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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?

Luciana
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1 Answers1

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$$G=UDU^T$$ $$GG'=UD^2U^T$$

$$(GG')^\frac12=UDU^T=G$$

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:

$$G = \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}$$

Siong Thye Goh
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  • Thank you for answering! I saw on the page (https://math.stackexchange.com/questions/1864772/lower-bounding-the-trace-of-a2-using-the-trace-of-at-a?rq=1) that there is a class of matrices that satisfy tr(A^2) = (trA)^2. But I don't know what kind of matrix this is. Do you know if this is true for real matrices, symmetric and positive-defined? And perhaps I can start from this premise to prove the equation above. – Luciana Oct 17 '17 at 16:52
  • I think my counter example is real, symmetric, and positive definite and it doesn't satisfy the equality as well. $tr(A^2)=1+2^2=5$ but $tr(A)^2=(1+2)^2=9$. – Siong Thye Goh Oct 17 '17 at 17:42
  • If the matrix is real, symmetric, positive definite, and the eigenvalues are from ${0,1}$, then I think we have $tr(A^2)=(tr(A))^2$. – Siong Thye Goh Oct 17 '17 at 17:46
  • Thank you for answering! – Luciana Oct 24 '17 at 17:22