# 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}

Find a $3×3$ matrix A that has eigenvalues $\lambda =0$ , 4 ,-4 with corresponding eigenvectors
$\left[\begin{array}{c}0\\ 1\\ -1\end{array}\right],\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]$

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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
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,

$P=\left[\begin{array}{ccc}0& 1& 0\\ 1& -1& 1\\ -1& 1& 1\end{array}\right]$
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 ${P}^{-1}=\frac{adj\left(P\right)}{|P|}$
Now,
$adj\left(P\right)={\left[\begin{array}{ccc}+\left(-1×1-1×1\right)& -\left(1×1-1×\left(-1\right)\right)& +\left(1×1-\left(-1\right)×\left(-1\right)\right)\\ -\left(1×1-0×\left(-1\right)\right)& +\left(0×1-0×\left(-1\right)\right)& -\left(0×1-1×\left(-1\right)\right)\\ +\left(1×1-0×\left(-1\right)\right)& -\left(0×1-0×1\right)& +\left(0×\left(-1\right)-1×1\right)\end{array}\right]}^{T}$
$={\left[\begin{array}{ccc}-2& -2& 0\\ -1& 0& -1\\ 1& 0& -1\end{array}\right]}^{T}$
$=\left[\begin{array}{ccc}-2& -1& 1\\ -2& 0& 0\\ 0& -1& -1\end{array}\right]$
And
$|P|=|\begin{array}{ccc}0& 1& 0\\ 1& -1& 1\\ -1& 1& 1\end{array}|$
$=0|\begin{array}{cc}-1& 1\\ 1& 1\end{array}|-|\begin{array}{cc}1& 1\\ -1& 1\end{array}|+0|\begin{array}{cc}1& -1\\ -1& 1\end{array}|=-2$
Hence,

Jeffrey Jordon