Complexity Explorer Santa Fe Institute

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Vector and Matrix Algebra

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3.1 Diagonalization and Powers » Quiz #17 Solution

Question 1:

Recall that the diagonalization of matrix A from Quiz #16 was given as: A=\begin{bmatrix} 1 & -\frac{2}{3} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 3 \end{bmatrix}.

It therefore follows that  A^{10}=\begin{bmatrix} 1 & -\frac{2}{3} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix} ^{10} \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{2}{3} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1024 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 3 \end{bmatrix} =\begin{bmatrix} 1024 & 341} \\ 0 & 1 \end{bmatrix}.