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You can be completely explicit in this matter. For $T_j$ in a commutative algebra $$T_1T_2\dots T_k=\frac{1}{2^k k!}\sum_{\epsilon_j=\pm 1} \epsilon_1\dots\epsilon_k(\epsilon_1T_1 +\dots+\varepsilon_{k}T_{k})^k.$$ The following lemma in available in the Euclidean case.

Lemma. Let $V$ be an Euclidean finite-dimensional vector space, and $A$ a symmetric $k$-multilinear form. We have $\sup_{\Vert T\Vert=1} \vert{A T^k}\vert =\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert.$

This lemma is a consequence of the 1928 paper by O.D. Kellogg [MR1544896]. This is not true in the non-Euclidean case where the inequality $$\sup_{\Vert T\Vert=1} \vert{A T^k}\vert \le \kappa_k\sup_{\Vert{T_j}\Vert=1} sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert. T_k}\vert\le \kappa_k \sup_{\Vert T\Vert=1} \vert{A T^k}\vert,$$ holds true in general with the best constant $\kappa_{k}= k^k/k!.$

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You can be completely explicit in this matter. For $T_j$ in a commutative algebra $$T_1T_2\dots T_k=\frac{1}{2^k k!}\sum_{\epsilon_j=\pm 1} \epsilon_1\dots\epsilon_k(\epsilon_1T_1 +\dots+\varepsilon_{k}T_{k})^k.$$ The following lemma in available one in the Euclidean case.

Lemma. Let $V$ be an Euclidean finite-dimensional vector space, and $A$ a symmetric $k$-multilinear form. We have  \sup_{\Vert T\Vert=1} \vert{A T^k}\vert =\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert. $This lemma is a consequence of the 1928 paper by O.D. Kellogg [MR1544896]. This is not true in the non-Euclidean case where the inequality $$\sup_{\Vert T\Vert=1} \vert{A T^k}\vert \le \kappa_k\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert.$$ holds true in general with the best constant$ \kappa_{k}= k^k/k!. $1 You can be completely explicit. For$T_j$in a commutative algebra $$T_1T_2\dots T_k=\frac{1}{2^k k!}\sum_{\epsilon_j=\pm 1} \epsilon_1\dots\epsilon_k(\epsilon_1T_1 +\dots+\varepsilon_{k}T_{k})^k.$$ The following lemma in available one the Euclidean case. Lemma. Let$V$be an Euclidean finite-dimensional vector space, and$A$a symmetric$k\$-multilinear form. We have $$\sup_{\Vert T\Vert=1} \vert{A T^k}\vert =\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert$$