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# Understand the problem

If A = \begin{bmatrix} 2&i\\i&0\end{bmatrix} , the trace of $$A^{10}$$ is
##### Source of the problem
Sample Questions (MMA):2019
Linear Algebra
Medium
##### Suggested Book
Introduction to Linear Algebra by Pearson

Do you really need a hint? Try it first!

A = \begin{pmatrix} 2&i\\i&0 \end{pmatrix} ; $$A^2$$ = \begin{pmatrix} 3&2i\\2i&-1 \end{pmatrix} $$A^3$$ = \begin{pmatrix} 4&3i\\i&-2 \end{pmatrix} Now , we can observe that the trace is always going to be 2 . Now we’re to seal the deal
One way to think about this problem is to go into eigen values as the tr(A) .( Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values and proper values, or latent roots .) (Eigenvectors are a special set of vectors associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic vectors, proper vectors, or latent vectors ), remains unaltered in between similar matrices.
| $$\lambda$$I – A | = 0 => \begin{matrix} \lambda -2&-i\\-i&\lambda \end{matrix} =0 => $$\lambda$$( $$\lambda$$ – 2) + 1 = 0 => $$\lambda^2$$ – 2$$\lambda$$ + 1 = 0 =>$$(\lambda – 1)^2$$ = 0 => $$\lambda$$ = 1,1 So , what will be the canonical form ?
So , the canonical matrix will be \begin{pmatrix} 1&0\\0&1 \end{pmatrix} = A’ , i.e $$\le P \epsilon GL_2(\mathbb{C})$$ o.t A’ = $$PAP^{-1}$$ $$(A’)^{10}$$ = \begin{pmatrix} 1^{10}&0\\0&1^{10}\end{pmatrix} tr($$A^{10}$$) = tr(($$P^{-1}A’P)^{10}$$) => tr($$P^{-1}(A’)^{10}P)$$ => tr(($$A’)^{10}$$) => ($$1^{10}$$) + ($$1^{10}$$) => 2(Ans)

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