Difference between revisions of "HW3.C Jeff Kubascik ECE301Fall2008mboutin" - Rhea

Revision as of 21:11, 17 September 2008

Some defines:

$ \,m=\left[ \begin{array}{ccc} x & y & z \end{array} \right] \, $ is the message

$ \,A=\left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right] \, $ is the encryption matrix

$ \,e=\left[ \begin{array}{ccc} s & t & u \end{array} \right] \, $ is the encrypted message

How can Bob Decrypt the Message?

We have the equation

$ \,e=mA\, $

which is how the message is being encrypted. If we multiply both sides by the inverse of $ \,A\, $, we get

$ \,eA^{-1}=mAA^{-1}=mI=m\, $

Therefore, we can get the original message back if we multiply the encrypted message by $ \,A^{-1}\, $, given that the inverse of $ \,A\, $ exists.

Can Eve Decrypt the Message Without Finding the Inverse of A?

Yes, since $ \,e=mA\, $ is linear.

Proof of Linearity

Say we have two inputs $ \,m_1\, $ and $ \,m_2\, $ yielding outputs

$ \,e_1=m_1A\, $ and

$ \,e_2=m_2A\, $, respectively.

thus,

$ \,ae_1+be_2=am_1A+bm_2A\, $

Now, apply $ \,am_1+bm_2\, $ to the system

$ \,(am_1+bm_2)A=am_1A+bm_2A\, $

Since the two results are equal

$ \,am_1A+bm_2A=am_1A+bm_2A\, $

the system is linear.

What is the Decrypted Message?