Linear transformation $T(X_1,X_2,...,X_n)=(0,X_1,X_2,...,X_{n-1}) $

Question

Let $T$ be a linear transformation from $F^n\to F^n$ defined by:
$$T(X_1,X_2,...,X_n)=(0,X_1,X_2,...,X_{n-1}) $$

I need to calculate $T^k$ for every $k\in N$ and to find the matrix represents $T$.

I have no idea how I should start to think about the solution.
Any help is welcome. thanks.

Answer 1

note that $$T^2(X_1,X_2,...,X_n)=T (0,X_1,X_2,...,X_{n-1}) = (0,0,X_1,X_2,...,X_{n-2})$$

so $$T^k(X_1,X_2,...,X_n)=0$$ whenever $k\ge n$

the matrix rep will have $T_{ij}=1$ when $j=i+1$ and $T_{ij}=0$ otherwise

Answer 2

Computing $T^2$: just apply $T$ twice. $$T^2(X_1,\ldots, X_n)=T(0,X_1,X_2,\ldots, X_{n-1}) = (0,0,X_1,X_2,\ldots, X_{n-2}).$$ Can you see the pattern for $T^k$? What happens when $k \ge n$?

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You want to find a matrix $A$ such that $$A \begin{bmatrix}X_1 \\ X_2 \\ X_3 \\ \vdots \\ X_n \end{bmatrix} =\begin{bmatrix}0 \\X_1 \\ X_2 \\ \vdots\\ X_{n-1} \end{bmatrix} $$

Just handle it one row of $A$ at a time. For example, the second row of $A$ satisfies $a_{21} X_1 + a_{22} X_2 + \cdots + a_{2n} X_n = X_1$, so the second row is $(1,0,0,\ldots,0)$.

A different way to think about this is as follows. If we apply $A$ to a standard unit vector $e_j$, we get the $j$th column of $A$. If we want $A e_j = T(e_j)$, this implies $T(e_j)$ is the $j$th column of $A$.

Answer 3

Basically this is shifting operator. Whenever you apply it to a set of n elements it shifts them by 1 position and make the first position as 0, now if you apply it again (i.e.you have applied $T^2$) you will get shifts them by 1 position and make the first position as $0$.

T 2 ( X 1 , X 2 , . . . , X n ) = T ( 0 , X 1 , X 2 , . . . , X n − 1 ) = ( 0 , 0 , X 1 , X 2 , . . . , X n − 2 )

Thus, proceeding in similar manner you will get $T^k$ =$0$ for all k$\ge n$

Answer 4

$T$ just shifts coordinates to the right. So $T^k(x_1,\ldots,x_n) = (0,\ldots,0,x_1,\ldots,x_{n-k})$ (you can check this by induction to be very formal).

On standard basis elements, we have $T(e_i) = e_{i+1}$ for $1 \leq i \leq n-1$ and $T(e_n) = 0$, so we find that the matrix representation is the identity matrix where each $1$ on the diagonal has been shifted downwards. In other words, it is zero everywhere except on the subdiagonal, where it is populated with $1$s.