Ce sont ceux où la première matrice a autant de colonnes que la seconde a de lignes : ${\displaystyle AB}$, ${\displaystyle BD}$, ${\displaystyle CA}$, ${\displaystyle CC}$ et ${\displaystyle DD}$.

${\displaystyle {\begin{aligned}AB&={\begin{pmatrix}0&1&3&0\\8&5&0&6\\0&0&1&1\end{pmatrix}}{\begin{pmatrix}0&1\\0&0\\3&3\\9&5\end{pmatrix}}\\&={\begin{pmatrix}0\times 0+1\times 0+3\times 3+0\times 9&0\times 1+1\times 0+3\times 3+0\times 5\\8\times 0+5\times 0+0\times 3+6\times 9&8\times 1+5\times 0+0\times 3+6\times 5\\0\times 0+0\times 0+1\times 3+1\times 9&0\times 1+0\times 0+1\times 3+1\times 5\end{pmatrix}}\\&={\begin{pmatrix}9&9\\54&38\\12&8\end{pmatrix}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}BD&={\begin{pmatrix}0&1\\0&0\\3&3\\9&5\end{pmatrix}}{\begin{pmatrix}1&3\\5&0\end{pmatrix}}\\&={\begin{pmatrix}0\times 1+1\times 5&0\times 3+1\times 0\\0\times 1+0\times 5&0\times 3+0\times 0\\3\times 1+3\times 5&3\times 3+3\times 0\\9\times 1+5\times 5&9\times 3+5\times 0\end{pmatrix}}\\&={\begin{pmatrix}5&0\\0&0\\18&9\\34&27\end{pmatrix}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}CA&={\begin{pmatrix}0&1&3\\8&5&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0&1&3&0\\8&5&0&6\\0&0&1&1\end{pmatrix}}\\&={\begin{pmatrix}0\times 0+8\times 1+0\times 3&0\times 1+1\times 5+3\times 0&0\times 3+1\times 0+3\times 1&0\times 0+1\times 6+3\times 1\\8\times 0+5\times 8+0\times 0&8\times 1+5\times 5+0\times 0&8\times 3+5\times 0+0\times 1&8\times 0+5\times 6+0\times 1\\0\times 0+0\times 8+1\times 0&0\times 1+0\times 5+1\times 0&0\times 3+0\times 0+1\times 1&0\times 0+0\times 6+1\times 1\end{pmatrix}}\\&={\begin{pmatrix}8&5&3&9\\40&33&24&30\\0&0&1&1\end{pmatrix}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}CC&={\begin{pmatrix}0&1&3\\8&5&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0&1&3\\8&5&0\\0&0&1\end{pmatrix}}\\&={\begin{pmatrix}0\times 0+1\times 8+3\times 0&0\times 1+1\times 5+3\times 0&0\times 3+1\times 0+3\times 1\\8\times 0+5\times 8+0\times 0&8\times 1+5\times 5+0\times 0&8\times 3+5\times 0+0\times 1\\0\times 0+0\times 8+1\times 0&0\times 1+0\times 5+1\times 0&0\times 3+0\times 0+1\times 1\end{pmatrix}}\\&={\begin{pmatrix}8&5&3\\40&33&24\\0&0&1\end{pmatrix}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}DD&={\begin{pmatrix}1&3\\5&0\end{pmatrix}}{\begin{pmatrix}1&3\\5&0\end{pmatrix}}\\&={\begin{pmatrix}1\times 1+3\times 5&1\times 3+3\times 0\\5\times 1+0\times 5&5\times 3+0\times 0\end{pmatrix}}\\&={\begin{pmatrix}16&3\\5&15\end{pmatrix}}.\end{aligned}}}$

Soit ${\displaystyle (e_{1},\dots ,e_{n})}$ la base canonique de ${\displaystyle K^{n}}$ et ${\displaystyle E_{k}=\operatorname {Vect} \left(e_{1},\dots ,e_{k}\right)}$, pour ${\displaystyle k}$ de ${\displaystyle 0}$ à ${\displaystyle n}$.

Pour tout ${\displaystyle i}$ de ${\displaystyle 1}$ à ${\displaystyle n}$, ${\displaystyle A(e_{i})\in E_{i-1}}$ donc pour tout ${\displaystyle k}$ de ${\displaystyle 1}$ à ${\displaystyle n}$, ${\displaystyle A(E_{k})\subset E_{k-1}}$ et a fortiori, ${\displaystyle A^{k}(E_{k})=A^{k-1}(A(E_{k}))\subset A^{k-1}(E_{k-1})}$. Par conséquent :

${\displaystyle A^{n}(K^{n})=A^{n}(E_{n})\subset A^{n-1}(E_{n-1})\subset \dots \subset A(E_{1})\subset E_{0}=\operatorname {Vect} (\varnothing )=\{0\}}$.