How To Find Minimal Polynomial

Question

$$A=\left(\begin{array}{ccccc} 4 & 1 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 9 & 0 \\ 0 & 0 & 0 & 0 & 9 \end{array}\right)$$

I know that the characteristic polynomial is $(\lambda-4)^3(\lambda-9)^2$ I know that the minimal polynomial can be a least $(\lambda-4)(\lambda-9)$ and $(\lambda-4)^3(\lambda-9)^2$ at most.

The matrix is $\text{diagonal}(J_2(4),J_1(4),J_1(9),J_1(9))$

How can I continue?

Answer 1

We have

$(A-4I)e_1 = 0$

$(A-4I)e_2 = e_1 \ne 0 $ and $(A-4I)^2 e_2=0$.

$(A-4I)e_3 = 0$

$(A-9I)e_4 = 0$

$(A-9I)e_5 = 0$

Therefore, $(A-4I)(A-9I)\ne0$ but $(A-4I)^2(A-9I)=0$.

Thus, the minimal polynomial of $A$ is $(\lambda-4)^2(\lambda-9)$.

Answer 2

The given matrix is a block diagonal matrix with blocks $\begin{bmatrix} 4&1 \\ &4\end{bmatrix}$, a $2\times 2$ block, followed by the $1\times 1$ blocks, $[4]$, $[9]$, $[9]$. Multiplication and addition can be done "on the blocks". A polynomial annihilates (by the obvious action) $A$, iff it does it for/on each block.

So we are searching for the minimal (polynomial,) divisor of the characteristic polynomial $(X-4)^3(X-9)^2$ which kills all blocks. The first block is killed by $(X-4)^2$. The next ones respectively by $(X-4)$, $(X-9)$, $(X-9)$. The lowest common multiple of them is $$(X-4)^2(X-9)\ .$$

Answer 3

The minimal polynomial has a factor $(x - \lambda)^m$ if the largest block with the eigenvalue $\lambda$ in the Jordan form is of the size $m\times m$.

In your case, the size of the largest block with the eigenvalue $4$ is $2\times 2$, and the size of the largest block with the eigenvalue $9$ is $1\times 1$.

Hence the minimal polynomial is $$(x-4)^2(x-1)$$