# Constructing a linear map given its minimal polynomial.

Let $$\mathbb{R^3}$$ be our vector space and over the field $$\mathbb{R}$$. Given that its minimal polynomial is $$x^3-x^2$$ construct a linear transformation $$T:\mathbb{R^3}\to\mathbb{R^3}$$.

My thoughts

Since $$m_T(x)=x^2(x-1)$$, we deduce that $$T$$ must be upper triangularisable and also $$\chi_T(x)=-m_T(x)$$. So I construct such $$T$$ based on the standard basis $$\{e_1,e_2,e_3\}$$ of $$\mathbb{R^3}$$ so that $$T(e_1)=0$$, $$T(e_2)=0$$ and $$T(e_3)=e_3.$$

My doubts

Is this a viable method to approach this problem? Is there a quicker way to approach this question?

Also what can we do if the minimal polynomial is less than the degree of our vector space? Is there an algorithm-ish that I can follow?

The $$T$$ you suggest is diagonal(isable) with respect to the standard basis, and its minimal polynomial has simple roots.
The easiest way to build a solution is to think in terms of Jordan forms. Here you should expect two blocks, $$\pmatrix{0&1\\ 0&0}$$ and $$\pmatrix{1}$$.
In general if the degree of the minimal polynomial is less than the dimension of the domain, it simply means that some eigenspaces are of dimension larger than $$1$$, and therefore that there is more than one Jordan block for the same eigenvalue.
The companion matrix of a polynomial $$p$$ has $$p$$ as both its minimal and its characteristic polynomial. This gives a systematic solution when the degree of $$p$$ equals the dimension of the space.