Given a real, symmetric and positive-definite matrix G we have:
Frobenius norm of G = [trace(GG')]^1/2
G' = transposed matrix of G
I need to prove that:
Frobenius norm of G = trace[(GG')^1/2]
Could someone help me please?
[Math] Frobenius norm and trace
trace
Best Answer
$$G=UDU^T$$ $$GG'=UD^2U^T$$
$$(GG')^\frac12=UDU^T=G$$
Hence $$\operatorname{trace}((GG')^\frac12)=\operatorname{trace}(G)=\operatorname{trace}(D)=\sum_{i=1}^nd_{ii}$$ but $$\operatorname{trace}(GG')=\sum_{i=1}^n d_{ii}^2$$ hence $$\left( \operatorname{trace}(GG')\right)^\frac12=\sqrt{\sum_{i=1}^n d_{ii}^2}$$
but we know that $1$-norm are $2$-norm need not be equal.
Example of counter example:
$$G = \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}$$