Proposition 7

Let $D_+^*$ be the group of diagonal matrices with diagonal elements $> 0$. The mapping $(U, D, T) \mapsto UDT$ is a homeomorphism of $\mathbf{U}(n, \mathbf{K}) \times D_+^* \times \mathbf{T}_1(n, \mathbf{K})$ onto $\mathbf{GL}(n, \mathbf{K})$.

Let $(\varepsilon_1, \ldots, \varepsilon_n)$ be the canonical basis of $K^n$. Let $X \in \mathbf{GL}(n, K)$. Then the $X \cdot \varepsilon_i = f_i$ form a basis of E. To this basis $(f_i)$ one can associate a basis $(e_i)$ as in Lemma 10. Let $U$ be the matrix of the unitary automorphism of E that transforms $\varepsilon_i$ into $e_i$. Then

with $\alpha_{ii} > 0$ for $i = 1, 2, \ldots, n$. Thus $X = UC$, where $C$ is the matrix

Moreover, $U$ and $C$ depend continuously on X by Lemma 10. On the other hand, formula (8) shows that $C$ may be put in the form $DT$ with $D \in \mathbf{D}_+^*$, $T \in \mathbf{T}_1(n, K)$, $D$ and $T$ depending continuously on $C$. The uniqueness of the decomposition $X = UDT$ follows from the uniqueness property of Lemma 10.

The homeomorphism of Prop. 7 is called the Iwasawa decomposition of $\mathbf{GL}(n, K)$.

The group $G = \mathbf{D}_+^* \cdot \mathbf{T}_1(n, K)$ is the set of upper triangular matrices over K whose diagonal elements are $> 0$. Let us identity the element $(z_{ij})$ of this group with the element

Arguing exactly as in Example 4, one finds as right Haar measure on this group the measure (when $K = \mathbf{R}$)

Then applying Prop. 13 of §2, No. 9, one sees that if $\mathbf{GL}(n, K)$ is identified with $\mathbf{U}(n, K) \times G$ by the mapping $(U, S) \mapsto US$, a Haar measure on $\mathbf{GL}(n, K)$ is given by (when $K = \mathbf{R}$)

where $\alpha$ denotes a Haar measure on $\mathbf{U}(n, K)$.