Jordan Normal Form of Matrix: 65 Characters

[Gregg]

$$\left(<br /> \begin{array}{ccc}<br /> 0 & 1 & 0 \\<br /> -1 & -1 & 1 \\<br /> -1 & 0 & 1<br /> \end{array}<br /> \right)$$

The answer is

$$\left\{\left(<br /> \begin{array}{ccc}<br /> 1 & -1 & 0 \\<br /> 0 & 1 & -1 \\<br /> 1 & 0 & 0<br /> \end{array}<br /> \right),\left(<br /> \begin{array}{ccc}<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 1 \\<br /> 0 & 0 & 0<br /> \end{array}<br /> \right)\right\}$$


The first one $$\left( \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 1<br /> \end{array} \right)$$ which is an eigenvector.

The eigenvalues are 0,0,0.

Looking at $$\text{Ker}(M-\lambda I)^2) =\text{Ker}(M^2)=\text{Ker}(\left(<br /> \begin{array}{ccc}<br /> -1 & -1 & 1 \\<br /> 0 & 0 & 0 \\<br /> -1 & -1 & 1<br /> \end{array}<br /> \right))$$

$$z=x+y$$

For the first eigenvector $$\left( \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 1<br /> \end{array} \right)$$ is in this generalised eigenspace $E_1 \subset E_2$ I thought about trying to find two vectors that produce the normal vector on that plane when crossed but that doesn't work and I'm not sure it's even the right thing to do. Can choose $$\left( \begin{array}{c}<br /> -1 \\<br /> 1 \\<br /> 0<br /> \end{array} \right)$$ for z=x+y. But $$\text{Ker}(M^3)=\text{Ker}(0)$$

I know there is the method $(M-\lambda I)\vec{v}_2=\lambda \vec{v_1}$ ? Do I have to use that? I'd prefer to be able to use kernels.

 

[Gregg]

I found by doing $$A \vec{v_i} = \vec{v_{i-1}}$$. But why didn't the nullspace approach work?

 

[HallsofIvy]

Yes, 0 is a triple eigenvalue. The characteristic equation is $x^3= 0$ and, since every matrix satisfies its own characteristic equation, $A^3v= 0$ for every vector. But, as you have found, only multiples of <1, 0, 1> are eigenvectors. That is, there is only a one dimensional subspace of eigevectors- that satisfy Av= 0. So we may have some vectors, u, such that Au is not 0 but $A^2u= 0$ which in turn means $A(Au)= 0$ which means Au must be a multiple of <1, 0, 1>.

We look for <x, y, z> such that
$$\begin{pmatrix}0 & 1 & 0 \\ -1 & -1 & 1\\ -1 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}y \\-x- y+ z\\ -x+ z\end{pmatrix}= \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}$$
which gives y= 1, -x- y+ z= 0, x+ z= 0. Since y= 1, the other two equations reduce to z- x= 1. If we take x= 0, we get z= 1 so <0, 1, 1> is a "generalized" eigenvalue.

But we must have some vectors, v, such that neither $Av$ nor $A^2v$ is 0 but $A^3v= 0$ which means $u= Av$ must be such that $A^2v= 0$ which is what we just found:
$$\begin{pmatrix}0 & 1 & 0 \\ -1 & -1 & 1\\ -1 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}y \\-x- y+ z\\ -x+ z\end{pmatrix}= \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$$
which gives the equations y= 0, -x- y+ z= -x+ 0+ z= 1, and -x+ z= 1. Those are really the same as we had before- taking a different vaue for x, say 1 instead of 0, z= 2. A second "generalized" eigenvalue is <1, 0, 2>.

Form matrix P having those three vectors as columns. Then $P^{-1}AP$ is in "Jordan Normal Form".