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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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.

Log in to reply

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.

Log in to reply

Can anyone help me?

Log in to reply