linear algebra - Determining a matrix given the characteristic and minimal polynomial

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 $7x7$ 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?

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

$$\begin{pmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 7 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 7 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 7 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}.$$