A Basis for a Jordan Normal Form

Question

In my assignment I have to find a Jordan normal form for this matrix:

Thank you for your help, and I'm sorry the question is pronunced with Latex.

Answer 1

The process is based on a clear understanding of generalized eigenvectors. Since you are only dealing with one Jordan block, not only the matrix $A-\lambda I$ is nilpotent, but also it is similar to $N$. In other words, the null space of $(A-\lambda I)^i$ increases by one dimension for each increase of the power, until it becomes the whole $\mathbb R^3$, i.e. $(A-\lambda I)^3=0$, the zero matrix. $$ 1=\dim\ker(A-\lambda I)\lneq \dim\ker(A-\lambda I)^2\lneq\dim\ker(A-\lambda I)^3=3 $$ To get the canonical form, start from the "most general" eigenvectors:

1. pick a vector $f_3$ such that $(A-\lambda I)^3f_3=0$ but $(A-\lambda I)^2f_3\neq 0$, so it is just any vector such that $(A-\lambda I)^2f_3\neq 0$.
2. Next, let $f_2=(A-\lambda I)f_3$, and $f_2\neq 0$ since $(A-\lambda I)^2f_3$ and hence $(A-\lambda I)^2f_3\neq 0$. This time you see that $$ Af_3=\lambda f_3+f_2 $$ and also that $(A-\lambda I)f_2\neq 0$ since $(A-\lambda I)f_2=(A-\lambda I)^2f_3$.
3. Finally, let $f_1=(A-\lambda I)f_2$, $f_1\neq 0$ from 2., and then: $$ Af_2=\lambda f_2+f_1 $$ Now $f_1$ is a non-general eigenvector of $A$ since $$(A-\lambda I)f_1=(A-\lambda I)^2f_2=(A-\lambda I)^3f_3=0$$ or $$ Af_1=\lambda f_1 $$

To sum it up: $$ A[f_1,f_2,f_3]=[\lambda f_1,\lambda f_2+f_1,\lambda f_3+f_2]=\begin{bmatrix}\lambda& 1& 0\\ 0 & \lambda& 1\\ 0 &0&\lambda\end{bmatrix}[f_1,f_2,f_3] $$

Answer 2

let $N = A - I = \pmatrix{0&-3&3\\-2&-7&13\\-1&-4&7}$ row reduce this matrix and find that it has rank 2 and null space is spanned by $u = \pmatrix{3\\-1\\0}.$ now row reduce $N^2$ and find that its rank is $1$ so the null space of $N^2$ has dimension $2.$ we can solve $Nv = u$ to find $v = \pmatrix{3\\-1\\0}.$ since $N^3$ is the zero matrix we can solve $Nw = v$ to find $w = \pmatrix{7\\0\\1.}$

now we have the jordan chain $B = \{u, v, w\}$ that spans $R^3$ and satisfies $$Au = u, Av = v + u, Aw = w + v$$ with respect to the basis $B, A$ has the representation $\pmatrix{1&1&0\\0&1&1\\0&0&1}.$