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Let

$$A= \begin{bmatrix} 2&2&3\\ 1&3&3\\ -1&-2&-2 \end{bmatrix} . $$

Find the Jordan Form, $J$, of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$. I have already found the Jordan Form of this matrix, that is,

$$J = \begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}.$$

The part that I am confused about is finding the matrix $Q$. I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - \lambda I$. The characteristic polynomial of $A$ is

$$p_A(\lambda) = \lambda^3 - 3\lambda^2 + 3\lambda + 1 = (\lambda - 1)^3.$$

I have found the eigenvector associated with $\lambda = 1$ to be

$$v = (-5, 1, 1).$$

However, $(A - I)^2 = 0$, so I am confused on how to find the generalized eigenvectors. Thanks in advance!

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  • 2
    $\begingroup$ This has been asked many times at this site, for example here, or here, or here, etc. I think, you could follow the steps and explanations given at this site. $\endgroup$ – Dietrich Burde Jan 7 at 20:32
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Hint:

You should proceed backwards:

  • take any vector $u_3$ in $\ker(A-I)^2\smallsetminus\ker(A-I)$, i.e. any vector in $\mathbf R^3$ which does not satisfy the equation $\;x+2y+3z=0$, e.g. $u_3=(1,0,0)$.
  • set $u_2=(A-I)u_3$. This vector is an eigenvector.
  • complete $u_2$ with a linearly independent vector $u_1$, so as to obtain a basis of the eigenspace.
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