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.

#Algebra

Easy Math Editor

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Comments

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 - 4 years, 10 months ago

Can anyone help me?

Indraneel Mukhopadhyaya - 4 years, 10 months 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.

Prasun Biswas - 4 years, 8 months ago

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

Indraneel Mukhopadhyaya - 4 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$