I am trying to write the matrix $A=\begin{pmatrix} 2 & 2 & -5 \\ 3 & 7 & -15 \\ 1 & 2 & -4 \end{pmatrix}$ in Jordan-Normal Form.
It's characteristic polynomial is $|A-xI|=x^3-5x^2+7x-3=(x-1)^2(x-3)=0$
So, it has eigenvalues 1 and 3. I calculated three eigenvectors $v_1=\left(\begin{smallmatrix} 1 \\ 3 \\ 1 \end{smallmatrix}\right), v_2=\left(\begin{smallmatrix}5\\0\\1\end{smallmatrix}\right), v_3=\left(\begin{smallmatrix}-2\\1\\0\end{smallmatrix}\right)$.
So $Z=\begin{pmatrix} 1&5&-2\\3&0&1\\1&1&0\end{pmatrix},\text{ }Z^{-1}=-\frac{1}{2}\begin{pmatrix}-1&-2&5\\1&2&-7\\3&4&-15\end{pmatrix}$
$Z^{-1}AZ=\begin{pmatrix}3&0&0\\0&1&0\\0&0&1\end{pmatrix}$, but shouldn't the answer be $Z^{-1}AZ=\begin{pmatrix}3&0&0\\0&1&1\\0&0&1\end{pmatrix}$?
Thanks.