Complexity Explorer Santa Fe Institute

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Vector and Matrix Algebra

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3.1 Diagonalization and Powers » Quiz #16 Solution

The eigenvalues of the matrix  A=\begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix} are 2 and -1 with corresponding eigenvectors: v_1= \langle 1,0 \rangle, v_2 =\langle -\frac{1}{3}, 1\rangle, respectively.

This gives the following diagonalization:  A=\begin{bmatrix} 1 & -\frac{1}{3} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & -\frac{1}{3} \\ 0 & 1\end{bmatrix}^{-1}=\begin{bmatrix} 1 & -\frac{1}{3} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1\end{bmatrix}.