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If $A = \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & i & i \\ 0 & 0 & 0 & -i \end{pmatrix}$ the matrix $A^4$, calculated by the use of Cayley-Hamilton theorem or otherwise, is _______

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Let λ be eighen value

Characteristic polynomial is

$(1-λ)(-1-λ)(i-λ)(-i-λ)$

$=\left ( \lambda ^{2}-1 \right )\left ( \lambda ^{2}+1 \right )$

$=\lambda ^{4}-1$

Characteristic equation is $\lambda ^{4}-1=0$

According to Cayley Hamilton theorem every matrix matrix satisfies its own characteristic equation

So, $A^{4}=$$I$

edited

Cayley- Hamilton Theorem is very useful to find:

•  Inverse, of the given matrix
• The higher power of  the given matrix

from the characteristic equation how we can conclude A^4=I ?

edited

see the above videos.

According to Cayley Hamilton theorem, every matrix satisfies its own characteristic equation and vice versa.

got it..thanks!

This is an upper triangular matrix, therefore its eigen values or characteristic roots will be all the diagonal elements (1, -1, i, -i)

therefore, (λ – 1)(λ + 1)(λ – i)(λ + i) = 0 will hold true

which will give characteristic equation as:

λ^4 – 1 = 0

and according to Cayley Hamilton theorem every matrix matrix satisfies its own characteristic equation

=> A^4 – I = 0

A^4 = I