Let k be an algebraically closed field of characteristic different from 2. Up to isomorphism, the algebra ${\mathrm{Mat}}_{n\times n}\left(k\right)$ can be endowed with a k-linear involution in one way if n is odd and in two ways if n is even. In this paper, we consider r-tuples $A_{\bullet }\in {\mathrm{Mat}}_{n\times n}(k{)}^r$ such that the entries of $A_{\bullet }$ fail to generate ${\mathrm{Mat}}_{n\times n}\left(k\right)$ as an algebra with involution. We show that the locus of such r-tuples forms a closed subvariety $Z(r;V)$ of ${\mathrm{Mat}}_{n\times n}(k{)}^r$ that is not irreducible. We describe the irreducible components and we calculate the dimension of the largest component of $Z(r;V)$ in all cases. This gives a numerical answer to the question of how generic it is for an r-tuple $(a_1,\dots ,a_r)$ of elements in ${\mathrm{Mat}}_{n\times n}\left(k\right)$ to generate it as an algebra with involution.

设k是特征不同于 2 的代数闭域。直到同构，代数${垫}_{n\times n}\left(k\right)$如果n是奇数，则可以以一种方式赋予k -线性对合，如果n是偶数，则可以以两种方式赋予。在本文中，我们考虑r -元组$A_{\bullet }\in {垫}_{n\times n}(k{)}^r$这样的条目$A_{\bullet }$无法生成${垫}_{n\times n}\left(k\right)$作为具有对合的代数。我们表明这种r -元组的轨迹形成了一个封闭的子变体$Z(r;V)$的${垫}_{n\times n}(k{)}^r$那不是不可简化的。我们描述了不可约成分，我们计算了最大成分的维数$Z(r;V)$在所有情况下。这给出了r元组的通用性问题的数值答案$(A_{\mathrm{1个}},\dots \dots ,A_r)$中的元素${垫}_{n\times n}\left(k\right)$将其生成为具有对合的代数。