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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
1 year, 3 months ago

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Yes,that works.Every square matrix satisfies its own characteristic equation and hence the result follows.

Indraneel Mukhopadhyaya - 1 year, 1 month ago

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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.

Prasun Biswas - 1 year, 1 month ago

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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.

Indraneel Mukhopadhyaya - 1 year, 3 months ago

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Can anyone help me?

Indraneel Mukhopadhyaya - 1 year, 3 months ago

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