can someone give an intuition for what Cyclical subspace means ?
also we saw the given a non zero vector x, a basis for Cyclical subspace is: $$W = span({x,T(x),T^2(x),...})$$ can you explain how this relates to Cyclical subspace and explain to why this a basis? Thanks you
Consider a finite-dimensional $k$-vector space $V$ with a linear operator $T : V \to V.$ Given any nonzero vector $v$ in $V,$ we define the $T$-cyclic subspace generated by $v$ to be the subspace of $V$ that consists of all vectors of the form $f(T)(v),$ where $f(x)$ is a polynomial in $k[x].$ One other way to see the $T$-cyclic subspace generated by $v$ is as the intersection of all $T$-invariant subspaces that contain $v.$ Particularly, if $W$ is such a subspace, then $W$ contains $v$ so that $W$ contains $T(v)$ so that $W$ contains $T(T(v)) = T^2(v),$ etc. so that $W$ contains all non-negative integer powers $T^i(v).$ Further, $W$ is a subspace, so it contains all linear combinations of these powers.
By hypothesis that $V$ is finite-dimensional (with $\dim V = n$), the vectors $v, T(v), T^2(v), \dots$ cannot be linearly independent. In fact, by the Cayley-Hamilton Theorem, the characteristic polynomial $\chi_T(x) = \det(xI - T)$ of $T$ satisfies $\chi_T(T) = 0.$ Observe that $\chi_T$ has degree $\dim V = n$ so that $0 = \chi_T(T) = T^n - a_{n - 1} T^{n - 1} + \cdots + a_1 T + a_0 I$ yields $T^n = -a_{n - 1} T^{n - 1} - \cdots - a_1 T - a_0 I,$ from which it follows that $T^n(v) = -a_{n - 1} T^{n - 1}(v) - \cdots - a_1 T(v) - a_0 v.$ We conclude therefore that the $T$-cyclic subspace generated by $v$ is spanned by $\{v, T(v), \dots, T^{n - 1}(v) \}.$
We can say more, as it turns out: the set of polynomials $\operatorname{ann} T(v) = \{g(x) \in k[x] \,|\, g(T)(v) = 0 \}$ in $k[x]$ is an ideal in $k[x].$ Considering that $k[x]$ is a principal ideal domain, $\operatorname{ann} T(v)$ is generated by some monic polynomial $p(x).$ Ultimately, the dimension of the $T$-cyclic subspace generated by $v$ is equal to the degree of $p(x).$ Consequently, if $\deg p = i,$ then a basis for the $T$-cyclic subspace generated by $v$ is $\{v, T(v), \dots, T^{i - 1}(v) \}.$ One can check that these are linearly independent (because $p(x)$ is the polynomial of minimal degree such that $p(T)(v) = 0$) and that they span (because $0 = p(T)(v) = T^i + a_{i - 1} T^{i - 1}(v) + \cdots + a_1 T(v) + a_0 v$).