Jordan basis and the Jordan canonical form illustrated with an example

Consider the matrix $A_epsilon = begin{bmatrix} 0 & 1 \ epsilon & 0end{bmatrix}$ and observe its Jordan form as $epsilon$ approaches $0$. The minimal polynomial of $A_epsilon$ is $x^2 – epsilon x$, leading to eigenvalues of $0$ and $epsilon$. Thus, the Jordan canonical form is found to be: $$begin{bmatrix} epsilon & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{bmatrix} $$ In the case where $epsilon neq 0$, the Jordan form is a single block of size $3$, which can be represented by the matrix: $$begin{bmatrix}0 & 0 & 0 \ 1 & 0 & 0 \ 0 & epsilon & 0end{bmatrix}$$


Determine the operator’s canonical forms, namely
Jordan form
jordan basis
, associated with the matrix $A$ representing the function $f$.

I understand that this is a typical instance of linear algebra. However, I recently learned this concept and would appreciate it if someone could verify whether my solution is accurate.

The solution involves verifying that
characteristic polynomial
is equivalent to $-(lambda+1)^3$, thus demonstrating that the operator $(f+text{id})$ is nilpotent. The matrix representing this operator is denoted as $A+E$ and is given by $$A+E=begin{bmatrix}
-1 & 0 & -1 \
-1 & 0 & -1 \
1 & 0 & 1
end{bmatrix}$$ with a rank of one. Further verification shows that $(A+E)^2=0$ with a rank of zero. Therefore, the
Jordan Canonical form
of $f+text{id}$ is composed of two cells, namely $J_2(0)$ and $J_1(0)$, where $J_d(lambda)$ is a square matrix of size $dtimes d$ with 1’s on
, $lambda$’s on the diagonal, and zero on all other elements. Consequently, the
Jordan canonical form
of $f$ is given by $$J=begin{bmatrix}
-1 & 1 & 0 \
0 & -1 & 0 \
0 & 0 & -1

To construct the Jordan basis, we can utilize the fact that $(f+text{id})^2=0$. Hence, $mathbb{R^3}$ is the kernel of $(f+text{id})^2$, and it can be verified that $langle (0,1,0),(-1,0,1)rangle$ is the kernel of $(f+text{id})$. For instance, we can select $(0,0,-1)$ as a vector in the kernel of $(f+text{id})^2$ but not in the kernel of $(f+text{id})$. Applying $(f+text{id})$ to this vector yields $(1,1,-1)$.

To complete the process, it is necessary to select a vector from $ker(f+text{id})$ that is not dependent on $(1,1,-1)$ and $(-1,1,1)$. A suitable vector for this purpose is $(-1,1,1)$.

We can assign labels to the vectors as follows: $e_1=(1,1,-1)$, $e_2=(0,0,-1)$, and $e_3=(-1,1,1)$. These vectors form a Jordan basis denoted by $e_1,e_2,e_3$ and the matrix representation of the operator $f$ in this basis is $J$.

Could someone please verify if everything is okay here? I may have overlooked certain calculations and clarifications.


All is well, but when choosing the basis for the eigenvalue with multiplicity, it’s recommended (but not obligatory) to keep it orthogonal. Therefore, for $e_3$, I suggest using $e_3=(-1,2,1)$. It’s a matter of choice for this particular problem, but if further work with these vectors is required, orthogonality may be helpful in reducing work.

Your method of locating the cells of
jordan matrix
functions effectively in 3d, but may not be applicable in higher dimensions. Take, for instance, the polynomials $(lambda+1)^4=0$ and $(f+id)^2=0″, which have two possible solutions: $(J_2,J_2)$ and $(J_2,J_1,J_1)`. You are likely aware of this fact.

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