# 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

## 2 Answers

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 $v_n$.

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@Dider: so, the basis I found is incorrect? – user6163 Apr 24 '11 at 11:32
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
@Arturo Thanks for your comment. Not sure I see what can go wrong starting from an eigenvector, even when the eigenspace is not 1-dimensional. (Unrelated: my given name has as many letters as yours.) – Did Apr 24 '11 at 12:30
@Didier: I'm really sorry about the name; I keep mistyping it, and have done it before (I touch type, and there are a number of names that I always seem to mistype; "Michael" always comes out "Michale"...). When there are blocks of different size, it tends to be easier to start at the "end", but you're right that you cannot run into blind alleys. – Arturo Magidin Apr 24 '11 at 12:39

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.

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