If $A\\in M_{n}(R)$ and $A$ is supertriangular then $A^{n}=0$

theorem: Let $R$ be commutative ring with identity.If an n-square matrix $A\\in Mat_{n}(R)$ is supertriangular then $A^{n}=0$ .

proof: Find the characteristic polynomial of $A$ by computing the determinant of $A-tI$ . The square matrix $A-tI$ is a triangular matrix. The determinant of a triangular matrix is the product of the diagonal element of the matrix. Therefore the characteristic polynomial is $p(t)=t^{n}$ and by the Cayley-Hamilton theorem the matrix $A$ satisfies the polynomial. That is $A^{n}=0$ .
QED

单词	IfAinMnRAndAIsSupertriangularThenAn0
释义	If $A\\in M_{n}(R)$ and $A$ is supertriangular then $A^{n}=0$ theorem: Let $R$ be commutative ring with identity.If an n-square matrix $A\\in Mat_{n}(R)$ is supertriangular then $A^{n}=0$ . proof: Find the characteristic polynomial of $A$ by computing the determinant of $A-tI$ . The square matrix $A-tI$ is a triangular matrix. The determinant of a triangular matrix is the product of the diagonal element of the matrix. Therefore the characteristic polynomial is $p(t)=t^{n}$ and by the Cayley-Hamilton theorem the matrix $A$ satisfies the polynomial. That is $A^{n}=0$ . QED
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If A∈Mn⁢(R) and A is supertriangular then An=0

If $A\\in M_{n}(R)$ and $A$ is supertriangular then $A^{n}=0$