# Finding the minimal polynomial of a matrix in the real vector space of continuous real-valued functions.

I'm studying for a linear algebra final and I'm working through some old test problems:

In the real vector space of continuous real-valued functions defined on $\mathbb{R}$ consider the following functions $p_i, i =0,1,2$ and $\exp$ defined as follows: $p_i(x)=x^i$ and $exp(x)=e^x$ for all $x\in \mathbb{R}$. Set $V=span_\mathbb{R}\{p_0,p_1,p_2,\exp\}$ and consider the endomorphism $\sigma: V\to V$ defined as $$(\sigma f)(x):= f(x-1) \text{ for all } x\in \mathbb{R}$$ Determine the matrix representation, characteristic polynomial, eigenspaces, and the minimal polynomial of $\sigma$. Is $\sigma$ diagonalizable?

I think I've got everything but the minimal polynomial (please correct me if I'm wrong). I've determined that the matrix representation is $$A:=[\sigma]_\beta^\beta = \begin{pmatrix} 1 & -1 & 1 & 0\\ 0 & 1 & -2 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{-1} \end{pmatrix}$$ Also, the characteristic polynomial $\chi_A = (x-1)^3(x-e^{-1})$ and I found the eigenspaces to be $$\left\langle \begin{pmatrix} p_0\\ 0\\ 0\\ 0 \end{pmatrix} \right\rangle \text{ and } \left\langle \begin{pmatrix} 0\\ 0\\ 0\\ \exp \end{pmatrix} \right\rangle$$ Thus since the eigenvalue $1$ has algebraic multiplicity $3$ but geometric multiplicity $1$, $\sigma$ is not diagonalizable.

I know that the minimal polynomial $M_A$ divides $\chi_A$, so $M_A$ is either $(x-1)(x-e^{-1})$, $(x-1)^2(x-e^{-1})$ or $(x-1)^3(x-e^{-1})$. In theory, I know that if the polynomial with smallest degree with $A$ as a root is the minimal polynomial, but to check this seems computationally cumbersome for a paper and pencil exam. Is there some better way?

• Let $\tau(f) = \sigma(f) - f$. Find the least power of $\tau$ such that $\tau^k(p_i) = 0$ for $0 \leqslant i \leqslant 2$. – Daniel Fischer Dec 13 '16 at 21:32
• I think I know what you mean, but I get that $\tau(p_1)(x) = (\sigma(p_1)-p_1)(x) = x-1-x=-1$, so $\tau^2(p_1)(x) = \tau(\tau(p_1(x))) = \tau(-1) =-1$ and thus also $\tau^3(p_1)(x) =-1$. So there is no least power of $\tau$ such that $\tau^k(p_1)=0$. What am I missing? – M47145 Dec 13 '16 at 22:46
• $\tau(-1) = -\tau(p_0) = -(\sigma(p_0) - p_0) = -(p_0 - p_0) = 0$. $\tau$ maps constant polynomials to $0$, and decreases the degree of non-constant polynomials by one. – Daniel Fischer Dec 13 '16 at 22:50
• Oh yes, thanks for the help! – M47145 Dec 13 '16 at 22:51
• Writing column vectors with $p_0$ and $\exp$ in their coordinates is wrong. You either write functions, or you write their coordinates on the chosen basis (which coordinates are just constant real numbers), mixing the two representations makes no sense. When you write down the eigenspaces you mean $\langle p_0\rangle$ and $\langle\exp\rangle$ rather than the column vectors you wrote. – Marc van Leeuwen Dec 23 '16 at 11:39

## 1 Answer

To understand which power of $x - 1$ you'll see in the minimal polynomial, it is enough to understand the minimal polynomial of the matrix

$$B = \begin{pmatrix} 0 & -1 & 1 \\ 0 & 0 & -2 \\ 0 & 0 & 0 \end{pmatrix}$$.

If we denote by $m_B(X)$ the minimal polynomial of $B$, the minimal polynomial of $A$ will be

$$m_A(X) = m_B(X - 1)(X - e^{-1}).$$

Now, $B$ is a $3 \times 3$ nilpotent matrix and $B \neq 0$ so we only need to check whether $B^2 = 0$ or not to deduce the minimal polynomial. Since $B^2 e_3 = B(e_1 - 2e_2) = 2e_1$, we see that $B^2 \neq 0$ and so $m_B(X) = X^3$ and $m_A(X) = (X - 1)^3(X - e^{-1})$.