Dimension of the vector subspace $\text{Span} (\{ T_{m} | m \in \mathbb{Z} \}) $

Solution 1:

Let me denote $\text{End}_{\mathbb{R}}(V)$ as $L(V,V)$. for easier notation. ( It stands for $\mathbb{R}-$linear maps from $V$ to $V$).

We can identify each such map $T$ with a matrix and $L(V,V)\cong M_{(n+1)}(\mathbb{R})$.

( Where $M_{n}(\mathbb{R})$ denotes the space of Real Square matrices of size $n$).

Now $T_{m}(f(x))=f(x+m)$.

So if $f(x)=\sum_{i=0}^{n}a_{i}x^{i}$.

Then $$T_{m}(f(x))=\sum_{i=0}^{n}a_{i}(x+m)^{i}=\sum_{i=0}^{n}\sum_{k=0}^{i}a_{i}\binom{i}{k}x^{k}m^{i-k}$$

In terms of the standard basis for polynomials , the matrix for $n=2$(for polynomials of degree less than equal to $2$) is

$$[T_{m}]=\begin{bmatrix} 1&m&m^{2}\\ 0&1&2m\\ 0&0&1\\ \end{bmatrix}$$.

So the subspace is $\text{span}(S)$ where $S=\{[T_{m}]\in M_{(n+1)}(\mathbb{R}):m\in\mathbb{Z}\}$

If you now examine the space $\text{span}(S)$ you see that you need one variable for each degree of $m$ that appears in the matrix representation. ( This can be figured using the constraints the coefficients of a general $(n+1)\times (n+1)$ matrix has to satisfy in order to lie in the span)

For example in $n=2$(Polynomials of degree less than equal to $2$) you get the degrees of $m$ are $\{0,1,2\}$ . So in general, we require $n+1$ arbitrary scalars to completely represent the span. That is you need the matrices $\left\{\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix},\begin{bmatrix} 0&1&0\\0&0&2\\0&0&0\end{bmatrix},\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}\right\}$ to form a basis for $\text{span}(S)$.

Like this you can generalize to $(n+1)\times (n+1)$ matrices and claim that you need $n+1$ such basis matrices to span the subspace.

To Summarize: -Yes you need $n+1$ such basis matrices to span the set. Hence $\dim(\text{span}(S))=n+1$