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
Difficulty Level
Suggested Book
Introduction to Linear Algebra by Pearson

Start with hints

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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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