# Linear transformations

1. Feb 3, 2005

### Felix83

Im a second semester engineering student and im a few weeks into a linear algebra class. I understand most of it, but my teacher has to work to speak english so she doesn't explain things very well. We just started linear transformations and a few things seem unclear to me.

Take a shear transformation for example. You multiply a vector by a matrix and end up with another vector - the 'image' of the original vector. There is an example in my book that considers the square formed by area under y=2 from x=0 to x=2. It considers every point as a vector, and then multiplies them by a matrix to get a transformation that looks like a parallelogram - the height of it is still two, but it is horizontally stretched. It shows calculations for a couple of the corner points. The first is the vector [0,2] ([x1,x2] - written vertical just hard to type that way :) ) multiplied by the 2x2 matrix ( [1,3] for the top row and [0,1] for the bottom ) to get [6,2] . Next the vector [2,2] is multiplied by the same matrix to get [8,2] .

What does the matrix mean - what do the coefficients actually represent? If you consider a shape such as a square and you want to transform it to a parallelogram, how do you know what coefficients to use in the matrix?

Last edited: Feb 4, 2005
2. Feb 4, 2005

### Galileo

Suppose you want a square whose sides have length a to be transformed into a parallelogram with height a, but (horizontal) length b, with the upper left vertex at (c,a). This uniquely defines the parallelogram.

So you want the vector [a,0] to be transformed into [b,0] (lenght b), and [0,a] must be transformed into [c,a] (still height a, and upperleft vertex (c,a)).

In general this can be done by forming a 2x2 matrix A and solving: $A{a \choose 0}={b \choose 0}$ and $A{0 \choose a}={c \choose a}$ for A.
This'll lead to 4 equations in 4 unknowns.

Since this example is alot easier than a general case, some insight might show you that since $A{a \choose 0}={b \choose 0}$ the first column must be $b/a \choose 0$.
Likewise, since $A{0 \choose a}={c \choose a}$, the second column must be $c/a \choose 1$. so that:

$$A=\left(\begin{array}{cc}b/a & c/a \\ 0& 1\end{array}\right)$$.

For a=2, b=2 and c=6 you get
$$A=\left(\begin{array}{cc}1 & 3 \\ 0 & 1\end{array} \right)$$.

which is the matrix in your example, giving the parallelogram with vertices (0,0),(2,0),(6,2) and (8,2).

Last edited: Feb 4, 2005
3. Feb 4, 2005

### Hurkyl

Staff Emeritus
Just like a 2-tuple is a representation of a vector with respect to a chosen basis, a 2x2 matrix is a representation of a linear transformation with respect to the chosen bases on the source and target spaces.

4. Feb 4, 2005

### mathwonk

the easiest case is to transform the square with sides given by the vectors (1,0) and (0,1). then to transform it into the parallelgram with sides given by vectors (a,b) and (c,d), just put (a,b) and (c,d) in as the columns of the matrix. try it.

5. Feb 4, 2005

### Felix83

ok thanks, so in a transformation, a matrix simply represents the numbers requred to transform a set of vectors to another set. this makes alot more sense now. i think the problem is the book doesnt emphasize enough when it changes what the matrix is defined as. it starts out as the coefficients of a set of linear equations, then a couple sections later it represents sets of vectors, and then with transformations it represents the set of numbers needed to transform a vector to another vector. books and teachers really need to point this out before all other things. if you were going through this and didnt know that this was changing, it would be very confusing and frustrating.

6. Feb 4, 2005

### mathwonk

suppose you have the equations ax+by = e, and cx+dy = f, and you want to solve them.

If you form the matrix of coefficients, you have first row (a b) and second row (c d).

Thus the first column is (a,c) and the second column is (b,d).

then this matrix can be viewed as a transformation that takes the vector (1,0) to the vector (a,c), and takes the vector (0,1) to the vector (b,d).

Then by asking for a solution to the equations above,

ax+by = e, and cx+dy = f,

you are asking whether any vector gets taken to the vector (e,f) by this matrix.

i.e. since (x,y) can be written as x(1,0) + y(0,1), and since (1,0) and (0,1) go to

(a,c) and (b,d), it follows that (x,y) goes to x(a,c) + y(b,d).

so you are asking whether it is possible to wqrite (e,f) in the form x(a,c) + y(b,d).

enough for now.