I am revising for a Linear Algebra exam and am quite stuck on this question-any help, guidance or tips appreciated!

So I have a mapping $T:V\rightarrow V$ where $V$ is a finite dimensional vector space. I am a told that there is some vector $v\in V$ such that the set of vectors $v, Tv, T^2v,...T^{n-1}v$ forms a basis for V. Then, I am asked to express $T^nv$ as a linear combination of these vectors. And from here to write down a matrix of T.

I thought that $T^nv$ could only be expressed arbitrarily in terms of the basis vectors, I couldn't know any specifics, so $T^nv=a_1v+...+a_nT^{n-1}v$.

I then get my matrix for T with respect to this basis as $$ \begin{matrix} 0 & 0 & 0 & ...&0&a_1 \\ 1 & 0 & 0 &...&0&a_2 \\ 0 & 1 & 0 &...&0&a_3 \\ ...&....&...&...&...&...\\ 0&0&0&...&1&a_n \end{matrix} $$

Then, from here, they want me to show that the minimal polynomial and the characteristic polynomial are equal. This is where I think I get stuck, unless I've missed something earlier, as I end up with a complicated expansion for the characteristic polynomial.

I think I want to show that the matrix has $n=dimV$ eigenvalues, as then, as the minimal polynomial has a root at every eigenvalue, they must be equal. I don't know if I want to get it into triangular form to do this, and if so how.


Proof minimal polynomial = characteristic polynomial

Let \begin{equation} \mu(t) = \sum_{i=0}^{\deg \mu(t)} c_i t^i \end{equation}

be the minimal polynomial, and let $f(t)$ be the characteristic polynomial of $A$. We know that $\mu(t)$ divides $f(t)$; so $\deg \mu(t) \leq \deg f(t)$. Once we show that the degrees are the same, the fact that both polynomials are monic forces them to be equal. So, to prove this, note that \begin{align} 0 &= \mu(T)(v) \\ &= \sum_{i=0}^{\deg \mu(t)}c_i T^i(v) \end{align}

Notice that if $\deg \mu(t) < n = \deg f(t)$, this would contradict the linear independence of the basis $\{ v, T(v), \dots, T^{n-1}(v) \}$. Hence, the degrees are the same, thus completing the proof.

Additional Info:

I'll modify your notation slightly. We know there exist constants $b_0, \dots, b_{n-1} \in F$ (the field we're working over) such that \begin{equation} T^n(v) + \sum_{i=0}^{n-1}b_i \cdot T^i(v) = 0 \end{equation} Then, the matrix of $T$ relative to the given basis will be \begin{equation} A= \begin{pmatrix} 0 & 0 & 0 & \dots & 0 & -b_0 \\ 1 & 0 & 0 & \dots & 0 & -b_1 \\ 0 & 1 & 0 & \dots & 0 & -b_2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 & -b_{n-1} \end{pmatrix} \end{equation}

A direct proof by induction and cofactor expansion will show you that the characteristic polynomial (and hence minimal polynomial) of $A$ is given by \begin{equation} f(t) = t^n + \sum_{i=0}^{n-1}b_i t^i. \end{equation} (assuming you defined characteristic polynomial as $\det (tI-A)$, otherwise there will be an overall sign of $(-1)^n$).

In the spirit of this question, you should read up about companion matrices of polynomials


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