In 19 and 20 verify the foregoing result for the given
Chapter 8, Problem 20E(choose chapter or problem)
Let P denote a matrix whose columns are eigenvectors \(\mathbf{K}_{1}, \quad \mathbf{K}_{2},\) . . . , \(\mathbf{K}_{n}\) corresponding to distinct eigenvalues \(\lambda_{1}, \lambda_{2}\), . . . , \(\lambda_{n}\) of an n x n matrix A. Then it can be shown that \(\mathbf{A}=\mathbf{P D P}^{-1}\), where D is a diagonal matrix defined by
\(\mathbf{D}=\left(\begin{array}{cccc}
\lambda_{1} & 0 & \cdots & 0 \\
0 & \lambda_{2} & \cdots & 0 \\
\cdot & \cdot & & \cdot \\
\cdot & \cdot & & \vdots \\
0 & 0 & \cdots & \lambda_{n}
\end{array}\right)
\) (9)
In Problems 19 and 20 verify the foregoing result for the given matrix.
Text Transcription:
mathbf K_1
mathbf K_2
mathbf K_n
lambda_1
lambda_2
lambda_n
mathbf A =mathbf P D P^-1
mathbf D = ({array} cccc lambda_1 & 0 & cdots & 0 0 & lambda_2 & cdots & 0 cdot & cdot & & cdot cdot & cdot & & vdots 0 & 0 & cdots & lambda_n {array})
mathbf A = ({array} ll 2 & 1 1 & 2 {array})
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer