Simple looking matrix limit

Question

I am trying to evaluate:

$$\lim_{V\rightarrow {\bf{0}}^{n\times n}}\frac{|\text{tr}(V^2)|}{\|V\|_{op}}$$

($\text{tr}(V^2)$ is the trace of the matrix $V^2$)

Hoping that the limit is $0$ to show a function is Frechet differentiable, it seems this should be the answer intuitively.

I am slightly stuck, my only idea at the moment is

$$\lim_{V\rightarrow {\bf{0}}^{n\times n}}{|\frac{1}{\|V\|_{op}}\text{tr}(VV)|} = \lim_{V\rightarrow {\bf{0}}^{n\times n}}{|\text{tr}(\hat{V}V)|} = \text{tr}({\bf{0}}^{n\times n}) = 0$$

Which feels incorrect.

Answer 1

Write $V = [v_{ij}]$ and notice that by the Cacuchy-Schwarz inequality

$$ \left| \operatorname{tr}(V^2) \right| = \left|\sum_{i,j=1}^{n} v_{ij}v_{ji} \right| \leq \sum_{i,j=1}^{n} |v_{ij}|^2 = \operatorname{tr}(V^* V) = \sum_{i=1}^{n} \lambda_i $$

where $V^*$ is the adjoint (conjugate transpose) of $V$ and $\lambda_i$'s are eigenvalues of $V^*V$. (Since $V^*V$ is positive-semidefinite Hermitian matrix, it follows that $\lambda_i \geq 0$.) We also notice that

$$ \|Vx\|^2 = \langle Vx, Vx\rangle = \langle x, V^* Vx \rangle $$

and hence by the spectral theorem we obtain

$$ \| V \|_{\text{op}} = \max_{\|x\|=1} \frac{\|Vx\|}{\|x\|} = \left( \max_{i=1,\cdots,n} \lambda_i \right)^{1/2}. $$

Combining altogether,

$$ \left| \operatorname{tr}(V^2) \right| = \sum_{i=1}^{n} \lambda_i \leq n \left( \max_{i=1,\cdots,n} \lambda_i \right) = n \| V\|_{\text{op}}^2. $$

Therefore the conclusion follows from the squeezing lemma.

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Let me also give a slightly different reasoning: Consider the Frobenius norm

$$\| V \|_F = \sqrt{\operatorname{tr}(V^*V)} = \left( \sum_{i,j=1}^{n} |v_{ij}|^2 \right)^{1/2}.$$

It is easy to check that $ (A, B) \mapsto \operatorname{tr}(A^* B) $ is an inner product and hence $\|V\|_F$ is indeed a norm. Now recall that any two norms on a finite-dimensional space are comparable to each other, so there exists an absolute constant $C > 0$ for which $ \| V\|_F \leq C \|V\|_{\text{op}} $ holds for any $n\times n$ matrix $V$. So by the Cauchy-Schwarz inequality,

$$ \left|\operatorname{tr}(V^2)\right| \leq \| V^* \|_F \| V \|_F = \| V \|_F^2 \leq C^2 \|V \|_{\text{op}}^2 $$

and again the limit is zero by the squeezing lemma.

Answer 2

Assume that $V$ approaches the zero matrix from a constant direction $X$, i.e. $$V=\beta X$$ Then $$\eqalign{ L &= \lim_{V\to 0} \frac{|{\rm tr}(V^2)|}{\|V\|} \cr &= \lim_{\beta\to 0} \frac{|\beta|^2\,\,|{\rm tr}(X^2)|}{|\beta|\,\,\|X\|} \cr &= \frac{|{\rm tr}(X^2)|}{\|X\|}\,\,\lim_{\beta\to 0}|\beta| \cr &= 0 \cr }$$