Steps to Diagonalize a Matrix
Step 5: Verify the Diagonalization
- To confirm, check:
\[\LARGE
A = \color{red}{P} \color{blue}{D} \color{green}{P^{-1}}
\]
- P−1 is the inverse of P.
\[\Large
A =
\begin{bmatrix}
4 & 1 \\
3 & 2
\end{bmatrix}
\]
\[\Large P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \]
\[\Large D = \begin{bmatrix} \textcolor{red}{5} & 0 \\ 0 & \textcolor{blue}{1} \end{bmatrix} \]
\[\Large P^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \]
\[\Large P D = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} \textcolor{red}{5} & 0 \\ 0 & \textcolor{blue}{1} \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 5 & -1 \end{bmatrix} \]
\[\Large P D P^{-1} = \begin{bmatrix} 5 & 1 \\ 5 & -1 \end{bmatrix} \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} = A \]
\[\Large P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \]
\[\Large D = \begin{bmatrix} \textcolor{red}{5} & 0 \\ 0 & \textcolor{blue}{1} \end{bmatrix} \]
\[\Large P^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \]
\[\Large P D = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} \textcolor{red}{5} & 0 \\ 0 & \textcolor{blue}{1} \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 5 & -1 \end{bmatrix} \]
\[\Large P D P^{-1} = \begin{bmatrix} 5 & 1 \\ 5 & -1 \end{bmatrix} \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} = A \]
Since A has satisfied the above equation we can conclude “A is Diagonalizable”
Which of the following statements verifies that a matrix A is diagonalizable?