in an econometrics lecture. Namely,
$$
textrm{Var}(hat{beta}^{textrm{ridge}}) = sigma^2 (X^{T}X + lambda I)^{-1} X^T X [(X^T X + lambda I)^{-1}]^T to mathbf{0}
$$
where $hat{beta}^textrm{ridge}$ is the ridge estimator in a linear model, $X in mathbb{R}^{n times p}$ is the design matrix, and the equality is known. The limit, however, wasn't justified.
linear-algebra matrices limits
linear-algebra matrices limits
asked Jan 24 at 17:22
zxmknzxmkn
340213
asked Jan 24 at 17:22
zxmknzxmkn
340213
asked Jan 24 at 17:22
zxmknzxmkn
340213
asked Jan 24 at 17:22
zxmknzxmkn
340213
asked Jan 24 at 17:22
zxmknzxmkn
340213
340213
4
$begingroup$
$A$ can be any matrix above. The point is, the inverse of a matrix is a continuous function in a neighbourhood of the identity, therefore since $A - lambda I$ is going to eventually be invertible, we may pass the limit inside the inverse by continuity, giving the desired result by the continuity of scalar multiplication.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 17:27
3
$begingroup$
If $|cdot|$ is a matrix norm, then the Neumann series guarantees that $A+lambda I$ is invertible with $$(A+lambda I)^{-1} = sum_{n=0}^{infty} frac{(-1)^n}{lambda^{n+1}}A^n, $$ which converges uniformly on the region $|lambda| geq |A|+delta$ for any given $delta > 0$. By the Weierstrass M-test, the limit as $lambdatoinfty$ can be evaluated term-wise, proving the desired claim.
$endgroup$
– Sangchul Lee
Jan 24 at 17:34
$begingroup$
@астонвіллаолофмэллбэрг Great! That completes my line of reasoning. For others looking on, here's why there is a neighborhood of $I$ in $M_n(mathbb{R})$ in which $(cdot)^{-1}$ is continuous: $(cdot)^{-1} : GL_n(mathbb{R}) to GL_n(mathbb{R})$ is continuous and $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$ (see: math.stackexchange.com/a/810675/369800). [To understand the proof just linked: determinant continuous (see: math.stackexchange.com/a/121834/369800) and adjoint continuous (see: math.stackexchange.com/a/2031642/369800)]
$endgroup$
– zxmkn
Jan 24 at 19:07
$begingroup$
Recall that the inverse matrix is the adjugate matrix divided by the discriminant. Thus a "singularity" of the inversion only happens when the discriminant vanishes.
$endgroup$
– Alexey
Jan 26 at 20:45
add a comment |
4
$begingroup$
$A$ can be any matrix above. The point is, the inverse of a matrix is a continuous function in a neighbourhood of the identity, therefore since $A - lambda I$ is going to eventually be invertible, we may pass the limit inside the inverse by continuity, giving the desired result by the continuity of scalar multiplication.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 17:27
3
$begingroup$
If $|cdot|$ is a matrix norm, then the Neumann series guarantees that $A+lambda I$ is invertible with $$(A+lambda I)^{-1} = sum_{n=0}^{infty} frac{(-1)^n}{lambda^{n+1}}A^n, $$ which converges uniformly on the region $|lambda| geq |A|+delta$ for any given $delta > 0$. By the Weierstrass M-test, the limit as $lambdatoinfty$ can be evaluated term-wise, proving the desired claim.
$endgroup$
– Sangchul Lee
Jan 24 at 17:34
$begingroup$
@астонвіллаолофмэллбэрг Great! That completes my line of reasoning. For others looking on, here's why there is a neighborhood of $I$ in $M_n(mathbb{R})$ in which $(cdot)^{-1}$ is continuous: $(cdot)^{-1} : GL_n(mathbb{R}) to GL_n(mathbb{R})$ is continuous and $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$ (see: math.stackexchange.com/a/810675/369800). [To understand the proof just linked: determinant continuous (see: math.stackexchange.com/a/121834/369800) and adjoint continuous (see: math.stackexchange.com/a/2031642/369800)]
$endgroup$
– zxmkn
Jan 24 at 19:07
$begingroup$
Recall that the inverse matrix is the adjugate matrix divided by the discriminant. Thus a "singularity" of the inversion only happens when the discriminant vanishes.
$endgroup$
– Alexey
Jan 26 at 20:45
4
4
$begingroup$
$A$ can be any matrix above. The point is, the inverse of a matrix is a continuous function in a neighbourhood of the identity, therefore since $A - lambda I$ is going to eventually be invertible, we may pass the limit inside the inverse by continuity, giving the desired result by the continuity of scalar multiplication.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 17:27
$begingroup$
$A$ can be any matrix above. The point is, the inverse of a matrix is a continuous function in a neighbourhood of the identity, therefore since $A - lambda I$ is going to eventually be invertible, we may pass the limit inside the inverse by continuity, giving the desired result by the continuity of scalar multiplication.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 17:27
3
3
$begingroup$
If $|cdot|$ is a matrix norm, then the Neumann series guarantees that $A+lambda I$ is invertible with $$(A+lambda I)^{-1} = sum_{n=0}^{infty} frac{(-1)^n}{lambda^{n+1}}A^n, $$ which converges uniformly on the region $|lambda| geq |A|+delta$ for any given $delta > 0$. By the Weierstrass M-test, the limit as $lambdatoinfty$ can be evaluated term-wise, proving the desired claim.
$endgroup$
– Sangchul Lee
Jan 24 at 17:34
$begingroup$
If $|cdot|$ is a matrix norm, then the Neumann series guarantees that $A+lambda I$ is invertible with $$(A+lambda I)^{-1} = sum_{n=0}^{infty} frac{(-1)^n}{lambda^{n+1}}A^n, $$ which converges uniformly on the region $|lambda| geq |A|+delta$ for any given $delta > 0$. By the Weierstrass M-test, the limit as $lambdatoinfty$ can be evaluated term-wise, proving the desired claim.
$endgroup$
– Sangchul Lee
Jan 24 at 17:34
$begingroup$
@астонвіллаолофмэллбэрг Great! That completes my line of reasoning. For others looking on, here's why there is a neighborhood of $I$ in $M_n(mathbb{R})$ in which $(cdot)^{-1}$ is continuous: $(cdot)^{-1} : GL_n(mathbb{R}) to GL_n(mathbb{R})$ is continuous and $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$ (see: math.stackexchange.com/a/810675/369800). [To understand the proof just linked: determinant continuous (see: math.stackexchange.com/a/121834/369800) and adjoint continuous (see: math.stackexchange.com/a/2031642/369800)]
$endgroup$
– zxmkn
Jan 24 at 19:07
$begingroup$
@астонвіллаолофмэллбэрг Great! That completes my line of reasoning. For others looking on, here's why there is a neighborhood of $I$ in $M_n(mathbb{R})$ in which $(cdot)^{-1}$ is continuous: $(cdot)^{-1} : GL_n(mathbb{R}) to GL_n(mathbb{R})$ is continuous and $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$ (see: math.stackexchange.com/a/810675/369800). [To understand the proof just linked: determinant continuous (see: math.stackexchange.com/a/121834/369800) and adjoint continuous (see: math.stackexchange.com/a/2031642/369800)]
$endgroup$
– zxmkn
Jan 24 at 19:07
$begingroup$
Recall that the inverse matrix is the adjugate matrix divided by the discriminant. Thus a "singularity" of the inversion only happens when the discriminant vanishes.
$endgroup$
– Alexey
Jan 26 at 20:45
$begingroup$
Recall that the inverse matrix is the adjugate matrix divided by the discriminant. Thus a "singularity" of the inversion only happens when the discriminant vanishes.
$endgroup$
– Alexey
Jan 26 at 20:45
add a comment |
2 Answers
2
active
oldest
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$begingroup$
The eigenvalues of $A+lambda I$ are of the form $lambda+mu$, where $mu$ is an eigenvalue of $A$ (necessarily real). Then, for $lambda$ sufficiently large, the eigenvalues of $A+lambda I$ are all $>1$.
Note that a matrix $S$ that diagonalizes $A$ also diagonalizes $A+lambda I$, let $A=SDS^{-1}$, with $D$ diagonal.
Then $(A+lambda I)^{-1}$ is diagonalizable with eigenvalues in $(0,1)$ and therefore
$$
lim_{lambdatoinfty}(A+lambda I)^{-1}=
SBigl(,lim_{lambdatoinfty}(D+lambda I)^{-1}Bigr)S^{-1}=0
$$
It is not necessary that $A$ is semipositive definite. Any symmetric matrix will do.
answered Jan 24 at 18:43
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
The answer I liked the best was left in the comments by астон вілла олоф мэллбэрг, since it shows that $A$ does not need any special structure. Here I'm pulling his answer down and including a bit more detail.
We have
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1},
$$
and we claim that $Big(frac{1}{lambda}A + I Big)^{-1} to I^{-1} = I quad (lambda to infty)$.
Therefore,
$$
(A + lambda I)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1} to 0 cdot I = mathbf{0} quad (lambda to infty),
$$
which was the desired result.
We complete the proof by showing the claim. Since $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$, we find some $epsilon > 0 $ such that the open ball $B(I, epsilon) subseteq GL_n(mathbb{R})$. Hence, for sufficiently large $lambda$, we know that $(A/lambda + I) in B(I, epsilon) subseteq GL_n(mathbb{R})$. Also knowing that $(cdot)^{-1} : GL_n to GL_n$ is continuous, we have
$$
lim_{lambda to infty}Big(frac{1}{lambda}A + I Big)^{-1} = Big(lim_{lambda to infty} frac{1}{lambda}A + I Big)^{-1}= (I)^{-1} = I,
$$
which completes the proof.
To understand the linked proof of the continuity of $(cdot)^{-1}$, see here for justification that the determinant operator is continuous and here for justification that the adjoint operator is continuous.
answered Jan 26 at 20:35
zxmknzxmkn
340213
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The eigenvalues of $A+lambda I$ are of the form $lambda+mu$, where $mu$ is an eigenvalue of $A$ (necessarily real). Then, for $lambda$ sufficiently large, the eigenvalues of $A+lambda I$ are all $>1$.
Note that a matrix $S$ that diagonalizes $A$ also diagonalizes $A+lambda I$, let $A=SDS^{-1}$, with $D$ diagonal.
Then $(A+lambda I)^{-1}$ is diagonalizable with eigenvalues in $(0,1)$ and therefore
$$
lim_{lambdatoinfty}(A+lambda I)^{-1}=
SBigl(,lim_{lambdatoinfty}(D+lambda I)^{-1}Bigr)S^{-1}=0
$$
It is not necessary that $A$ is semipositive definite. Any symmetric matrix will do.
answered Jan 24 at 18:43
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
The eigenvalues of $A+lambda I$ are of the form $lambda+mu$, where $mu$ is an eigenvalue of $A$ (necessarily real). Then, for $lambda$ sufficiently large, the eigenvalues of $A+lambda I$ are all $>1$.
Note that a matrix $S$ that diagonalizes $A$ also diagonalizes $A+lambda I$, let $A=SDS^{-1}$, with $D$ diagonal.
Then $(A+lambda I)^{-1}$ is diagonalizable with eigenvalues in $(0,1)$ and therefore
$$
lim_{lambdatoinfty}(A+lambda I)^{-1}=
SBigl(,lim_{lambdatoinfty}(D+lambda I)^{-1}Bigr)S^{-1}=0
$$
It is not necessary that $A$ is semipositive definite. Any symmetric matrix will do.
answered Jan 24 at 18:43
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
The eigenvalues of $A+lambda I$ are of the form $lambda+mu$, where $mu$ is an eigenvalue of $A$ (necessarily real). Then, for $lambda$ sufficiently large, the eigenvalues of $A+lambda I$ are all $>1$.
Note that a matrix $S$ that diagonalizes $A$ also diagonalizes $A+lambda I$, let $A=SDS^{-1}$, with $D$ diagonal.
Then $(A+lambda I)^{-1}$ is diagonalizable with eigenvalues in $(0,1)$ and therefore
$$
lim_{lambdatoinfty}(A+lambda I)^{-1}=
SBigl(,lim_{lambdatoinfty}(D+lambda I)^{-1}Bigr)S^{-1}=0
$$
It is not necessary that $A$ is semipositive definite. Any symmetric matrix will do.
answered Jan 24 at 18:43
egregegreg
184k1486205
$endgroup$
The eigenvalues of $A+lambda I$ are of the form $lambda+mu$, where $mu$ is an eigenvalue of $A$ (necessarily real). Then, for $lambda$ sufficiently large, the eigenvalues of $A+lambda I$ are all $>1$.
Note that a matrix $S$ that diagonalizes $A$ also diagonalizes $A+lambda I$, let $A=SDS^{-1}$, with $D$ diagonal.
Then $(A+lambda I)^{-1}$ is diagonalizable with eigenvalues in $(0,1)$ and therefore
$$
lim_{lambdatoinfty}(A+lambda I)^{-1}=
SBigl(,lim_{lambdatoinfty}(D+lambda I)^{-1}Bigr)S^{-1}=0
$$
It is not necessary that $A$ is semipositive definite. Any symmetric matrix will do.
answered Jan 24 at 18:43
egregegreg
184k1486205
answered Jan 24 at 18:43
egregegreg
184k1486205
answered Jan 24 at 18:43
egregegreg
184k1486205
answered Jan 24 at 18:43
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
$begingroup$
The answer I liked the best was left in the comments by астон вілла олоф мэллбэрг, since it shows that $A$ does not need any special structure. Here I'm pulling his answer down and including a bit more detail.
We have
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1},
$$
and we claim that $Big(frac{1}{lambda}A + I Big)^{-1} to I^{-1} = I quad (lambda to infty)$.
Therefore,
$$
(A + lambda I)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1} to 0 cdot I = mathbf{0} quad (lambda to infty),
$$
which was the desired result.
We complete the proof by showing the claim. Since $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$, we find some $epsilon > 0 $ such that the open ball $B(I, epsilon) subseteq GL_n(mathbb{R})$. Hence, for sufficiently large $lambda$, we know that $(A/lambda + I) in B(I, epsilon) subseteq GL_n(mathbb{R})$. Also knowing that $(cdot)^{-1} : GL_n to GL_n$ is continuous, we have
$$
lim_{lambda to infty}Big(frac{1}{lambda}A + I Big)^{-1} = Big(lim_{lambda to infty} frac{1}{lambda}A + I Big)^{-1}= (I)^{-1} = I,
$$
which completes the proof.
To understand the linked proof of the continuity of $(cdot)^{-1}$, see here for justification that the determinant operator is continuous and here for justification that the adjoint operator is continuous.
answered Jan 26 at 20:35
zxmknzxmkn
340213
$endgroup$
add a comment |
$begingroup$
The answer I liked the best was left in the comments by астон вілла олоф мэллбэрг, since it shows that $A$ does not need any special structure. Here I'm pulling his answer down and including a bit more detail.
We have
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1},
$$
and we claim that $Big(frac{1}{lambda}A + I Big)^{-1} to I^{-1} = I quad (lambda to infty)$.
Therefore,
$$
(A + lambda I)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1} to 0 cdot I = mathbf{0} quad (lambda to infty),
$$
which was the desired result.
We complete the proof by showing the claim. Since $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$, we find some $epsilon > 0 $ such that the open ball $B(I, epsilon) subseteq GL_n(mathbb{R})$. Hence, for sufficiently large $lambda$, we know that $(A/lambda + I) in B(I, epsilon) subseteq GL_n(mathbb{R})$. Also knowing that $(cdot)^{-1} : GL_n to GL_n$ is continuous, we have
$$
lim_{lambda to infty}Big(frac{1}{lambda}A + I Big)^{-1} = Big(lim_{lambda to infty} frac{1}{lambda}A + I Big)^{-1}= (I)^{-1} = I,
$$
which completes the proof.
To understand the linked proof of the continuity of $(cdot)^{-1}$, see here for justification that the determinant operator is continuous and here for justification that the adjoint operator is continuous.
answered Jan 26 at 20:35
zxmknzxmkn
340213
$endgroup$
add a comment |
$begingroup$
The answer I liked the best was left in the comments by астон вілла олоф мэллбэрг, since it shows that $A$ does not need any special structure. Here I'm pulling his answer down and including a bit more detail.
We have
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1},
$$
and we claim that $Big(frac{1}{lambda}A + I Big)^{-1} to I^{-1} = I quad (lambda to infty)$.
Therefore,
$$
(A + lambda I)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1} to 0 cdot I = mathbf{0} quad (lambda to infty),
$$
which was the desired result.
We complete the proof by showing the claim. Since $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$, we find some $epsilon > 0 $ such that the open ball $B(I, epsilon) subseteq GL_n(mathbb{R})$. Hence, for sufficiently large $lambda$, we know that $(A/lambda + I) in B(I, epsilon) subseteq GL_n(mathbb{R})$. Also knowing that $(cdot)^{-1} : GL_n to GL_n$ is continuous, we have
$$
lim_{lambda to infty}Big(frac{1}{lambda}A + I Big)^{-1} = Big(lim_{lambda to infty} frac{1}{lambda}A + I Big)^{-1}= (I)^{-1} = I,
$$
which completes the proof.
To understand the linked proof of the continuity of $(cdot)^{-1}$, see here for justification that the determinant operator is continuous and here for justification that the adjoint operator is continuous.
answered Jan 26 at 20:35
zxmknzxmkn
340213
$endgroup$
The answer I liked the best was left in the comments by астон вілла олоф мэллбэрг, since it shows that $A$ does not need any special structure. Here I'm pulling his answer down and including a bit more detail.
We have
$$
(A + lambda I)^{-1} = Big(lambda( frac{1}{lambda}A + I ) Big)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1},
$$
and we claim that $Big(frac{1}{lambda}A + I Big)^{-1} to I^{-1} = I quad (lambda to infty)$.
Therefore,
$$
(A + lambda I)^{-1} = frac{1}{lambda} Big(frac{1}{lambda}A + I Big)^{-1} to 0 cdot I = mathbf{0} quad (lambda to infty),
$$
which was the desired result.
We complete the proof by showing the claim. Since $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$, we find some $epsilon > 0 $ such that the open ball $B(I, epsilon) subseteq GL_n(mathbb{R})$. Hence, for sufficiently large $lambda$, we know that $(A/lambda + I) in B(I, epsilon) subseteq GL_n(mathbb{R})$. Also knowing that $(cdot)^{-1} : GL_n to GL_n$ is continuous, we have
$$
lim_{lambda to infty}Big(frac{1}{lambda}A + I Big)^{-1} = Big(lim_{lambda to infty} frac{1}{lambda}A + I Big)^{-1}= (I)^{-1} = I,
$$
which completes the proof.
To understand the linked proof of the continuity of $(cdot)^{-1}$, see here for justification that the determinant operator is continuous and here for justification that the adjoint operator is continuous.
answered Jan 26 at 20:35
zxmknzxmkn
340213
answered Jan 26 at 20:35
zxmknzxmkn
340213
answered Jan 26 at 20:35
zxmknzxmkn
340213
answered Jan 26 at 20:35
zxmknzxmkn
340213
340213
add a comment |
add a comment |
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$begingroup$
$A$ can be any matrix above. The point is, the inverse of a matrix is a continuous function in a neighbourhood of the identity, therefore since $A - lambda I$ is going to eventually be invertible, we may pass the limit inside the inverse by continuity, giving the desired result by the continuity of scalar multiplication.
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– астон вілла олоф мэллбэрг
Jan 24 at 17:27
3
$begingroup$
If $|cdot|$ is a matrix norm, then the Neumann series guarantees that $A+lambda I$ is invertible with $$(A+lambda I)^{-1} = sum_{n=0}^{infty} frac{(-1)^n}{lambda^{n+1}}A^n, $$ which converges uniformly on the region $|lambda| geq |A|+delta$ for any given $delta > 0$. By the Weierstrass M-test, the limit as $lambdatoinfty$ can be evaluated term-wise, proving the desired claim.
$endgroup$
– Sangchul Lee
Jan 24 at 17:34
$begingroup$
@астонвіллаолофмэллбэрг Great! That completes my line of reasoning. For others looking on, here's why there is a neighborhood of $I$ in $M_n(mathbb{R})$ in which $(cdot)^{-1}$ is continuous: $(cdot)^{-1} : GL_n(mathbb{R}) to GL_n(mathbb{R})$ is continuous and $GL_n(mathbb{R})$ is open in $M_n(mathbb{R})$ (see: math.stackexchange.com/a/810675/369800). [To understand the proof just linked: determinant continuous (see: math.stackexchange.com/a/121834/369800) and adjoint continuous (see: math.stackexchange.com/a/2031642/369800)]
$endgroup$
– zxmkn
Jan 24 at 19:07
$begingroup$
Recall that the inverse matrix is the adjugate matrix divided by the discriminant. Thus a "singularity" of the inversion only happens when the discriminant vanishes.
$endgroup$
– Alexey
Jan 26 at 20:45