Let matrix $A \in \mathbb{R}^{n\times n}$ 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.

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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. $\endgroup$ Jan 24, 2019 at 17:27
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    $\begingroup$ 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 $\lambda\to\infty$ can be evaluated term-wise, proving the desired claim. $\endgroup$ Jan 24, 2019 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, 2019 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, 2019 at 20:45
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    $\begingroup$ @Mah I would like to think we can do so, but the argument is likely to be more convoluted. I think because $B^TB$ is positive definite, we can lower bound the smallest eigenvalue of $A+\lambda B^TB$ so that it goes to infinity with $\lambda$, then we can be done. $\endgroup$ Feb 22, 2021 at 4:09

2 Answers 2


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 $$ \lim_{\lambda\to\infty}(A+\lambda I)^{-1}= S\Bigl(\,\lim_{\lambda\to\infty}(D+\lambda I)^{-1}\Bigr)S^{-1}=0 $$ It is not necessary that $A$ is semipositive definite. Any symmetric matrix will do.


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.

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    $\begingroup$ I may not be астон вілла олоф мэллбэрг anymore, but Teresa is very pleased that her idea was useful, even if it was two years ago! $\endgroup$ Feb 22, 2021 at 4:49

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