# Urgent help needed with linear algebra problem

• June 12th 2005, 10:28 PM
Mathman24
Urgent help needed with linear algebra problem
Hi

I have a linear algebra problem which I need urgent assistance with.

I'm present with something called a translation $(x,y) \rightarrow (x+3,y+5)$. I'm tasked with showing that this "translation" can't be done as a linear mapping using regular coordinants.

Any hints/idear on how I do that ????

Hope and pray to hear from You.

Sincerely and Best Regards,

Fred
• June 13th 2005, 07:39 AM
hpe
Suppose this were a linear mapping. Then if $(a_1,b_1) \to (c_1,d_1)$ as well as $(a_2,b_2) \to (c_2,d_2)$, it would follow that $(a_1+a_2,b_1+b_2) \to (c_1+c_2,d_1+d_2)$. (Check the definition of linearity in your notes.)

Now try this for your mapping:
$(1,1) \to (4,6)$ and $(2,2) \to (5,7)$, but $
(1+2,1+2) = (3,3) \to (6,8) \ne (4+5,6+7)$
.

Thus the mapping is not linear.
• June 13th 2005, 09:32 AM
Mathman24

How do I then prove that the mapping can be done using homogeneous coordinants ??

Sincerely

Fred

Quote:

Originally Posted by hpe
Suppose this were a linear mapping. Then if $(a_1,b_1) \to (c_1,d_1)$ as well as $(a_2,b_2) \to (c_2,d_2)$, it would follow that $(a_1+a_2,b_1+b_2) \to (c_1+c_2,d_1+d_2)$. (Check the definition of linearity in your notes.)

Now try this for your mapping:
$(1,1) \to (4,6)$ and $(2,2) \to (5,7)$, but $
(1+2,1+2) = (3,3) \to (6,8) \ne (4+5,6+7)$
.

Thus the mapping is not linear.