# Linear Algebra, Finding matrix for transformation

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 $$$$\mu(t) = \sum_{i=0}^{\deg \mu(t)} c_i t^i$$$$

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.

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 $$$$T^n(v) + \sum_{i=0}^{n-1}b_i \cdot T^i(v) = 0$$$$ Then, the matrix of $$T$$ relative to the given basis will be $$$$A= \begin{pmatrix} 0 & 0 & 0 & \dots & 0 & -b_0 \\ 1 & 0 & 0 & \dots & 0 & -b_1 \\ 0 & 1 & 0 & \dots & 0 & -b_2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 & -b_{n-1} \end{pmatrix}$$$$
A direct proof by induction and cofactor expansion will show you that the characteristic polynomial (and hence minimal polynomial) of $$A$$ is given by $$$$f(t) = t^n + \sum_{i=0}^{n-1}b_i t^i.$$$$ (assuming you defined characteristic polynomial as $$\det (tI-A)$$, otherwise there will be an overall sign of $$(-1)^n$$).