Let $S$ be the left shift operator over the space of infinite sequences of complex numbers. Describe the kernels of the operators $S-I$, $(S-I)^2$, $(S-I)^3$.
I started by doing the matrices of $S$ and $S-I$ and found that the following must apply: $(S-I)(a,b,c,d,.....)=(-a+b,-b+c,-c+d,...)$, and hence the kernel is a one dimensional sunspace of the form $(a,a,a,...)$. Is this correct, or is the space more than one dimensional?
For $(S-I)^2$, I squared the matrix $S-I$ found above and concluded that: $(S-I)^2(a,b,c,d,e,f,...)=(a-2b+c,b-2c+d,c-2d+e,...)$. And for the vector to be in the kernel each of the components must equal zero. Can somebody please help me solve this?
For $(S-I)^3$, I got $(S-I)^3(a,b,c,d,e,...)=(-a+3b-3c+d,-b+3c-3d+e,....)$. Can somebody please help me find the kernel of this?
If we restrict $S$ to each of the above kernels what will be the minimal polynomial? State the Jordan canonical form for each of these restrictions and give a Jordan basis.
Am I right in thinking that when restricted to the kernel the minimal polynomials will just be $S-I$, $(S-I)^2$ and $(S-I)^3$.
Since the operator is infinite dimensional I am confused as to how to construct a Jordan canonical form for the above as they are only finite dimensional?
To fix notation, let us denote the sequence space in question - the seqences over $\def\C{\mathbf C}\def\N{\mathbf N}\C$ - with $V := \C^{\N}$. We have the map $S \colon V \to V$, given by $$ Sx = (x_1,x_2, x_3, \ldots), \qquad x = (x_0, x_1, \ldots) \in V $$ As you state correctly, if $x \in \ker (S - \def\I{\mathrm{Id}}\I)$, we have for each $k \in \N$: $$ 0 = [(S-\I)x]_k = x_{k+1} - x_k \tag{$*$}$$ That is, $x$ has to be constant and $\ker (S - \I) = \C \cdot (1, 1, \ldots)$ is one-dimensional.
Now, let's have a look at $(S- \I)^2$. Note that $x \in \ker (S-\I)^2$ iff $(S-\I)x \in \ker (S - \I)$. As, we noted in $(*)$, $S - \I$ maps a sequence $x \in V$ to the sequence of differences $(x_{k+1} - x_k)_k$. Now $(S - \I)x$ belongs to $\ker (S - \I)$ iff it is constant, that is $x \in \ker (S -\I)^2$ iff it has constant differences, that is, we have $$ x_k = x_0 + k(x_1 - x_0), \qquad k \in \N $$ Hence, $\ker (S -\I)^2$ is two-dimensional $$ \ker (S - \I)^2 = \C \cdot (1, 1, \ldots) + \C \cdot (0, 1, 2, \ldots) $$
For $(S- \I)^3$ we argue as above, we have $x \in \ker (S - \I)^3$ iff $(S-\I) x \in \ker (S- \I)^2$, that is, for some $b, c \in \C$, we can write $$ x_{k+1} - x_k = b + kc, \quad k \in \N $$ To find $x$, let $x_0 =a \in \C$, than we find \begin{align*} x_1 - x_0 &= b &\iff x_1 &= a+b \\ x_2 - x_1 &= b+c &\iff x_2 &= a + 2b + c\\ x_3 - x_2 &= b+2c &\iff x_3 &= a + 3b + 3c \end{align*} So we guess $x_k = a + kb + \frac 12 k(k-1)c$, and check this by induction: \begin{align*} x_k + b + kc &= a + kb + \frac 12 k(k-1)c + b + kc\\ &= a + (k+1)b + \frac 12(k+1)kc \end{align*} That is $$ \ker (S -\I)^3 = \C \cdot (1, 1, \ldots) + \C \cdot (0,1,2,3, \ldots) + \C \cdot (0,0,1,3,6,\ldots) $$ Let's denote $v_1 := (1,1,1,\ldots)$, $v_2 := (0,1,2,3, \ldots)$, $v_3 := (0,0,1,3,6,\ldots)$. Than $\ker (S -\I) = \def\span{\mathop{\rm span}}\span \{v_1\}$, $\ker (S-\I)^2 = \span \{v_1, v_2\}$ and $\ker(S -\I)^3 = \span\{v_1, v_2, v_3\}$.
Now \begin{align*} S v_1 &= v_1\\ S v_2 &= (1,2,3\ldots) = v_1 + v_2\\ S v_3 &= (0,1,3,6,\ldots) = v_2 + v_3 \end{align*} That is, in the bases we gave above, we have that \begin{align*} S|_{\ker(S - \I)} &\text{ has the matrix } \begin{pmatrix} 1 \end{pmatrix}\\ S|_{\ker(S - \I)^2} &\text{ has the matrix } \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\\ S|_{\ker(S - \I)^3} &\text{ has the matrix } \begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix} \end{align*} Note that these matrices are already in Jordan form, I'm sure you can calculate the minimal polynomials from these.