Finding two distinct vectors

**DrNerj** · September 23rd 2012, 09:05 AM

Hi everyone. I'm hoping someone can help me with this:

I need to find two distinct vectors, x and y, that each have a length of 2 and when each is multiplied by the 2X2 matrix A, they yield the same result modulo 26.

A=
(9 5)
(7 3)

Any help with this is much appreciated.

**johnsomeone** · September 23rd 2012, 09:23 AM

What kind of values are permitted for the coordinates of x and y? Real numbers? Rationals? Integers?
By "length 2", do you mean in the ordinary sense, or in the "modulo 26" sense?

**DrNerj** · September 23rd 2012, 09:25 AM

The values for x and y have to be integers modulo 26.
And in regards to length, I mean in the ordinary sense.

**johnsomeone** · September 23rd 2012, 05:37 PM

There are only four vectors in $\mathbb{R}^2$ that have integer coordinates and length 2:

$\left\{\begin{pmatrix} 2 \\0 \end{pmatrix}, \begin{pmatrix} -2 \\0 \end{pmatrix}, \begin{pmatrix} 0 \\2 \end{pmatrix}, \begin{pmatrix} 0 \\-2 \end{pmatrix} \right\} = \left\{ \pm \begin{pmatrix} 2 \\0 \end{pmatrix}, \pm \begin{pmatrix} 0 \\2 \end{pmatrix} \right\}$ .

Then matrix A operates on that set via:

$\begin{pmatrix} 9& 5 \\7& 3 \end{pmatrix} \left(\pm \begin{pmatrix} 2 \\0 \end{pmatrix} \right) = \pm \begin{pmatrix} 9& 5 \\7& 3 \end{pmatrix} \begin{pmatrix} 2 \\0 \end{pmatrix} = \pm \begin{pmatrix} 18 \\14 \end{pmatrix}$ , and

$\begin{pmatrix} 9& 5 \\7& 3 \end{pmatrix} \left(\pm \begin{pmatrix} 0 \\2 \end{pmatrix}\right) = \pm \begin{pmatrix} 9& 5 \\7& 3 \end{pmatrix} \begin{pmatrix} 0 \\2 \end{pmatrix} = \pm \begin{pmatrix} 10\\ 6 \end{pmatrix}$ .

So A applied to the set of vectors that have integer coordinates and length 2 produces the four vectors: $\left\{ \pm \begin{pmatrix} 18 \\14 \end{pmatrix}, \pm \begin{pmatrix} 10\\ 6 \end{pmatrix}\right\}$ .

When looked at mod 26, those four vectors are:

$\left\{\begin{pmatrix} 18 \\14 \end{pmatrix}, \begin{pmatrix} -18 \\-14 \end{pmatrix}, \begin{pmatrix} 10\\ 6 \end{pmatrix}, \begin{pmatrix} -10\\ -6 \end{pmatrix} \right\}$

$= \left\{\begin{pmatrix} 18 \\14 \end{pmatrix}, \begin{pmatrix} 8 \\12 \end{pmatrix}, \begin{pmatrix} 10\\ 6 \end{pmatrix}, \begin{pmatrix} 16\\ 20 \end{pmatrix}\right\}$ .

No two of those are equal. Thus there does not exist two distinct vectors, both having integer coordinates and length two, such that the result of applying that matrix to them is equal modulo 26.

**DrNerj** · September 23rd 2012, 08:45 PM

Thanks for your help!

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