# The dimension of space of polynomials with matrix.

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 $$\dim W.$$

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.

• What is your mathematical background? Have you seen abstract algebra? Jul 4 at 7:44

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 $$\dim(W)= 2$$

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.