# Abstract Linear Transformation Question

Let $T: \mathbb{R}^n\rightarrow \mathbb{R}^n$ a linear transformation defined by:

$\forall \begin{bmatrix}x_1\\x_2\\ \vdots \\ x_n \end{bmatrix} \in \mathbb{R}^n \ T \begin{bmatrix}x_1\\x_2\\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix}0\\x_1\\ \vdots \\ x_{n-1} \end{bmatrix}$

1. Calculate $T^k$ for every $0 \leq k < n$.
2. Prove $T^n=0$. (That is $T^n(\vec v)=\vec 0$ for every $\vec v \in \mathbb{R}^n$.
3. Prove $T^k=0$ for every $n \leq k$.
• Is that $x_n-1$ or $x_{n-1}$? – Casteels Dec 17 '14 at 15:28
• Do you know what $T^k$ means? Isn't it clear what happens each time you apply $T$? – rschwieb Dec 17 '14 at 15:36
• $x_{n-1}$ , sorry. – G-kan Dec 17 '14 at 15:44

Can you describe in words what $T$ does to the components of a vector? If you can do this, then you can answer the first two questions easily.

If you know that statement 2 holds, then for statement 3, if $k \ge n$, then $T^k=T^{k-n} T^n = T^{k-n} 0 = 0.$

A linear transform is completely determined by what it does on a basis, hence, let's see what your transform does over the standard basis:

$$E = \{(1, 0, 0, ..., 0, 0), (0, 1, 0, ..., 0, 0), ..., (0, 0, 0, ..., 0, 1)\}$$

Where we call each of these ones $e_i$ ($e_i$ is a vector with a $1$ in the i-th position). Notice that the matrix associated to $T$ over $E$ is of the form

$$[T]_{EE} = \begin{pmatrix} (Te_1)_E | (Te_2)_E | ... | (Te_n)_E \end{pmatrix}$$

However, notice that $Te_i = e_{i+1}$ except for the $i = n$ case, in which $Te_n = 0$ (that is, it moves the one down one position). Hence

$$[T]_{EE} = \begin{pmatrix} 0 & 0 & 0 & ... & 0 & 0 \\ 1 & 0 & 0 & ... & 0 & 0 \\ 0 & 1 & 0 & ... & 0 & 0\\ 0 & 0 & 1 & ... & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \dots & 1 & 0 \\ \end{pmatrix}$$

Now, try doing some examples when $n = 2$ and $n = 3$, taking exponents of the matrix, that should give you some feel for what's happening.

You have a typo: it should be $x_{n-1}$ and not $x_n-1$.

Assuming $x_{n-1}$ is correct, let's denote by $p_i(x)$ the $i$-th component of a vector $x$, so $$x=\begin{bmatrix} p_1(x)\\ p_2(x)\\ \vdots\\ p_n(x) \end{bmatrix}$$ Let's also, for convenience, set $p_i(x)=0$ if $i\le0$. Thus your $T$ can be described by $$p_i(T(x))=p_{i-1}(x)\qquad(i=1,2,\dots,n)$$ Let's prove by induction that $$p_i(T^k(x))=p_{i-k}(x)\qquad(i=1,2,\dots,n)$$ The base step ($k=0$) is obvious, because $T^0$ is the identity map. Suppose the result holds for $k$; then $$p_i(T^{k+1}(x))=p_i(T(T^{k}(x)))=p_{i-1}(T^k(x))\overset*=p_{i-1-k}(x) =p_{i-(k+1)}(x)$$ for $i=1,2,\dots,n$. The use of the induction hypothesis is marked by $\overset*=$. Thus the thesis is proved.

If $k\ge n$, then $p_{i-k}(x)=0$ for all $x$, because $i-k\le0$ when $1\le i\le n$. Therefore $T^k=0$ for $k\ge n$.