Linear Transformation and their Matrices

June 17, 2019

Projection is a linear transformation, for example. Its linear transformation $T$ is $\mathbb R^2\rightarrow \mathbb R^2$ . Linear transformation can happen without any coordinates or matrices.

Rules for linear transformation:

$T(v+w)=T(v)+T(w)$
$T(cv)=cT(v)$

It’s like what we can do with vectors. Shifting a vector $v$ by $v+v_0$ is not linear since $T(2v)=2v+v_0$ does not end up with $2T(v)=2v+2v_0$ . Taking a vector's norm is also not linear since $T(-2v)=2\Vert v\Vert$ .

In fact, all transformation produced by matrix are linear transformations.

$T(v)=Av\tag{1}$

This satisfies the rules above easily. Some easy matrices:

Flip a graph along x-axis: $\begin{bmatrix}1 & 0\\0&-1 \end{bmatrix}$
Rotate a graph counterclockwise 90 degrees: $\begin{bmatrix}0 &-1\\1&0 \end{bmatrix}$

How much do we need to know about a transformation to know $T(v)$ ? We only need to know the effect of $T(v_1), T(v_2),…,T(v_n)$ for any input basis $v_1,…,v_n$ . We do not need to know its effect to all vectors but only the basis.

Write a transformation into Matrix

In fact, all the things we need to write a transformation with matrix is

Basis for the input $v_1,…v_n$
Basis for the output $w_1,…w_m$

Then a matrix with size $m\times n$ , its numbers are, in fact $T(v_1)=a_{11}w_1+a_{21}w_2+…+a_{m1}w_m$ the a’s are the elements in the matrix (more details in next lectures). Picking the right basis is important. The linear combination of the basis gives the coordinates $(c_1,c_2,…,c_n)$ . But different bases give different coordinates.

Pick the good basis

Let’s take projection as an example. If we want a projection in $\mathbb R^2$ that project the vectors into the 45 degrees line (but they are both in $\mathbb R^2$ ), if we use a regular basis (for input and output basis) $\begin{bmatrix}1 & 0\\0&1 \end{bmatrix}$ , then the transformation matrix will be $\begin{bmatrix}1/2 & 1/2\\1/2&1/2 \end{bmatrix}$ . But if we just pick the (it’s like cheating but ok) 45 degree line and its perpendicular one as basis, then’ll be left with $\begin{bmatrix}1 & 0\\0&0 \end{bmatrix}$ because we’re like clearing the vector’s y coordinates.