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 $$ \sup_{\Vert T\Vert=1} \vert{A T^k}\vert \le \kappa_k\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert. $$ holds true in general with the best constant $ \kappa_{k}= k^k/k!. $

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 $$ \sup_{\Vert T\Vert=1} \vert{A T^k}\vert \le \sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert\le \kappa_k \sup_{\Vert T\Vert=1} \vert{A T^k}\vert, $$ holds true in general with the best constant $ \kappa_{k}= k^k/k!. $

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 $$ \sup_{\Vert T\Vert=1} \vert{A T^k}\vert =\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert $$

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 $$ \sup_{\Vert T\Vert=1} \vert{A T^k}\vert \le \kappa_k\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert. $$ holds true in general with the best constant $ \kappa_{k}= k^k/k!. $