1. ## Linear transformation problem

I apologize in advance for my inability to type out matrices. Anyway the

The linear transformation is T [x] [z]
[y] = [x]
[z] [y]

and the question states find T^2013. I'm sorry but I have no idea how we are supposed to solve this, short of taking an eternity to apply it 2013 times. Is there some sort of pattern that will be apparent? I'm just not seeing it, Thank you very much

2. ## Re: Linear transformation problem

sorry that should be x y z in the first column equal to z x y in the other column

3. ## Re: Linear transformation problem

Originally Posted by grandmarquis84
sorry that should be x y z in the first column equal to z x y in the other column
To solve a problem like this, you try it and hope to find a pattern. Send a vector through the transformation once, twice, three times, and eventually, you should see a pattern.

To get it to display nicely, quote this response and you will see how I get this to display:

$T \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}z \\ x \\ y\end{bmatrix}$

4. ## Re: Linear transformation problem

Originally Posted by SlipEternal
To solve a problem like this, you try it and hope to find a pattern. Send a vector through the transformation once, twice, three times, and eventually, you should see a pattern.

To get it to display nicely, quote this response and you will see how I get this to display:

$T \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}z \\ x \\ y\end{bmatrix}$
Ah thank you for the help, I really appreciate it. I must admit I'm a little confused on how to send the vector through twice and three times. The examples we've done in class had only 2 rows and we multiplied the matrix by a constant a, and this one with 3 columns and 3 separate variables is confusing me

5. ## Re: Linear transformation problem

\begin{align*}T\begin{bmatrix}x \\ y \\ z\end{bmatrix} & = \begin{bmatrix}z \\ x \\ y\end{bmatrix} \\ T^2\begin{bmatrix}x \\ y \\ z\end{bmatrix} & = T\left(T\begin{bmatrix}x \\ y \\ z \end{bmatrix} \right) \\ & = T\begin{bmatrix}z \\ x \\ y\end{bmatrix} \\ & = \begin{bmatrix}y \\ z \\ x\end{bmatrix} \\ T^3\begin{bmatrix}x \\ y \\ z\end{bmatrix} & = T\left(T^2\begin{bmatrix}x \\ y \\ z\end{bmatrix}\right) \end{align*}

Think you can finish from here?

6. ## Re: Linear transformation problem

Originally Posted by SlipEternal
\begin{align*}T\begin{bmatrix}x \\ y \\ z\end{bmatrix} & = \begin{bmatrix}z \\ x \\ y\end{bmatrix} \\ T^2\begin{bmatrix}x \\ y \\ z\end{bmatrix} & = T\left(T\begin{bmatrix}x \\ y \\ z \end{bmatrix} \right) \\ & = T\begin{bmatrix}z \\ x \\ y\end{bmatrix} \\ & = \begin{bmatrix}y \\ z \\ x\end{bmatrix} \\ T^3\begin{bmatrix}x \\ y \\ z\end{bmatrix} & = T\left(T^2\begin{bmatrix}x \\ y \\ z\end{bmatrix}\right) \end{align*}

Think you can finish from here?
Ah thank you! Does this mean that it essentially alternates between the two columns? so the transformation $T^3 \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}z \\ x \\ y\end{bmatrix}$

7. ## Re: Linear transformation problem

Originally Posted by grandmarquis84
Ah thank you! Does this mean that it essentially alternates between the two columns? so the transformation $T^3 \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}z \\ x \\ y\end{bmatrix}$
No. The transformation cycles through the elements. The top element moves to the middle. The middle element moves to the bottom. And the bottom element moves to the top. You went the wrong direction.