Let $p_a=(x-2)^2(x-7)^4x$ be the characteristic polynomial of the matrix $A$ and $(x-2)^2(x-7)x$ the minimal polynomial. Determine the matrix $A$.
My work: I know the matrix has to be $7×7$ and in its diagonal it must have two $2$, four $7$ and one $0$, so:
\begin{bmatrix}{}
2& & & & & & \\
& 2& & & & &\\
& & 7 & & & &\\
& & & 7 & & &\\
& & & & 7& & \\
& & & & & 7 &\\
& & & & & & 0\\ \end{bmatrix}
I don't know how to follow, what information gives me the minimal polynomial?
Best Answer
The minimal polynomial in this case gives you the information about the relevant Jordan blocks. Since it has $(x-2)^2$ as a factor, you must have one $2 \times 2$ Jordan block associated to the eigenvalue $2$ (and not two $1 \times 1$ Jordan blocks). To see why, note that the minimal polynomial of
$$ \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} $$
is $(x - 2)$ while the minimal polynomial of
$$ \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} $$
is $(x - 2)^2$.
Similarly, since the minimal polynomial has $(x-7)$ as a factor, al the Jordan blocks associated to the eigenvalue $7$ must be $1 \times 1$. Hence, $A$ is similar to the matrix
$$ \begin{pmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 7 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 7 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 7 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}. $$