Limit involving norm of matrices

Question

How can I show that $$ \lim_{V\to 0}| \text{Tr}(V^2)|/||V|| =0$$ where $||\cdot || $ is any norm $V$ is an $n\times n $ real matrix?

Answer 1

Let $M$ denote the set of $n\times n$ matrices of the reals. Then, the mapping $$\langle A,B \rangle = Tr(A^T B) $$ is an inner product on $M$, which induces the Frobenius norm $$ \|A\|_F^2 = \langle A, A \rangle. $$

1. By Cauchy Schwarz you obtain $|Tr(A^2)| \le \|A\|^2_F$. (notice it is really $\le$ and not $=$.)

2. Since $M$ is finite dimensional, every pair of norms are equivalent. Thus, there exist constants $c, C >0$ with $$ c \|A\|_F \le \|A\| \le C\|A\|_F $$ for every $A\in M$.

3. Both together yield the claimed statement.

Answer 2

Let $V=(V_{ij})_{n\times n}$. Fix $M(V)= \max_{ij}|V_{ij}|$. Note that $V\to 0\Leftrightarrow M(V)\to 0$. Notice that $$ V^2=V\cdot V= (V_{ij})_{n\times n}\cdot (V_{ij})_{n\times n}=\left(\sum_{k=1}V_{ik}\cdot V_{kj}\right)_{n\times n} $$ implies $$ \mathrm{trace}(V^2) = \mathrm{trace}\left(\left(\sum_{k=1}^nV_{ik}\cdot V_{kj}\right)_{n\times n}\right) =\sum_{\ell=1}^{n}\left(\sum_{k=1}^nV_{\ell k}\cdot V_{k\ell}\right) = \sum_{\ell=1}^{n}\sum_{k=1}^nV_{\ell k}\cdot V_{k\ell} $$ and $$ |\mathrm{trace}(V^2)|= M^2(V)\cdot \sum_{\ell=1}^{n}\sum_{k=1}^n\frac{V_{\ell k}}{M(V)}\cdot \frac{V_{k\ell}}{M(V)} \leq M^2(V)\sum_{\ell=1}^{n}\sum_{k=1}^n 1\cdot 1\leq M^2(V) n^2 $$ for $M(V)=\max_{\ell, k}|V_{\ell k }|$. Since $\|V\|=\sqrt[2\,]{\mathrm{trace}(V^TV)}$, we have that similarly to the calculations above that $$ \|V\|= M(V)\cdot \sqrt[2\,]{\underbrace{\sum_{\ell=1}^{n}\sum_{k=1}^n\frac{V_{\ell k}}{M(V)}\cdot \frac{V_{k\ell}}{M(V)}}_{>1}}\geq M(V)\sqrt{1}=M(V). $$ So we can conclude that $$ 0\leq {|\mathrm{Trace}(V^2)|\over ||V|| }\leq \frac{M(V)^2\cdot n^2}{M(V)}= M(V)\cdot n^2\to 0 \mbox{ for } V\to 0 $$