# Linear maps

• Oct 6th 2010, 09:20 AM
SyNtHeSiS
Linear maps
Let S: $\displaystyle \mathbb{R}^2 \to \mathbb{R}^2$ be the function defined by S(x, y) = (x - y, y) for all (x,y) $\displaystyle \varepsilon \mathbb{R}^2$

Show that S is a linear map

Now I know that a linear map preserves the shape and takes lines to lines and points to points, but how would you show that S is a linear map using a formula such as
$\displaystyle S(\alpha \mathbf{x} + \beta \mathbf{y})$?
• Oct 7th 2010, 04:05 AM
HallsofIvy
Quote:

Originally Posted by SyNtHeSiS
Let S: $\displaystyle \mathbb{R}^2 \to \mathbb{R}^2$ be the function defined by S(x, y) = (x - y, y) for all (x,y) $\displaystyle \varepsilon \mathbb{R}^2$

Show that S is a linear map

Now I know that a linear map preserves the shape and takes lines to lines and points to points, but how would you show that S is a linear map using a formula such as
$\displaystyle S(\alpha \mathbf{x} + \beta \mathbf{y})$?

Here, your x is a pair, $\displaystyle (x_1, y_1)$ and your y is a pair, $\displaystyle (x_2, y_2)$. $\displaystyle \alpha\mathbf{x}+ \beta\mathbf{y}= \alpha(x_1, y_1)+ \beta(x_2, y_2)= $$\displaystyle (\alpha x_1+ \beta x_2, \alpha y_1+ \beta y_2). What to you get when you apply S to that pair? (It might be less confusing to do the sum and scalar multiplication separately. That is, find \displaystyle S(x_1+ x_2, y_1+ y_2) and \displaystyle S(\alpha x_1, \alpha x_2).) • Oct 7th 2010, 11:15 AM SyNtHeSiS You get \displaystyle S((\alpha x1 + \beta x2) - (\alpha y1 + \beta y2 ), \alpha y1 + \beta y2), but I dont know what to to do next. • Oct 8th 2010, 07:22 AM HallsofIvy Quote: Originally Posted by SyNtHeSiS You get \displaystyle S((\alpha x1 + \beta x2) - (\alpha y1 + \beta y2 ), \alpha y1 + \beta y2), but I dont know what to to do next. No, you get \displaystyle S(\alpha(x_1, y_1)+ \beta(x_2, y_2))= (\alpha x1 + \beta x2) - (\alpha y1 + \beta y2 ), \alpha y1 + \beta y2). What you have still has the "S" but S has already been performed. Now, separate those: \displaystyle (\alpha x1 + \beta x2) - (\alpha y1 + \beta y2 ), \alpha y1 + \beta y2)= (\alpha x1-\alpha y1+ \beta x2- \beta y2, \alpha y1+ \beta y2)$$\displaystyle = (\alpha (x1- y1)+ \beta(x2- y2), \alpha y1+ \beta y2)= (\alpha (x1- y1), \alpha y1)+ (\beta(x2- y2), \beta y2)= \alpha(x1- y1, y1)+ \beta(x2- y2, y2)$.
• Oct 8th 2010, 10:00 AM
SyNtHeSiS
Quote:

Originally Posted by HallsofIvy
= \alpha(x1- y1, y1)+ \beta(x2- y2, y2)[/tex].

How would you get from this step to $\displaystyle \alpha S(x1, y1) + \beta S(x2,y2)$? (I didnt understand this part while looking at the answer in my notes).
• Oct 9th 2010, 03:42 AM
HallsofIvy
Quote:

Originally Posted by SyNtHeSiS
How would you get from this step to $\displaystyle \alpha S(x1, y1) + \beta S(x2,y2)$? (I didnt understand this part while looking at the answer in my notes).

$\displaystyle \alpha(x1- y1, y1)+ \beta(x2-y2, y2)= \alpha S(x1, y1)+ \beta S(x2, y2)$

Well, you are told that S(x1, y1)= (x1- y1, y1) and S(x2, y2)= (x2- y2, y2), aren't you?