# Obtaining a lower bound of a matrix norm

I was wondering (on a setting where $$\vec X_i \sim \mathcal{N}(\vec\mu, \mathbb{I})$$ are $$n$$ random $$d$$-dimensional multivariate normal vectors with unknown mean $$\vec\mu$$) how I could obtain a lower bound, with high probability, on the spectral norm:

$$\left\| \frac{1}{n} \sum_{i=1}^n \vec X_i \vec X_i^T - \hat{\vec\mu} \hat{\vec\mu}^T - \mathbb{I} \right\|$$

without having to depend my bound on $$\|\vec\mu\|$$.

In the above quantity, $$\hat{\vec\mu}$$ denotes an estimate of the mean value, for which I have that, with high probability and $$n=O(\text{something})$$ it holds $$\| \hat{\vec\mu} - \vec\mu \| \leq \epsilon$$ i.e. I can get a "good" approximation to $$\vec\mu$$.

Given that the covariance matrix of $$\vec X_i$$ is merely $$\mathbb{I}$$, this should be enough for me to lower bound the above quantity without $$\|\vec\mu\|$$.

It seems to me intuitive that I should be able to get the above quantity a lower bound (with high probability) without relying on some bound of $$\|\vec\mu\|$$ but I'm not sure how to obtain that. Trying to break the $$\frac{1}{n} \sum_{i=1}^n \vec X_i \vec X_i^T$$ into $$\frac{1}{n} \sum_{i=1}^n (\vec X_i - \vec\mu) (\vec X_i - \vec\mu)^T$$ hasn't led me somewhere (?), because then I need to additionally bound terms like $$\left\|\vec\mu \left( \frac{1}{n} \sum_{i=1}^n \vec X_i - \vec\mu \right)^T \right\|$$ which would introduce into my bound the terms $$\|\vec\mu\|$$ that I do not desire.

What could I do to lower bound the above quantity and avoid any dependence on $$\|\vec\mu\|$$?

In the above, $$\|\cdot\|$$ denotes the Euclidean norm for vectors and the spectral 2-norm for matrices.

• Is each $X_i$ a column vector? May 15 '21 at 18:55
• @NealYoung yes, it is a $d$-dimensional column vector.
– Jay
May 15 '21 at 22:21