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Show that the matrix A is diagonalisable. Also find the diagonal form and a diagonalising matrix $A = \begin{bmatrix} \ -9 & 4 & 4 \\ \ -8 & 3 & 4 \\ \ -16 & 8 & 7 \\ \end{bmatrix}$
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First, find the eigenvalues and eigenvectors

Eigenvalue: 3, eigenvector: $\left[\begin{array}{c}\frac{1}{2} \\ \frac{1}{2} \\ 1\end{array}\right]$.

Eigenvalue: $-1$, eigenvector: $\left[\begin{array}{l}\frac{1}{2} \\ 1 \\ 0\end{array}\right]$.

Eigenvalue: $-1$, eigenvector: $\left[\begin{array}{l}\frac{1}{2} \\ 0 \\ 1\end{array}\right]$.

Form the matrix $P$, whose column $i$ is eigenvector no. $i: P=\left[\begin{array}{ccc}\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & 1 & 0 \ 1 & 0 & 1\end{array}\right]$. Form the diagonal matrix $D$ whose element at row $i$, column $i$ is eigenvalue no. $i$ : $D=$ $\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$

The matrices $P$ and $D$ are such that the initial matrix $\left[\begin{array}{ccc}-9 & 4 & 4 \ -8 & 3 & 4 \ -16 & 8 & 7\end{array}\right]=P D P^{-1}$. $$ P=\left[\begin{array}{ccc} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & 1 & 0 \ 1 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0.5 & 0.5 & 0.5 \ 0.5 & 1 & 0 \ 1 & 0 & 1\ \end{array}\right] \text { A } $$

$$ D=\left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right] \text { A } $$

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