The dimension of space of polynomials with matrix.

Question

Let $V:=M_{3\times 3}\ (\mathbb C)$, i.e., $V$ is a set of $3\times 3$ matrices of complex number.

Let $A=\begin{pmatrix}0&-2&0\\1&3&0\\0&0&2\end{pmatrix}$, $W:=\{p(A)\mid p(t)\in \mathbb C [t]\}$, where $\mathbb C[t]$ is the set of polynomials whose cooefficients are complex numbers.

Then, $W$ is a subspace of $V$.

Calculate $\dim W.$

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I think the characteristic polynomial of $A$ is necessary so I calculated it : $(x-2)^2(x-1)$.

And from Cayley-Hamilton, I get $(A-2I)^2(A-I)=O.$

I don't know what should I do next.

For this $A$,

・ $A$ is not a nilpotent matrix

・ $A$ doesn't seem to have periodicity. ($n\in \mathbb N$ s.t. $A^n=A$ doesn't seem to exist.)

So I'm having difficulty finding what $W$ is like.

Thanks for any help.

Answer 1

Compute the minimal polynomial for $A$.

It is easy to see that it will be of the form $(x-2)^{a}(x-1)$ , where $1\leq a\leq 2$.

So if we take $(x-2)(x-1)$ then we see that this annihilates $A$ and hence $m_{A}(x)=x^{2}-3x+2$

Now $\Bbb{C}[t]$ is an Euclidean Domain and hence any polynomial $p(t)\in\Bbb{C}[t]$ can be written as$p(t)=m_{A}(t)q(t)+r(t)$ , where $0\leq\deg(r(t))<2$ .

Thus you have for any polynomial $p(t)\in\Bbb{C}[t]$ you have $p(A)=r(A)=a_{0}I+a_{1}A$ for some $a_{0},a_{1}\in\Bbb{C}$

Thus $W=\{a_{0}I+a_{1}A:a_{0},a_{1}\in\Bbb{C}\}$ . Thus $W=\text{span}\{I,A\}$

Now it is easy enough to see that $\{I,A\}$ are two linearly independent elements of the vector space $V$. (As $A$ is not a scalar multiple of $I$).

Thus $\dim(W)= 2$

Answer 2

Hint

$$I= \{p(t) \in \mathbb C[t] \mid p(A) = 0\}$$ is an ideal of $\mathbb C[t]$. As $\mathbb C[t]$ is a principal ideal ring, it is generated by a polynomial $\mu$. Namely the minimal polyomial of $A$. Then the dimension of $W$ is equal to the degree of $\mu$. Do you see why?

So all you have to do is to find $\mu$, which divides the characteristic polynomial $\chi_A$ of $A$ and has the same irreducible factors. So $\mu$ can only be $\chi_A$ itself or $q(x) = (x-2)(x-1)$. And the answer to this last question is simple... just compute $(A-2I)(A-I)$ to see what it is.