# Jordan Form of a 3x3 Matrix with an eigenvalue of multiplicity 3…

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!

• 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. – Dietrich Burde Jan 7 at 20:32

• 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.