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Let $W = \{ p(B) : p \text{ is a polynomial with real coefficients}\}$, where $$B= \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{pmatrix}$$

The dimension $d$ of the vector space $W$ satisfies

  1. $4 ≤ d ≤ 6$
  2. $6 ≤ d ≤ 9$
  3. $3 ≤ d ≤ 8$
  4. $3 ≤ d ≤ 4$
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Could you phrase your post as a question, and also tell us what you have tried so far? –  Trevor Wilson Dec 19 '12 at 6:42
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1 Answer

Since the characteristic polynomial of $B$ is $x^3-1$, we have (also by direct computation) $B^3 = I$. Furthermore, a quick calculation shows that $I,B,B^2$ are linearly independent. Hence $W = \text{sp} \{I,B,B^2\}$, from which the answer follows.

The dimension of a subspace is the maximum number of linearly independent vectors (or on this case, matrices, which can also be viewed as vectors). It should be clear that $I,B,B^2,B^3,...$ form a basis for $W$ (any polynomial of $B$ can be expressed as a finite linear combination of these matrices). Since $B^3=I$, this means $B^3$ and any higher power can be expressed in terms of $I,B,B^2$. So to finish, we need to show that $I,B,B^2$ are linearly independent.

To elaborate what I mean by $I,B,B^2$ being linearly independent: Note that $B^2 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$. Now suppose $\alpha_0 I+\alpha_1 B + \alpha_2 B^2 = 0$. Since $B_{11} = (B^2)_{11} = 0$, it follows that $\alpha_0 = 0$. Since $I_{31} = (B^2)_{31} = 0$, it follows that $\alpha_1 = 0$, and since $B^2 \neq 0$, it follows that $\alpha_2 =0$. Hence they are linearly independent.

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I suspect the OP may not quite understand what it means that two matrices are linearly independent. –  Andres Caicedo Dec 19 '12 at 6:43
    
Then dimension of $W$ would be out of reach? (But, point taken, I will elaborate.) –  copper.hat Dec 19 '12 at 6:43
    
That's what I would imagine. –  Andres Caicedo Dec 19 '12 at 6:45
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