Linear Transformations

linear transformation

A transformation $T:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is called a linear transformation if for all $x,y\in {\mathbb{R}}^n$ and $a\in \mathbb{R}$

$$\begin{array}{rl}{T}_1:& T(x+y)=T(x)+T(y)\\ & \\ T_2:& T(ax)=aT(x)\end{array}$$

Taking a=0 and a=01 in T_2 gives T(0) = 0 and T(-1x) = -T(x)

def

A vector y in ${\mathbb{R}}^n$ is called a linear combination of vectors $x_1,x_2,\dots ,x_k$ if y has the form [$y=a_1{x}_1+a_2{x}_2+\cdots +a_k{x}_k$]

Theorem

If $T:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is a linear transformation then $T(a_1{x}_1+\cdots +a_k{x}_k)$ = [ $a_{1}T(x_1)+\cdots +a_{k}T(x_k)$]

Proof

$$\begin{array}{rl}k=1,& T(a_1{x}_1)\\ & \\ =a_1& T(x_1)\\ & \\ k=2,& T(a_1{x}_1+a_2{x}_2)\\ & \\ =a_1& T(x_1)+a_{2}T(x_2)\\ & \end{array}$$

etc until you get to the theorem.

Example

If $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is a linear transformation, $T\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ -3\end{array}\right]$ and

$$\begin{gathered} t[1 \\ 2]=[5 \\ 1] \end{gathered}$$, find $$\begin{gathered} T[4 \\ 3] \end{gathered}$$.

Solution

Write $$\begin{gathered} z=[4 \\ 3] \end{gathered}$$, $$\begin{gathered} x=[1 \\ 1] \end{gathered}$$ and $$\begin{gathered} y=[1 \\ -2] \end{gathered}$$.

We don’t actually know that z = ax+by, but we’re assuming it for the purposes of this exercise. If it doesn’t work out, that means that z is not a linear combination of x and y. This would represent itself as an inconsistent matrix when we get to Reduced row-eschelon form.

Our goal is to write Z as z=ax+by so that T[4\\ 3] = T(z) = aT(x)+bT(y).

$$\begin{array}{r}z=ax+by\left[\begin{array}{c}4\\ 3\end{array}\right]=a\left[\begin{array}{c}1\\ 1\end{array}\right]+b\left[\begin{array}{c}1\\ -2\end{array}\right]a+b=4a-2b=3\end{array}$$ $$\left[\begin{array}{cccc}1& 1& |& 4\\ & & & \\ 1& -2& |& 3\end{array}\right]$$

r2 - r1

$$\left[\begin{array}{cccc}1& 1& |& 4\\ & & & \\ 0& -3& |& -1\end{array}\right]$$

r2/-3

$$\left[\begin{array}{cccc}1& 1& |& 4\\ & & & \\ 0& 1& |& \frac{1}{3}\end{array}\right]$$

r1 - r2

$$\left[\begin{array}{cccc}1& 0& |& \frac{11}{3}\\ & & & \\ 0& 1& |& \frac{1}{3}\end{array}\right]$$

a = 11/3 b = 1/3 z = 11/3[1\\ 1] + 1/3[1\\ -2]

b/c we’re looking for $$\begin{gathered} T[4 \\ 3]=T(z) \end{gathered}$$

$$\begin{array}{r}=11/3T(x)+1/3T(y)=11/3[2\\ -3]+1/3[5\\ 1]=1/3\cdot (11[2\\ -3]+[5\\ 1])=1/3\cdot [27\\ -32]\end{array}$$

Example

if A is m*n then the matrix transformation $T_A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is a linear transformation.

Sol.

We know $T_A(x)=Ax$ for $x\in {\mathbb{R}}^n$.

$$\begin{array}{rlr}{T}_A(x+y)& =A(x+y)& \\ & & \\ & =Ax+Ay& \\ & & \\ & =T_A(x)+T_A(y)& T_1\\ & & \end{array}$$ $$\begin{array}{rlr}{T}_A(ax)& =A(ax)& \\ & & \\ & =aA(x)& \\ & & \\ & =a{T}_A(X)& T_2\end{array}$$

Converse

The converse is also true: $T:{\mathbb{R}}^n\to {\mathbb{R}}^m$

Standard base of ${\mathbb{R}}^n$

[2\\ 5] = [2\\ 0] + [0\\ 5] = 2[1\\ 0] + 5[0\\ 1]

[-5\\ 18] = -5[1\\ 0] + 18[0\\ 1]

E notation

There’s some notation called e_1 which defines the 1st row of the n*1 matrix is a 1 [1\\ 0]. The rest are zeros. e_2 is [0\\ 1]. e_n would be [0\\ … \\ n].

This also ties to transformation space. So if e_1 is in R^n space,

$$\begin{array}{rl}T(x)& =T(x_1{e}_1+x_n{e}_n)\\ & \\ & =x_{1}T(e_1)+\cdots +x_{n}T(e_n)\\ & \\ & =[T(e_1),\dots ,T(e_n)]\cdot \left[\begin{array}{c}{x}_1\\ \dots \\ x_n\end{array}\right]\end{array}$$

e’s are n*1 I don’t quite understand why the last term above is m*n.

<2022-10-24 Mon>

Any linear transfromation $T:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is actually a matrix transformation. To see why, we define the standard basis of ${\mathbb{R}}^n$ to be the set of columns ${e}_1,e_2,\dots ,e_n$ of the identity matrix $I_n$.

Then, each $e_i\in {\mathbb{R}}^n$ and every vector x=[x_1\\ x_2\\ … \\ x_n] in ${\mathbb{R}}^n$ is a linear combination of the $e_i$.

In fact, x can be rewritten as $x_1{e}_1+x_2{e}_2+\cdots +x_n{e}_n$.

Now: $T(x)=T(x_1{e}_1+x_2{e}_2+\cdots +x_n{e}_n)$ which can be factored to be: $T(x)=x_{1}T(e_1)+x_{2}T(e_2)+\cdots +x_{n}T(e_n)$

Observe that each T(e_i) is a column in ${\mathbb{R}}^m$, so A = [T(e_1) T(e_2) … T(e_n)]

So we know that there are n entries in A (columns) and each of the columns have m entries, because the T statement at the top says we’re mapping fro n-space to m-space. So the shape is m*n space.

Hence, $T(x)=x_{1}T(e_1)+x_{2}T(e_2)+\cdots +x_{n}T(e_n)$

which is:

$$\begin{array}{r}=\left[\begin{array}{cccc}T(e_1)& T(e_2)& \dots & T(e_n)\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ \dots \\ x_n\end{array}\right]\end{array}$$

This shows that T is a the matrix transformation induced by A.

Theorem

Let $T:{\mathbb{R}}^n\to {\mathbb{R}}^m$ be a trnasformation.

1. T is linear iff it is a matrix transformation.
2. In this case & $T=T_A$ is the matrix trnasformation induced by a unique mxn matrix A given in terms of its columns by $A=\left[\begin{array}{ccccc}T(e_1)& amp;T(e_2)& amp;& hellip;& amp;T(e_n)\end{array}\right]wheree$\{e_1 & e_2 & \dots & e_n\}$isthestandardbasisof$\mathbb{R}^n$.

Proof

We only have to verify that the matrix A is unique. Suppose T is induced by another matrix B. Then,

T(x) = Bx for all x in R^n but T(x) = Ax for each x, so Bx=Ax for each x therefore A=B.

Ex

Define T: R^3 -> R^2 by T[x_1\\ x_2\\ x_3] = [x_1\\ x_2] for all [x_1\\ x_2\\ x_3] in R^3

Show that T is a linear transformation and use the previous theroem to find it’s matrix.

Sol:

let x = [x_1\\ x_2\\ x_3] and y = [y_1\\ y_2\\ y_3]

Then:

1. T(x+y) = T([x_1\\ x_2\\ x_3] + [y_1\\ y_2\\ y_3])

   = T([x_1 + y_1\\ x_2 + y_2\\ x_3+y_3]) = [x_1 + y_1\\ x_2 + y_2] ; only top 2 entries, by definition = [x_1\\ x_2] + [y_1\\ y_2] = T([x_1\\ x_2\\ x_3]) + T([y_1\\ y_2\\ y_3]) ; by definition = T(x) + T(y)

1. Let $a\in \mathbb{R}$ be a scalar. Then T(ax) = T([ax_1\\ ax_2\\ ax_3]) = [ax_1\\ ax_2] = a[x_1\\ x_2] = a T([x_1\\ x_2\\ x_3] = aT(x)

Do to prove that something is a linear transformation, we have do 1 & 2.

By theorem 2.6.2, The matrix of T is given by A=[T(e_1) T(e_2) T(e_3)]

$$\begin{array}{r}A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\end{array}$$

Since we go from R^3 -> R^2, the dimension of the transformation matrix is 2x3

Vector rotation

If we have a vector [1,0] (horizontal along the x-axis) and need to rotate it to [0,1] (vertical along y). This is counter-clockwise rotation.

Counterclockwise rotatin through $\pi /2$ about the origin is given by the matrix:

$$\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$$

In general, it should be:

$$\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ & \\ \sin \theta & \cos \theta \end{array}\right]$$

Where $\theta$ is the angle you’re putting in. I have no clue what this means.

If you’re rotating across a line, it’s [0,1\\ 1,0]

But that assumes that you’re rotating across y=x. If you’re flipping across a multiple of that, the formula is different:

y=mx

$$\begin{array}{r}\frac{1}{1+m^2}\left[\begin{array}{cc}1-m^2& 2m\\ 2m& m^2-1\end{array}\right]\end{array}$$

Q: I would have assumed that, instead of this forumla, we might have arrived at an answer by a combination of matrix multiplication. Is there a way to do that w/ matrix multiplication instead?

Yes, it’s in the book.

Q: If y=mx+b isn’t a linear transformation.. what word would you give for that?

We call those “non-linear transformations”.