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# Power of square matrix vanishes

A square matrix $$A$$ of order $$n$$ has all its principle diagonal elements and the elements lying below the principle diagonal equal to 0.

Prove that $$A^n$$ is a 0 matrix.

Note: Solve it using eigenvalues method.I already have solved using induction,so please do not give a solution using induction.

Note by Indraneel Mukhopadhyaya
9 months, 2 weeks ago

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Yes,that works.Every square matrix satisfies its own characteristic equation and hence the result follows. · 7 months, 2 weeks ago

Recalling that the eigenvalues of a triangular matrix are the elements on the principal diagonal (which are all $$0$$ for $$A$$), this is now trivial by the Cayley-Hamilton theorem. · 7 months, 2 weeks ago

By definition, the eigen values of A are the roots of the polynomial in $$k$$, $$|A-kI|$$$$=0$$.This can be simplified to get $$(-1)^n$$$$k^n$$$$=0$$.Hence, we get that, all the n eigen values of A are identically equal to 0.Hence, the n eigen values of $$A^n$$ are also each equal to 0.Also, when k is an eigen value of $$A^n$$ , then for the vector X, $$A^n$$$$X$$$$=kX$$$$=0$$.Since, every eigen value of $$A^n$$ is 0, so we get, $$A^n$$$$X$$$$=0$$, which means that the matrix operation $$A^n$$ is scaling the vector X , into a zero vector.Does this imply $$A^n$$$$=0$$ ? I am not able to understand from this step onwards. · 9 months, 1 week ago