Urgent help needed with linear algebra problem

**Mathman24** · June 12th 2005, 10:28 PM

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

**hpe** · June 13th 2005, 07:39 AM

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 $<br /> (1+2,1+2) = (3,3) \to (6,8) \ne (4+5,6+7)$ .

Thus the mapping is not linear.

**Mathman24** · June 13th 2005, 09:32 AM

Hello hpe and many thanks for Your answer,

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

Sincerely

Fred

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 $<br /> (1+2,1+2) = (3,3) \to (6,8) \ne (4+5,6+7)$ .

Thus the mapping is not linear.

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