4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ What is your mathematical background? Have you seen abstract algebra? $\endgroup$
    – sadman-ncc
    Jul 4 at 7:44

2 Answers 2

2
$\begingroup$

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 $\dim(W)= 2$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.