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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?

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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.

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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.

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