So in my Linear Algebra course I am given the following eigenvalues with their eigenvectors and I have to find out if they belong to a matrix, and if yes, to find the matrix. So this is given:
$\lambda_1=-4$ and $\vec{v_1} =\begin{bmatrix} 0\\ 0 \\ -6\\5 \end{bmatrix} \;$ $\lambda_2=2$ and $\vec{v_2} =\begin{bmatrix} 0\\ 0 \\ 0\\1 \end{bmatrix} \;$ $\lambda_3=2$ and $\vec{v_3} =\begin{bmatrix} 2\\ 1 \\ 0\\0 \end{bmatrix} \;$ $\lambda_4=1$ and $\vec{v_4} =\begin{bmatrix} -1\\ -1 \\ 0\\0 \end{bmatrix} \;$ So I have to find out if they belong to a matrix, and if yes to find that matrix. I first write the Matrix $X$ with the column vectors: $\begin{bmatrix} 0 & 0 & 2 & -1\\ 0 & 0 & 1 & -1\\ -6& 0 & 0 & 0\\ 5 & 1 & 0& 0\\ \end{bmatrix}$
I find that the determinant of this matrix is 6, which means it is invertible, now I find $X^{-1}$ $\begin{bmatrix} 0 & 0 & -\frac{1}{6} & 0\\ 0 & 0 & \frac{5}{6} & 1\\ 1 & -1 & 0 & 0\\ 1 & -2 & 0 & 0\\ \end{bmatrix}$
Lasty I Construct the diagonal matrix with the eigenvalues :
D= $\begin{bmatrix} -4 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0& 0& 2 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$
Now to find $A$ I use this property $A=XDX^{-1}$ When I do this computation, I get the following matrix for A
A= $\begin{bmatrix} 3 & -2& 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & -4 & 0\\ 0 & 0 & -\frac{5}{3} & 2\\ \end{bmatrix}$
However, using this matrix, when I do $(A+4I)$ so using $\lambda_1=-4$ I find an eigenvector
$v_1= \begin{bmatrix} 0\\ 0\\ 3.6\\ 1 \end{bmatrix} $ which is not the same as $\begin{bmatrix} 0\\0\\ -6\\5 \end{bmatrix}$ from the original problem.
So now I am a bit lost, did I make a mistake somewhere, or does this mean that the second, third and fourth are eigenvectors and eigenvalues of matrix $A$ but the first one is not?