Prove that $ g $ commutes with $ f $ if and only if $ g \in \text{span}\big(\text{Id} , f, \ldots, f^{n-1}\big) $

Question

Let $ E $ be a vector space of dimension $ n $, and let $ f \in \mathcal{L}(E)$ be a linear operator such that $ f^n = 0 $ and $f^{n-1} \neq 0 \ $.

From previous question i was able to prove that the family $ \big(x, f(x), \ldots, f^{n-1}(x)\big) $ is a basis of $E $.

Now here is where got stuck :

2. Let $ g \in \mathcal{L}(E) \ $. Prove that $ g $ commutes with $ f $ (i.e. $ f \circ g = g \circ f $) if and only if $ g \in \text{span}\big(\text{Id} , f, \ldots, f^{n-1}\big) $

Now the $"\Leftarrow"$ (if $ g $ in the span then $ g $ commutes with f) is easy but the other direction $"\Rightarrow"$ is what got me, I would appreciate if anyone can help me.

Answer 1

Hint: Show that the matrix representation of $f$ relative to the basis $(x, f(x), \dots,f^{n-1}(x))$ is given by $$ A = \begin{pmatrix} 0 & & & \cdots& 0 \\ 1 & 0 & & & \vdots\\ 0 & 1 & \ddots & & \\ \vdots & \ddots & \ddots & & \\ 0 & \cdots& 0 & 1 & 0 \end{pmatrix} $$ (using the basis in its reverse order also works and produces the transpose of this matrix and changes the rest of the approach correspondingly)

What does the matrix of $f^k$ (i.e. $A^k$) look like? Now, show that if $B$ is a matrix such that $AB = BA$, then $B$ must be of the form $$ B = \pmatrix{b_0 \\ b_1 & b_0\\ \vdots & \ddots & \ddots\\ b_n & \cdots & b_1 & b_0}, $$ where the unwritten entires are zero. That is, $B$ must be lower-triangular and Toeplitz. Once you know what the matrix of $A^k$ looks like, it should be easy to see that $B$ can be written as a linear combination of the powers of $A$, which is to say it can be written in the form $B = p(A)$ for some polynomial $p$, so that we correspondingly have $g = p(f)$.

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Here's a proof for the what the powers of $A$ look like

Claim: $$ [A^k]_{i,j} = \begin{cases} 1 & i-j = k\\0 & \text{otherwise} \end{cases} $$ where $[M]_{i,j}$ denotes the $i,j$ entry of $M$.

Proof: The base case ($k=0$) holds because $A^0 = I$. For the inductive step: we note that if $i,j$ are between $1$ and $n+1$ $$ [A^{k+1}]_{i,j} = [A A^{k}]_{i,j} = \sum_{p=1}^{n+1} A_{ip}[A^k]_{pj} $$ We note that $A_{ip}[A^k]_{pj}$ is only non-zero if $A_{ip} \neq 0$ and $[A^k]_{pj} \neq 0$. By our definition of $A$, $A_{ip}$ will only be non-zero if $p = i-1$. On the other hand: by our inductive hypothesis, $[A^k]_{pj}$ will only be non-zero if $p = j+k$. These can only be simultaneously true if $i-1 = j+k$, which is to say that $i-j = k+1$. Thus, we conclude that $[A^{k+1}]_{i,j} = 0$ whenever $i-j \neq k+1$.

Whenever $i-j = k+1$, we compute $$ [A^{k+1}]_{i,j} = \sum_{p=1}^{n+1} A_{ip}[A^k]_{pj} = A_{i,(i-1)}[A^k]_{(j+k),j} = 1 \cdot 1 = 1 $$ The conclusion follows.

Answer 2

Case $\impliedby$: Suppose $g \in \mathrm{span} (\mathrm{Id}, f, \dots, f^{n-1})$. That is $$g = a_0 \mathrm{Id} + a_i f + \cdots + a_{n-1} f^{n-1}$$ for constants $a_i$.

Then, for any $x \in E$, \begin{align*} (f \circ g)(x) &= f \left( a_0 \mathrm{Id}(x) + a_i f(x) + \cdots + a_{n-1} f^{n-1}(x) \right) \\ &= a_0 (f \circ \mathrm{Id})(x) + a_i (f \circ f)(x) + \cdots + a_{n-1} (f \circ f^{n-1})(x) \\ &= a_0 (\mathrm{Id} \circ f)(x) + a_i (f \circ f)(x) + \cdots + a_{n-1} (f^{n-1} \circ f)(x) \\ &= \left( (a_0 \mathrm{Id} + a_i f + \cdots + a_{n-1} f^{n-1}) \circ f \right)(x) \\ &= (g \circ f)(x) \text{.} \end{align*} Therefore, $f \circ g = g \circ f$.

(In short: turn $g$ into a linear combination of compositions of $f$, then use that a composition of $f$s can absorb a leading $f$ and emit a trailing $f$ (with the usual fact the identity commutes with everything).)

Case $\implies$: Suppose $g$ commutes with $f$ and let $e \in E$. We know $B = (e, f(e), \dots, f^{n-1}(e))$ is a basis of $E$, so it is sufficient to study $g$ on this basis. We find \begin{align*} g(e) &= \beta_0 \mathrm{Id}(e) + \beta_1 f(e) + \cdots + \beta_{n-1}f^{n-1}(e) \\ &\therefore \quad \left. g \right|_{\mathrm{span}\left( e \right)} \in \left. \mathrm{span} (\mathrm{Id}, f, \dots, f^{n-1}\;) \right|_{\mathrm{span}\left( e \right)} \text{,} \end{align*} for constants $\beta_i$. Then \begin{align*} (g \circ f)(e) &= (f \circ g)(e) \\ &= f(\beta_0 \mathrm{Id}(e) + \beta_1 f(e) + \cdots + \beta_{n-1}f^{n-1}(e)) \\ &= \beta_0 f(e) + \beta_1 f^2(e) + \cdots \beta_{n-1}f^{n-1}(e) + 0 \\ &\therefore \quad \left. g \right|_{\mathrm{span}\left( e, f(e) \right)} \in \left. \mathrm{span} (\mathrm{Id}, f, \dots, f^{n-1}\;) \right|_{\mathrm{span}\left( e \right)} \text{,} \\ (g \circ f^2)(e) &= (f^2 \circ g)(e) \\ &= \beta_0 f^2(e) + \beta_1 f^3(e) + \cdots + \beta_{n-1}f^{n-1}(e) + 0 + 0 \\ &\therefore \quad \left. g \right|_{\mathrm{span}\left( e, f(e), f^2(e) \right)} \in \left. \mathrm{span} (\mathrm{Id}, f, \dots, f^{n-1}\;) \right|_{\mathrm{span}\left( e \right)} \text{,} \\ &\vdots \\ (g \circ f^{n-1})(e) &= (f^{n-1} \circ g)(e) \\ &= \beta_0 f^{n-1}(e) + 0 + \cdots + 0 + 0 + 0 \\ &\therefore \quad \left. g \right|_{\mathrm{span}\left( e, f(e), f^2(e), \dots, f^{n-1}\;(e) \right)} \in \left. \mathrm{span} (\mathrm{Id}, f, \dots, f^{n-1}\;) \right|_{\mathrm{span}\left( e \right)} \text{.} \end{align*}

So $g$ restricted to a basis of $E$ is in $\mathrm{span} (\mathrm{Id}, f, \dots, f^{n-1}\;)$; therefore, $g \in \mathrm{span} (\mathrm{Id}, f, \dots, f^{n-1})$.