# Jordan form and base for a $n \times n$ Matrix

Given this matrix in size $n \times n$

$\begin{pmatrix} 1 &1 &1&\cdots & 1 & 1 & \\ 0 &1 & & & \\ . & &.& & & \\ . & & &.& & \\ . & & & &. & \\ 0& & & & &1 \end{pmatrix}$

EDIT: the matrix has all 1's above the diagonal.

I need to find Jordan form and base for this matrix.

I'd love your help understanding if I'm on the right track.

the characteristic polynomial is $f_{A}=(x-1)^{n}$ and the minimal one is $M_{A}=(x-1)$ (EDIT: it's clearly not right). So I can know that the dimension of the biggest Jordan block is 1, thus there are n Jordan blocks of size $1 \times 1$, the $\ker$ of eigenvalue 1 is all the dimension, So I simply assuemd that Jordan base is: $\left(\begin{smallmatrix} 1\\ 0\\ .\\ .\\ 0\\ \end{smallmatrix}\right)$, $\left(\begin{smallmatrix} 0\\ 1\\ .\\ .\\ 0\\ \end{smallmatrix}\right),\dots, \left(\begin{smallmatrix} 0\\ 0\\ .\\ .\\ 1\\ \end{smallmatrix}\right)$ .(If it's correct I will be happy to hear an explanation for that).

EDIT: For some stupid reason I first call this matrix the "Identity", Thanks for the comment below I changed it.

Trying to correct myself: So now I realize that I should sleep more and even much more significant fact that this matrix is nilpotent of rank n, so the minimal polynomial is $(x-1)^{n}$, and so finding the base will be a little more complicated. actually I'll need to find all the $\ker$s of $(x-1)^{i}$ for i=1,...n.

And After a while I think that Jordan base Will be $\left(\begin{smallmatrix} 1\\ 0\\ .\\ .\\ 0\\ \end{smallmatrix}\right)$, $(A-I)$ $\left(\begin{smallmatrix} 0\\ 1\\ .\\ .\\ 0\\ \end{smallmatrix}\right),\dots,(A-I)^{n-1}$ $\bigl(\begin{smallmatrix} 0\\ 0\\ .\\ .\\ 1\\ \end{smallmatrix}\bigr)$

Thank you, Have a great week.

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1. This is not the identity matrix. 2. The minimal polynomial is not $x-1$. It would have been that if the matrix had been the identity matrix, but it isn't. 3. it's not clear what the "..." represent - does this matrix have all 1's above the diagonal? Or just in the first row and the rest is zeros? –  Alon Amit Apr 24 '11 at 7:37
@Alon :I edited it. Thank you. –  user6163 Apr 24 '11 at 7:44
What is "Nylfotntiat"? The closest I can come to a guess is "nilpotent". –  Raeder Apr 24 '11 at 9:41
@Raeder: You right, sorry. :-) –  user6163 Apr 24 '11 at 9:52

As already mentioned, the minimal polynomial of $A$ is $(x-1)^n$. Hence the Jordan form of $A$ has $n$ ones on the main diagonal and $n-1$ ones on the first diagonal above the main diagonal and zeroes everywhere else. Now one looks for a basis $(v_1,\ldots,v_n)$ such that $Av_1=v_1$ and $Av_k=v_k+v_{k-1}$ for $k\ge2$. Hint: finding $v_1$ should be easy (no choice here), then one could try to find a suitable $v_2$, then a suitable $v_3$, then a suitable $v_4$, and so on until $v_n$.
I understand that you propose as linear basis the family $W=(w_k)_{1\le k\le n}$ with $w_k=(A-I)^{k-1}e_k$, where $(e_k)_{1\le k\le n}$ is the canonical basis. Then $w_1=e_1$ but already $w_2=w_1$ hence $W$ cannot be a basis. Let us keep your $w_1=e_1$, what $w_2$ could solve $Aw_2=w_2+w_1$? After that, what $w_3$ could solve $Aw_3=w_3+w_2$? –  Did Apr 24 '11 at 12:00
@Dider: It's possibly better to start by finding an element that is in the nullspace of $(A-I)^n$ but not in the nullspace of $(A-I)^{n-1}$ to get the cycle started "at the end". In this case it doesn't matter, because the eigenspace is 1-dimensional, but for other situations when the eigenspace is not 1-dimensional, you don't want to "pick the wrong $\mathbf{v}_1$". –  Arturo Magidin Apr 24 '11 at 12:03
You are right about $f_A(x)$, but clearly $m_A(x)=f_A(x)=(x-1)^n$ (check on lower order matrices). Hence there would be one eigenvector associated to $1$, and $n-1$ generalized eigenvectors.