What is the limit of $n(nI + A)^{-1}$ for $A$ p.s.d. matrix?

Question

Let $A$ be $K \times K$ positive definite symmetric matrix with known inverse $A^{-1}$ and set $I$ the identity matrix of dimension $K$.

Can one show that (generally or under some conditions)

$$\lim_{n \rightarrow \infty} n \left(nI + A\right)^{-1} = I$$

Answer 1

This limit holds for any matrix. Any $k \times k$ matrix $A$ cannot have eigenvalues of arbitrary magnitude, so $(n I + A)^{-1}$ is defined for all $n$ in some neighborhood of infinity (and minus infinity). In this neighborhood we can state that $$n\left(nI + A\right)^{-1}=n\left(n(I + \frac{1}{n}A)\right)^{-1} = \left(I + \frac{1}{n}A\right)^{-1}.$$ So if we define $f(h) = \left(I + hA\right)^{-1}$, we are interested in the limit $$\lim_{h\to 0} f(h).$$ We have seen that $f$ is indeed defined in a neighborhood around $0$ and since taking inverses is continuous we find that $f$ is continuous and thus $$\lim_{n \to \infty} n\left(nI + A\right)^{-1} = \lim_{h \to 0} f(h)= f(0) = (I+ 0 \cdot A)^{-1} = I.$$

---

I think of this argument as a sort of generalization of how I would prove that $$\lim_{n \to \infty} \frac{n}{n + a} = 1$$ for any real (or indeed complex) number $a$.

Answer 2

Another solution using Neumann series.

$n\left(nI + A\right)^{-1}== \left(I - \left(- \frac{1}{n}A \right)\right)^{-1}.$

$\sum_{k=0}^\infty \left(- \frac{1}{n}A \right)^k$ is converging in operator norm for any matrix $A$ for $n$ big enough, since for such n, $\frac{|||A|||}n \leq 1$ . Then, for $n$ big enough :

$$\left(I - \left(- \frac{1}{n}A \right)\right)^{-1}=\sum_{k=0}^\infty \left(- \frac{1}{n}A \right)^k=I+\sum_{k=1}^\infty \left(- \frac{1}{n}A \right)^k=I+S_n.$$

Finally, $S_n$ tends to $0$ in operator norm as $n \to \infty$. Indeed :

$$|||S_n|||\leq \sum_{k=1}^\infty \left(\frac{1}n\right)^k |||A|||^k=\sum_{k=1}^\infty \left(\frac{|||A|||}n\right)^k.$$

But, for n big enough $\frac{|||A|||}n \leq 1$, then :

$$|||S_n|||\leq \sum_{k=1}^\infty \left(\frac{|||A|||}n\right)^k = \frac{|||A|||}n \frac 1 {1-\frac{|||A|||}n}\to 0$$

Answer 3

See $A$ as a linear bounded operator on $\mathbb{C}^K$, then for any continous function $f$ on the spectrum $\sigma(A)$ of $A$, $$\lVert f\rVert=\lVert f(A)\rVert_{op}.$$

Define continuous function $f_n=n(n+t)^{-1}, f=1$ on $\sigma(A)$,then $f_n$ converges to $f$ uniformly, so $\lVert f_n(A)-f(A)\rVert_{op}=\lVert f_n-f\rVert\to 0$.