Limit of matrix inverse: $\lim_{\lambda \to \infty} (A + \lambda I)^{-1} = \mathbf{0}$? 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.
 A: 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.
A: 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.  
