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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.

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Everything you have calculated is correct. A Jordan-Normal-Form for your matrix is $$\pmatrix{3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}$$

Another solution would be for example $$\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3}$$

The Jordan-Normal-Form do not have to be unique and depends on how you arange the vectors in your matrix $\mathcal{Z}$

But $$\pmatrix{3 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1}$$ cannot be a Jordan-Normal-Form of your matrix. For the eigenvalue $1$ the dimension of the corresponding eigenspace is $2$ so your matrix is even diagonalizable!

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