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24 votes
24 votes
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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Best answer
56 votes
56 votes
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 by
8 votes
8 votes
This is an upper triangular matrix, therefore its eigen values or characteristic roots will be the diagonal elements $(1, -1, i, -i)$

therefore, $(\lambda - 1)(\lambda + 1)(\lambda - i)(\lambda + i) = 0$ will hold true

which will give the characteristic equation as:

$\lambda ^{4} - 1 = 0$

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

$\Rightarrow A^{4} - I = 0$

$\Rightarrow A^{4} = I$
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