# Jordan form of a matrix confusion

I was just wondering how I would go about finding the Jordan form of a matrix. I have this matrix here...

$\left( \begin{array}{ccc} 1 & 0 & -4\\ 0 & 3 & 0 \\ -2 & 0 & -1 \\ \end{array} \right)$

I found the Eigen values already. They are $\lambda=-3,3,3$ and the corresponding eigenvectors, respectively, are $v_1=\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$, $v_2=\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$ and $v_3=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, for $\lambda=-3,3,3$.

How do I get the Jordan form from this though? I know it has something to do with dimension and rank but I am not sure. If someone could point me in the right direction, that would be awesome thanks!

• Your matrix is diagonalizable since the algebraic multiplicity of each and every eigenvalue equals its geometric multiplicity... Commented Feb 3, 2018 at 10:16
• These are not the eigenvalues and eigenvectors of the given matrix... Commented Feb 3, 2018 at 10:17
• Please, if you are ok, you can accept the answer and set it as solved. Thanks! cdn.sstatic.net/img/faq/faq-accept-answer.png
– user
Commented Feb 5, 2018 at 23:35

Form the invertible (why?) matrix

$$P:=\begin{pmatrix}1&\!-2&0\\ 0&0&1\\ 1&1&0\end{pmatrix}$$

and now just check that

$$P^{-1}AP=\begin{pmatrix}\!-3&0&0\\ 0&3&0\\ 0&0&3\end{pmatrix}$$

Note that we have one eigenvalue with algebraic multiplicity 1, two eigenvalues with algebraic multiplicity 2 and a complete set of eigenvectors (i.e. 3 linearly independent eigenvectors) thus the matrix is diagonalizable.

Indeed since

$$Av_1=-3v_1 \quad Av_2=3v_2 \quad Av_3=3v_3$$

thus if we consider the matrices

$$P [v_1 \quad v_2 \quad v_3] \quad \quad D=\begin{pmatrix}\!-3&0&0\\ 0&3&0\\ 0&0&3\end{pmatrix}$$

we have

$$AP=PD\implies P^{-1}AP=D$$

note that for the existence of $P^{-1}$ is crucial that the three eigenvectors are linearly indepedent.