Minimal Polynomial and Jordan Basis

Question

Claim: Assume $A:V\rightarrow V$ is an endomorphism with $\dim V=d$. The minimal polynomial of this linear transformation is $m(t)=(t-\lambda)^d$. Choose $v$ such that $(A-\lambda)^{d-1} v\neq 0$. Then, $V$ has basis $B=\{v, (A-\lambda)v, \cdots, (A-\lambda)^{d-1} v\}$ such that the representation of $A$ on this basis is in Jordan Forms, e.g.:

\begin{pmatrix} \lambda & 1 &0\\ 0 & \lambda &1\\ 0 & 0 & \lambda \end{pmatrix}

I tried to prove this statement by induction. When $d=1$, we get $v$ as the only element in basis $B$, and the corresponding Jordan Form will be a single block consisting of $\lambda$. However, I am not sure how to proceed later. Also, I do not know why the representation of $A$ on this basis is in the Jordan Form when $d\geq 2$.

Any help will be appreciated.

Answer 1

Let $v$ with $(A-\lambda )^{d-1}v\ne 0$. Denote $$ v_i := (A-\lambda)^i v, \quad i=0\dots d-1. $$ Then it holds for $i=0\dots d-2$ $$ A v_i = (A-\lambda) v_i + \lambda v_i = \lambda v_i + v_{i+1}. $$ Hence the $i$-th row of the matrix representation contains $\lambda$ in the $i$-th columns and $1$ in column $i+1$. Pretty much the matrix you asked about.

For $i=d-1$, we have $Av_{d-1} = (A-\lambda)v_{d-1} + \lambda v_{d-1}=\lambda v_{d-1}$, hence the $\lambda$ entry in the lower-right corner.

Answer 2

The only part of the statement that is not obvious from the definitions (notably the definition of what it means to have a matrix for an endomorphism expressed on a given basis) is that $B$ is a basis of$~V$ in the first place. Since it has the right number of elements to be a basis, it will suffice to show that the vectors of $B$ are independent. Suppose for a contradiction there is a non-trivial relation $0=\sum_{i=0}^{d-1}c_i(A-\lambda I)^iv$, and choose one that has as many initial coefficients$~c_i$ to be zero as possible. So let $k\geq0$ be the index of the first nonzero coefficient; since all vectors of $B$ are nonzero by construction there are at least two nonzero coefficients in the relation (so $k\leq d-2$). Now apply the endomorphism $A-\lambda I$ to the relation $0=\sum_{i=k}^{d-1}c_i(A-\lambda I)^iv$ to obtain $$ 0=\sum_{i=k}^{d-1}c_i(A-\lambda I)^{i+1}v = \sum_{i=k+1}^{d-1}c_{i-1}(A-\lambda I)^iv $$ (the final term of the first sum is$~0$ and was dropped). This is a non-trivial relation starting at index $k+1$, contradicting the maximal choice of $k$.