Find a 3 times 3 matrix A that has eigenvalues lambda=0 , 4 ,-4 with corresponding eigenvectorsbegin{bmatrix}0 1-1 end{bmatrix},begin{bmatrix}1 -11 end{bmatrix},begin{bmatrix}0 11 end{bmatrix}

Step 1
We need to find a matrix A which has eigenvalues $\lambda =0,4,-4$ and eigenvectors as
$\left[\begin{array}{c}0\\ 1\\ -1\end{array}\right],\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]$
Now, a matrix is similar to a diagonal matrix D if there exist matrices P and $P^{-1}suchthatA=PD{P}^{-1}$
Here, diagonal matrix D consists of eigenvalues as the diagonal entries.
And the matrix P is the matrix that has eigenvectors as its columns.
Hence,
$D=\left[\begin{array}{ccc}0& 0& 0\\ 0& 4& 0\\ 0& 0& -4\end{array}\right]\text{ and }$
$P=\left[\begin{array}{ccc}0& 1& 0\\ 1& -1& 1\\ -1& 1& 1\end{array}\right]$
Step 2
Hence, the matrix A is
$A=PD{P}^{-1}$
$=\left[\begin{array}{ccc}0& 1& 0\\ 1& -1& 1\\ -1& 1& 1\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 0& 4& 0\\ 0& 0& -4\end{array}\right]{\left[\begin{array}{ccc}0& 1& 0\\ 1& -1& 1\\ -1& 1& 1\end{array}\right]}^{-1}$
Before finding A we need to find $P^{-1}={\left[\begin{array}{ccc}0& 1& 0\\ 1& -1& 1\\ -1& 1& 1\end{array}\right]}^{-1}$
Now, inverse of a matrix P is $P^{-1}=\frac{adj(P)}{|P|}$
Now,
$adj(P)={\left[\begin{array}{ccc}+(-1\times 1-1\times 1)& -(1\times 1-1\times (-1))& +(1\times 1-(-1)\times (-1))\\ -(1\times 1-0\times (-1))& +(0\times 1-0\times (-1))& -(0\times 1-1\times (-1))\\ +(1\times 1-0\times (-1))& -(0\times 1-0\times 1)& +(0\times (-1)-1\times 1)\end{array}\right]}^T$
$={\left[\begin{array}{ccc}-2& -2& 0\\ -1& 0& -1\\ 1& 0& -1\end{array}\right]}^T$
$=\left[\begin{array}{ccc}-2& -1& 1\\ -2& 0& 0\\ 0& -1& -1\end{array}\right]$
And
$|P|=\left|\begin{array}{ccc}0& 1& 0\\ 1& -1& 1\\ -1& 1& 1\end{array}\right|$
$=0\left|\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right|-\left|\begin{array}{cc}1& 1\\ -1& 1\end{array}\right|+0\left|\begin{array}{cc}1& -1\\ -1& 1\end{array}\right|=-2$
Hence,

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