Let $W = \{p(B) : p \text{ is a polynomial with real coefficients}\},$ where $$B = \begin{bmatrix}0&1&0\\ 0&0&1\\ 1&0&0 \end{bmatrix}.$$ Then the dimension $d$ of the vector space $W$ satisfies

1) $4 \le d \le 6$,

2) $6 \le d \le 9$,

3)$3 \le d \le 8$,

4)$3 \le d \le 4$.

Let $A$ be an $n \times n$ real matrix and consider its powers $A^0=I, A^1, A^2,\dots,A^m$

The minimal spanning set is the basis and hence we need to find the smallest degree polynomial satisfied by $A$. This is nothing but the minimal polynomial of $A$. Hence $$\text{dimension $d$ of W} = \text{degree of the minimal polynomial }m_A(x).$$

Caylay-Hamilton Theorem: Every $n \times n$ matrix satisfies its characteristic polynomial $p_A(x)$ which is a monic polynomial of degree $n$.

This shows that, from the above discussion, the matrices, $A^0, A^1, A^2, \dots, A^n$ are linearly dependent. Therefore $d \le n$. In our problem, the given matrix is of order $3$ and hence we have $$1 \le d \le 3.$$ Next we will prove that $d$ is actually equal to $3$. We need to show that it's minimal polynomial $m_A(x)$ also has degree $3$.

Let $p(x)$ be an annihilating polynomial of $A$. We have $p(x) = q(x) m_A(x) + r(x)$ with degree of $r(x)$ strictly less than the degree of $m_A(x)$ by the reminder theorem. This implies that $p(A) = q(A)m_A(A) + r(A)$ and we have $r(A) = 0$. Hence $r(x)$ is also an annihilating polynomial of $A$. This is possible only if $r(x)$ is the zero polynomial because the degree of $r(x)$ is strictly smaller then the degree of the minimal polynomial. Therefore $$p(x) = q(x) m_A(x).$$ In particular, degree of the minimal polynomial is less than or equal to $n$.

The characteristic polynomial of the given matrix $A$ is equal to $x^3 - 1$ which has three distinct roots. All these roots have to be roots of the minimal polynomial by the above result. So the degree of the minimal polynomial is equal to at least $3$ and hence by the previous paragraph, it should be equal to $3$. This shows that $$d = 3.$$

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1) $4 \le d \le 6$,

2) $6 \le d \le 9$,

3)$3 \le d \le 8$,

4)$3 \le d \le 4$.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me. Visit my blog every day and share the solutions with friends. Follow the blog by email. Thank you.****Solution**:**We will calculate the dimension $d$ explicitly.**We start with the following simple observation. Consider the polynomial $p(x) = x^3 + 3x^2 -2x +12$ then $p(A) = A^3 + 3A^2 - 2 A + 12 I$. In particular $p(A)$ is a linear combination of the matrices $A^0 = I, A^1, A^2, \dots$. So the set of all non-negative powers of $A$ spans $W$. We will find a basis inside this spanning set whose cardinality will be our required $d$.Let $A$ be an $n \times n$ real matrix and consider its powers $A^0=I, A^1, A^2,\dots,A^m$

**for some $m>0$**. These matrices are vectors in the vector space $M_n(\Bbb R)$. So it is possible to talk about the linear independence of these matrices. We say these matrices are linearly dependent if, as usual, there exist real scalars $\alpha_0,\alpha_1,\dots,\alpha_m$**not all zero**such that $\alpha_0 A^0 + \alpha_1 A^1 + \cdots + \alpha_m A^m = 0$. Equivalently, there exists a non-zero polynomial $p(x) = \alpha_0+\alpha_1 x+ \alpha_2 x^2 + \cdots + \alpha_m x^m$ with real coefficients**of degree atmost $m$**(some coefficient is non-zero and this coefficient need not be the leading coefficient) such that $p(A) = 0$. Such polynomials are known as the annihilating polynomials of $A$.**Hence, $A^0,A^1,A^2,\dots, A^m$ are linearly dependent if and only if there is a polynomial of degree almost $m$ satisfied by $A$.**The minimal spanning set is the basis and hence we need to find the smallest degree polynomial satisfied by $A$. This is nothing but the minimal polynomial of $A$. Hence $$\text{dimension $d$ of W} = \text{degree of the minimal polynomial }m_A(x).$$

Caylay-Hamilton Theorem: Every $n \times n$ matrix satisfies its characteristic polynomial $p_A(x)$ which is a monic polynomial of degree $n$.

This shows that, from the above discussion, the matrices, $A^0, A^1, A^2, \dots, A^n$ are linearly dependent. Therefore $d \le n$. In our problem, the given matrix is of order $3$ and hence we have $$1 \le d \le 3.$$ Next we will prove that $d$ is actually equal to $3$. We need to show that it's minimal polynomial $m_A(x)$ also has degree $3$.

**Result**: Every eigenvalue of $A$ is a root of the minimal polynomial $m_A(x)$.**Proof**: We will show that the minimal polynomial divides all the annihilating polynomials. In particular, it will divide the characteristic polynomial and hence the result follows.Let $p(x)$ be an annihilating polynomial of $A$. We have $p(x) = q(x) m_A(x) + r(x)$ with degree of $r(x)$ strictly less than the degree of $m_A(x)$ by the reminder theorem. This implies that $p(A) = q(A)m_A(A) + r(A)$ and we have $r(A) = 0$. Hence $r(x)$ is also an annihilating polynomial of $A$. This is possible only if $r(x)$ is the zero polynomial because the degree of $r(x)$ is strictly smaller then the degree of the minimal polynomial. Therefore $$p(x) = q(x) m_A(x).$$ In particular, degree of the minimal polynomial is less than or equal to $n$.

The characteristic polynomial of the given matrix $A$ is equal to $x^3 - 1$ which has three distinct roots. All these roots have to be roots of the minimal polynomial by the above result. So the degree of the minimal polynomial is equal to at least $3$ and hence by the previous paragraph, it should be equal to $3$. This shows that $$d = 3.$$

**only option (3) and (4) are correct.**Click

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