Question

$\color{green} \rm{If \: A= \begin{bmatrix} \rm{1}&0&0\\ \rm{ i}& \rm{\frac{ - 1 + i \sqrt{3} }{2}}&0 \\0& \rm{1 + 2i}& \rm{\frac{ - 1 - i \sqrt{3} }{2}}\end{bmatrix}} \\ \color{blue} \rm{then \: the \: trace \: of \:A^{102} \: is }$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$A= \begin{bmatrix} \rm{1}&0&0\\ \rm{ i}& \rm{\frac{ - 1 + i \sqrt{3} }{2}}&0 \\0& \rm{1 + 2i}& \rm{\frac{ - 1 - i \sqrt{3} }{2}}\end{bmatrix}$

can be rewritten as

$A= \begin{bmatrix} \rm{1}&0&0\\ \rm{ i}& \rm{ \omega }&0 \\0& \rm{1 + 2i}& \rm{ {\omega }^{2} }\end{bmatrix}$

where,

$\rm \: \omega = \frac{ - 1 + \sqrt{3}i }{2}$

$\rm \: {\omega }^{2} = \frac{ - 1 - \sqrt{3}i }{2}$

Let first find the eigen values of matrix A.

We know, to find eigen values of matrix A,

$\rm \: |A - kI| = 0$

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}1 - k&0&0\\1&\omega - k&0\\0&1 + 2i& {\omega - k}^{2} \end{array}\right | \end{gathered} = 0$

$\rm\implies \:(1 - k)(\omega - k)( {\omega }^{2} - k) = 0$

$\rm\implies \:k = 1, \: \omega , \: {\omega }^{2}$

$\rm\implies \: 1, \: \omega , \: {\omega }^{2} \: are \: eigen \: values \: of \: matrix \: A$

Now, we know that

$\rm \: k_i \: are \: eigen \: values \: of \: matrix \: A$

$\rm\implies \: k_i^{n} \: are \: eigen \: values \: of \: matrix \: {A}^{n}$

So,

$\rm\implies \: {1}^{102}, {\omega }^{102}, {\omega }^{204} \: are \: eigen \: values \: of \: matrix \: {A}^{102}$

$\rm\implies \: 1, {( {\omega}^{3})}^{34}, {( {\omega }^{3} )}^{68} \: are \: eigen \: values \: of \: matrix \: {A}^{102}$

$\rm\implies \: 1, 1, 1 \: are \: eigen \: values \: of \: matrix \: {A}^{102}$

We know, trace of a matrix is sum of the eigen values.

$\rm\implies \:tr( {A}^{102} ) = 1 + 1 + 1 = 3$