For a matrix $A \in \mathbb R^{n \times n}$, prove that $|tr(A)|\leq \sqrt{n}||A||_{F}$, where $tr(A)$ denotes the trace of the matrix A, and $||.||_{F}$ denotes the Frobenius norm.
We know that $|tr(A)|=|\sum_{i=1}^{n}a_{i,i}|$, whereas, $||A||_{F}=\sqrt{\sum_{i=1}^{n}\sum_{j=1}^{n}|a_{i,j}|^{2}}$.
$\implies \sqrt{n}||A||_{F} = \sqrt{n(\sum_{i=1}^{n}\sum_{j=1}^{n}|a_{i,j}|^{2})} \geq \sqrt{n(\sum_{i=1}^{n}|a_{i,i}|^{2})} = \sqrt{\sum_{i=1}^{n}e_{i}^{2}}\sqrt{\sum_{i=1}^{n}|Ae_i|^{2}}$, where $\{e_1,…,e_n\}$ forms the standard basis of $\mathbb R^n$.
But by Cauchy Schwarz Inequality, $\sqrt{\sum_{i=1}^{n}e_{i}^{2}}\sqrt{\sum_{i=1}^{n}|Ae_i|^{2}} \geq |\sum_{i=1}^{n} \langle e_i, Ae_i \rangle| = |tr(A)|$.
Hence, the proof.
Is there any mistake in my proof? I will be grateful if someone could grade my proof and correct it if necessary. I am using this result to prove the Backward stability of Cholesky's method. Thanks.
Best Answer
Your proof is correct, although somewhat unclear in notation. (What is $e_i$? is it a vector? So why are you squaring it?). Anyway, you can use Cauchy-Schwarz inequality a bit earlier, by noting that $$\sqrt{n\sum_{i=1}^n|a_{i,i}|^2}\geq\sum_{i=1}^n|a_{i,i}|\geq |tr(A)|.$$