Find the eigenvalues, and if necessary the corresponding eigenvectors, of A and determine whether A is diagonalizable.
$A=\left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 \end{array}\right]$

Solution

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A=\left[ \begin{array}{llll}{0} & {0} & {0} & {0} \\ {1} & {0} & {1} & {0} \\ {0} & {1} & {0} & {1} \\ {1} & {1} & {1} & {1}\end{array}\right]

The eigenvalues will be solutions of characteristic equation. Since, we know the characteristic equation form is det $(A-\lambda I)=0$ thus,

det\left[ \begin{array}{llll}{-\lambda} & {0} & {0} & {0} \\ {1} & {-\lambda} & {1} & {0} \\ {0} & {1} & {-\lambda} & {1} \\ {1} & {1} & {1} & {1-\lambda}\end{array}\right]=0

-\lambda\left[ \begin{array}{lll}{-\lambda} & {1} & {0} \\ {1} & {-\lambda} & {1} \\ {1} & {1} & {1-\lambda} \end{array}\right]=0

\lambda^{4}-\lambda^{3}-2\lambda^{2}=0

\lambda=0,-1,2

Eigenvalues are $0,-1$ and $2$ with multiplicity $2$ , the eigenvector of corresponding eigenvalues for $\lambda =0$

\left[ \begin{array}{llll}{0} & {0} & {0} & {0} \\ {1} & {0} & {1} & {0} \\ {0} & {1} & {0} & {1} \\ {1} & {1} & {1} & {1}\end{array}\right]\left[ \begin{array}{rrrr}{x} \\ {y} \\{z}\\{u}\end{array}\right]=\left[ \begin{array}{rrrr}{0} \\ {0}\\{0}\\{0} \end{array}\right]

x=-z

y=-u

So, eigenvector will be

v_{1}=\left\{t\left[ \begin{array}{rrrr}{1} \\ {0}\\{-1} \\{0}\end{array}\right]\hskip 0.5em t\epsilon R \hskip 0.5em s\left[ \begin{array}{rrrr}{0} \\ {1}\\{0} \\{-1}\end{array}\right]\hskip 0.5em s\epsilon R \right\}

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