Proposition 10

One has the equality

for every $n\in I_E$.

Lemma 3, c) proves that $\|\widehat{\wedge}^{n+1}u\|=\|\widehat{\wedge}^{n+1}|u|\|$; moreover, since $\alpha_i(u) =\alpha_i(|u|)$ for every $i\in I_E$ (remark 5 of IV, p. 157, (2)), it is enough to prove the assertion for $|u|$. This makes it possible to suppose that F = E and that $u$ is positive.

Let then $B = (e_j)_{j\in J}$ be an orthonormal basis of E such that $u$ is diagonalizable in the basis B (theorem 1 of IV, p. 149), and let $(\lambda_j)_{j\in J}$ be the family of eigenvalues of $u$ in the basis B. Endow J with a total order, and denote by $J_n$ the set of strictly increasing families $(j_0, . . . , j_n)\in J^{n+1}$ of elements of J. The vectors

for $\iota = (j_0, . . . , j_n)\in J_n$ form an orthonormal basis $B_n$ of $\widehat{\wedge}^{n+1}E$ (EVT, V, p. 34, prop. 5). For every $\iota = (j_0, . . . , j_n)\in J_n$, let

Then $(\widehat{\wedge}^{n+1}u)e_{\iota}=\lambda_{\iota}e_{\iota}$, hence $\widehat{\wedge}^{n+1}u$ is diagonal in the basis $B_n$. Consequently, one has

$\|\widehat{\wedge}^{n+1}u\|=$ sup$_{\iota\in I_n}\lambda_{\iota}$

(lemma 1 of IV, p. 147), which is equal to the product $\lambda_0(u)\cdots \lambda_n(u)$ of the $n+ 1$ greatest eigenvalues of $u$. The desired formula follows, since $\lambda_i(u) =\alpha_i(u)$ for every $i\in I_E$ when $u$ is positive (remark 5 of IV, p. 157, (3)).

In particular, if $\alpha_1(u)< \alpha_0(u) =\|u\|$, one sees that the inequality $\|\widehat{\wedge}^n(u)\|\leqslant \|u\|^n$ (EVT, V, p. 34, formula (29)) is not an equality in general if $n\geqslant 2$.