1 votes 1 votes If $A =\begin{bmatrix} 2 &i \\ i & 0 \end{bmatrix}$ , the trace of $A^{10}$ is $2$ $2(1+i)$ $0$ $2^{10}$ Linear Algebra isi2018-mma engineering-mathematics linear-algebra determinant + – akash.dinkar12 asked May 11, 2019 akash.dinkar12 913 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
7 votes 7 votes $\begin{vmatrix} 2-\lambda & i\\ i& -\lambda \end{vmatrix}$=0 $-(2-\lambda)*\lambda+1=0$ $(\lambda-1)^2=0$ $\lambda=1,1$ Eigen values of A is 1,1 Eigen values of $A^{10}$ will be $1^{10}, 1^{10}$=1,1 Trace of $A^{10}$=Sum of Eigen values=1+1=2 aditi19 answered Sep 23, 2019 aditi19 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes By matrix multiplication , $A^2$ = $\begin{bmatrix} 3 &2i \\ 2i & -1 \end{bmatrix}$ whose trace is 2. $A^3$ = $\begin{bmatrix} 4 & 3i\\ 3i & -2 \end{bmatrix}$ whose trace is 2. So we see the trace of the original matrix remains the same irrespective of repeated multiplication with itself. Therefore trace of $A^{10}$ = 2 Sayan Bose answered May 11, 2019 Sayan Bose comment Share Follow See all 0 reply Please log in or register to add a comment.