kevinlui's site

October 11

Announcements

• Section 2.3, 3.1 due next Thursday

• Write down name if you did worksheet 2

• Midterm next week

• Worksheet 3 will be posted tonight, it'll have some practice exam problems

3.1 Linear transformation

Theorem: Let $S=\{a_1,\ldots,a_n\}$ with $a_i\in \mathbb{R}^n$, $A=[a_i]$, and $T(x)=Ax$. (So $A$ is square). Then the following are equivalent:

• $S$ spans $\mathbb{R}^n$

• $S$ is linearly independent

• $Ax=b$ has a unique solution for all $b\in \mathbb{R}^n$

• $T(xs)=b$ has a unique solution for all $b\in \mathbb{R}^n$

• $T$ is onto

• $T$ is one-to-one.

Geometry of linear transformations from R^2 to R^2

Lines go to lines (or points)! Why? $T((1-s)u+sv)=(1-s)T(u)+sT(v)$.

The columns of the matrix tells you where the standard basis goes. Once you know this, you should know everything.

Let's see what happens to the square $\{(x,y):0\leq x,y\leq 1\}$ under the following transforms $$\begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ $$\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$

Piecing things together

Theorem: Let $S=\{v_1,\ldots,v_n\}$. Let $A$ be the matrix with the elements of $S$ as columns. Let $B$ be an echelon matrix equivalent to $A$. Let $T$ be a linear transform with $T(x)=Ax$. Then the following are equivalent

• The set $S$ is linearly independent.

• The linear equation $x_1v_1+\ldots+x_nv_n=0$ has only the trivial solution.

• Every columns of $B$ has a pivot. (computationally useful)

• For any $b\in \mathbb{R}^n$, the equation $x_1v_1+\ldots+x_nv_n=b$ has a unique solution.

• The homogenous equation $Ax=0$ has only the trivial solution.

• For any $b\in \mathbb{R}^n$, the equation $Ax=b$ has at most one solution.

• For any $b\in \mathbb{R}^n$, $b$ can be expressed as a linear combination of elements in $S$ in at most one way.

• The zero vector can be expressed as a linear combination of elements in $S$ in only one way.

• $T$ is a one-to-one linear transformation.

• The only solution to $T(x)=0$ is $x=0$. If $T(x)=0$, then $x=0$.

• There is at most one solution to $T(x)=b$.

Theorem: Let $S=\{v_1,\ldots,v_n\}$ be a set of vectors in $\mathbb{R}^m$. Let $A$ be the matrix with the elements of $S$ as columns. Let $B$ be an echelon matrix equivalent to $A$. Let $T$ be a linear transform with $T(x)=Ax$. Then the following are equivalent

• The set $S$ spans $\mathbb{R}^m$.

• The linear equation $x_1v_1+\ldots+x_nv_n=b$ always has a solution.

• Every row of $B$ has a pivot. (computationally useful)

• For any $b\in \mathbb{R}^n$, the equation $Ax=b$ has at least one solution.

• For any $b\in \mathbb{R}^n$, $b$ can be expressed as a linear combination of elements in $S$ in at least one way.

• $T$ is a onto linear transformation.

• There is always a solution to $T(x)=b$.

Examples:

Kristin DeVleming exam: Let $u_1=(4,4,2)$ and $u_2=(8,5,-3)$. Let $v=(26,17,-8)$. Write $v$ as a linear combination of $u_1,u_2$. Write a vector $w$ that is not in the span of $u_1,u_2$.

Josh Swanson exam: Are the following sets spanning?

• $\{(1,2,3),(-1,-1,2),(-1,0,7)\}$

• $\{(1,-1,1), (0,1,2),(-2,0,2),(1,3,1)\}$.