Jordan's Canonical Form of a Matrix

Question

I'm a little tied up about Jordan's canonical form of a square matrix. How would Jordan's Canonical form of the following square matrix be obtained?

$$A = \begin{pmatrix} 2 & 1 & 1 \\ -2 & -1 & -2\\ 1 & 1 & -2 \end{pmatrix}$$

Answer 1

Compute the eigenvalues, which are the roots of the characteristic polynomial. You'll get two roots: $1$ (a simple root) and $-1$ (a double one). Use this to prove that the Jordan normal form is$$\begin{pmatrix}1&0&0\\0&-1&1\\0&0&-1\end{pmatrix}.$$

Answer 2

You have to find a Jordan basis for $A$. As $-1$ is a double eigenvalue, to determine whether $A$ is diagonalisable, we have toknow whether the eigenspace for this value has dimension $2$ or not, and in the latter case determine a basis for the generalised eigenspace.

Now $\ker(A+I)$ has dimension $1$ since $$A+I= \begin{pmatrix} 3 & 1 & 1 \\ -2 & 0 & -2\\ 1 & 1 & -1 \end{pmatrix}$$ has rank $2$. The last two rows are linearly independent, and an eigenvector for the eigenvalue $-1$ satisfy the equations $x+z=0$, $\;x+y-z=0$.

$\ker (A+I)^2$ has dimension $2$ since $$(A+I)^2= \begin{pmatrix} 8 & 4 & 0 \\ -8 & -4 & 0\\ 0 & 0 & 0 \end{pmatrix}$$has rank $1$, and is defined by the equation $2x+y=0$.

To have a Jordan basis , we begin with taking $v_3\in\ker (A+I)^2 \setminus\ker (A+I)$. The simplest is to choose $v_3=(0,0,1)$.

Next set $v_2=(A+I)v_3=(1,-2,-1)$. This is an eigenvector for the eigenvalue $-1$, and by construction we have: $$Av_3=-v_3+v_2.$$ Last step: find an eigenvector for the simple eigenvalue $1$, i.e. a vector in the kernel of $$ A-I= A = \begin{pmatrix} 1 & 1 & 1 \\ -2 & -2 & -2\\ 1 & 1 & -3 \end{pmatrix}.$$ It has to satisfy the independent linear equations: $$x+y+z=0,\enspace x+y_3z=0 \iff z=0,\enspace x+y =0, $$ so we may choose $v_1=(1,-1,0)$.

By construction, in the basis $(v_11,v_2,v_3)$ the matrix of the endomorphism associated to $A$ is $$J(A)= \begin{pmatrix}1&0&0\\0&-1&1\\0&0&-1\end{pmatrix}.$$