Diagonalize $$$\left[\begin{array}{ccc}2 & 1 & 4\\0 & 2 & 0\\1 & 1 & 2\end{array}\right]$$$

The calculator will diagonalize (if possible) the square $$$3$$$x$$$3$$$ matrix $$$\left[\begin{array}{ccc}2 & 1 & 4\\0 & 2 & 0\\1 & 1 & 2\end{array}\right]$$$, with steps shown.
A

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Diagonalize $$$\left[\begin{array}{ccc}2 & 1 & 4\\0 & 2 & 0\\1 & 1 & 2\end{array}\right]$$$.

Solution

First, find the eigenvalues and eigenvectors (for steps, see eigenvalues and eigenvectors calculator).

Eigenvalue: $$$4$$$, eigenvector: $$$\left[\begin{array}{c}2\\0\\1\end{array}\right]$$$.

Eigenvalue: $$$2$$$, eigenvector: $$$\left[\begin{array}{c}4\\-4\\1\end{array}\right]$$$.

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

Form the matrix $$$P$$$, whose column $$$i$$$ is eigenvector no. $$$i$$$: $$$P = \left[\begin{array}{ccc}2 & 4 & -2\\0 & -4 & 0\\1 & 1 & 1\end{array}\right]$$$.

Form the diagonal matrix $$$D$$$ whose element at row $$$i$$$, column $$$i$$$ is eigenvalue no. $$$i$$$: $$$D = \left[\begin{array}{ccc}4 & 0 & 0\\0 & 2 & 0\\0 & 0 & 0\end{array}\right]$$$.

The matrices $$$P$$$ and $$$D$$$ are such that the initial matrix $$$\left[\begin{array}{ccc}2 & 1 & 4\\0 & 2 & 0\\1 & 1 & 2\end{array}\right] = P D P^{-1}$$$.

$$$P^{-1} = \left[\begin{array}{ccc}\frac{1}{4} & \frac{3}{8} & \frac{1}{2}\\0 & - \frac{1}{4} & 0\\- \frac{1}{4} & - \frac{1}{8} & \frac{1}{2}\end{array}\right]$$$ (for steps, see inverse matrix calculator).

Answer

$$$P = \left[\begin{array}{ccc}2 & 4 & -2\\0 & -4 & 0\\1 & 1 & 1\end{array}\right]$$$A

$$$D = \left[\begin{array}{ccc}4 & 0 & 0\\0 & 2 & 0\\0 & 0 & 0\end{array}\right]$$$A

$$$P^{-1} = \left[\begin{array}{ccc}\frac{1}{4} & \frac{3}{8} & \frac{1}{2}\\0 & - \frac{1}{4} & 0\\- \frac{1}{4} & - \frac{1}{8} & \frac{1}{2}\end{array}\right] = \left[\begin{array}{ccc}0.25 & 0.375 & 0.5\\0 & -0.25 & 0\\-0.25 & -0.125 & 0.5\end{array}\right]$$$A


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