## Announcements:

Read Markov chains part of Section 3.7 for next class. Work through recommended homework questions (and check for updates).

Extra Midterm Review: Today, 4:30-6:00pm, MC110. Bring questions.

Midterm location is based on first letter of last name: HSB35 A-H, HSB236 I-Q, HSB240 R-Z. Be sure to write in the correct room!

Midterm: Saturday, March 1, 6:30pm-9:30pm. It will cover the material up to and including Monday's lecture, but not electrical networks. Practice midterms are on the website. See the missed exam section of the course web page for policies, including for illness.

Tutorials: Quiz next week covers 3.5 and 3.6.

Help Centers: Monday-Friday 2:30-6:30 in MC106.

### New material: Section 3.5: Coordinates

Suppose $S$ is a subspace of $\R^n$ with a basis $\cB = \{ \vv_1, \ldots, \vv_k \}$, so $S$ has dimension $k$. Then we can assign coordinates to vectors in $S$, using the following theorem:

Theorem 3.29: For every vector $v$ in $S$, there is exactly one way to write $v$ as a linear combination of the vectors in $\cB$: $$\vv = c_1 \vv_1 + \cdots + c_k \vv_k$$

Proof: Try to work it out yourself! It's a good exercise.$\quad\Box$

We call the coefficients $c_1, c_2, \ldots, c_k$ the coordinates of $\vv$ with respect to $\cB$, and write $$[\vv]_{\cB} = \ccollll {c_1} {c_2} {\vdots} {c_k}$$

We already intuitively understood this theorem in the case where $S$ is a plane through the origin in $\R^3$. Here's an example of this case:

Example: Let $S$ be the plane through the origin in $\R^3$ spanned by $\vv_1 = \colll 1 2 3$ and $\vv_2 = \colll 4 5 6$, so $\cB = \{ \vv_1, \vv_2 \}$ is a basis for $S$. Let $\vv = \colll 6 9 {12}$. Then $$\vv = 2 \vv_1 + 1 \vv_2 \qqtext{so} [\vv]_{\cB} = \coll 2 1$$ Note that while $\vv$ is a vector in $\R^3$, it only has two coordinates with respect to $\cB$.

We already know how to find the coordinates. For this example, we would solve the system $$\bmat{rr} 1 & 4 \\ 2 & 5 \\ 3 & 6 \\ \emat \, \coll {c_1} {c_2} = \ccolll 6 9 {12}$$

Example: Let $\cB = \{ \ve_1, \ve_2, \ve_3 \}$ be the standard basis for $\R^3$, and consider $\vv = \colll 6 9 {12}$. Then $$\vv = 6 \ve_1 + 9 \ve_2 + 12\ve_3 \qqtext{so} [\vv]_{\cB} = \colll 6 9 {12}$$ We've implicitly been using the standard basis everywhere, but often in applications it is better to use a basis suited to the problem.

### Section 3.6: Linear Transformations

Given an $m \times n$ matrix $A$, we can use $A$ to transform a column vector in $\R^n$ into a column vector in $\R^m$. We write: $$T_A(\vx) = A \vx \quad\text{for \vx in \R^n}$$

Example: If $A = \bmat{rr} 0 & 1 \\ 2 & 3 \\ 4 & 5 \emat$ then $$\kern-9ex T_A\left(\coll {\!-1} 2\right) = A \coll {\!-1} 2 = \bmat{rr} 0 & 1 \\ 2 & 3 \\ 4 & 5 \emat \coll {-1} 2 = -1 \colll 0 2 4 + 2 \colll 1 3 5 = \colll 2 4 6$$ In general (omitting parentheses), $$\kern-9ex T_A \coll x y = A \coll x y = \bmat{rr} 0 & 1 \\ 2 & 3 \\ 4 & 5 \emat \coll x y = x \colll 0 2 4 + y \colll 1 3 5 = \colll y {2x+3y} {4x + 5y}$$ Note that the matrix $A$ is visible in the last expression.

Any rule $T$ that assigns to each $\vx$ in $\R^n$ a unique vector $T(\vx)$ in $\R^m$ is called a transformation from $\R^n$ to $\R^m$ and is written $T : \R^n \to \R^m$.

For our $A$ above, we have $T_A : \R^2 \to \R^3$. $T_A$ is in fact a linear transformation.

Definition: A transformation $T : \R^n \to \R^m$ is called a linear transformation if:
1. $T(\vu + \vv) = T(\vu) + T(\vv)$ for all $\vu$ and $\vv$ in $\R^n$, and
2. $T(c \vu) = c \, T(\vu)$ for all $\vu$ in $\R^n$ and all scalars $c$.

You can check directly that our $T_A$ is linear. For example, $$\kern-9ex T_A \left( c \coll x y \right) = T_A \coll {cx} {cy} = \colll {cy} {2cx + 3cy} {4cx + 5cy} = c \colll y {2x+3y} {4x + 5y} = c \, T_A \left( \coll x y \right)$$ Check condition (1) yourself, or see Example 3.55.

In fact, every $T_A$ is linear:

Theorem 3.30: Let $A$ be an $m \times n$ matrix. Then $T_A : \R^n \to \R^m$ is a linear transformation.

Proof: Let $\vu$ and $\vv$ be vectors in $\R^n$ and let $c \in \R$. Then $$T_A(\vu + \vv) = A(\vu + \vv) = A \vu + A \vv = T_A(\vu) + T_A(\vv)$$ and $$T_A(c \vu) = A(c \vu) = c \, A \vu = c \, T_A(\vu) \qquad\Box$$

Example 3.56: Let $F : \R^2 \to \R^2$ be the transformation that sends each point to its reflection in the $x$-axis. Show that $F$ is linear.

Solution: Give a geometrical explanation on the board.

Algebraically, note that $F(\coll x y) = \coll x {-y}$, from which you can check directly that $F$ is linear. (Exercise.)

Or, observe that $F(\coll x y) = \bmat{rr} 1 & 0 \\ 0 & -1 \emat \coll x y$, so $F = T_A$ where $A = \bmat{rr} 1 & 0 \\ 0 & -1 \emat$.

Example: Let $N : \R^2 \to \R^2$ be the transformation $$N \coll x y := \coll {xy} {x+y}$$ Is $N$ linear?

It turns out that every linear transformation is a matrix transformation.

Theorem 3.31: Let $T : \R^n \to \R^m$ be a linear transformation. Then $T = T_A$, where $$A = [\, T(\ve_1) \mid T(\ve_2) \mid \cdots \mid T(\ve_n) \,]$$

Proof: We just check: \kern-4ex \begin{aligned} T(\vx) &= T(x_1 \ve_1 + \cdots + x_n \ve_n) \\ &= x_1 T(\ve_1) + \cdots + x_n T(\ve_n) \qtext{since T is linear} \\ &= [\, T(\ve_1) \mid T(\ve_2) \mid \cdots \mid T(\ve_n) \,] \colll {x_1} {\vdots} {x_n} \\ &= A \vx = T_A(\vx) \qquad\qquad\Box \end{aligned}

The matrix $A$ is called the standard matrix of $T$ and is written $[T]$.

Example 3.58: Let $R_\theta : \R^2 \to \R^2$ be rotation by an angle $\theta$ counterclockwise about the origin. Show that $R_\theta$ is linear and find its standard matrix.

Solution: A geometric argument shows that $R_\theta$ is linear. On board.

To find the standard matrix, we note that $$\kern-6ex R_\theta \coll 1 0 = \coll {\cos \theta} {\sin \theta} \qqtext{and} R_\theta \coll 0 1 = \coll {-\sin \theta} {\cos \theta}$$ Therefore, the standard matrix of $R_\theta$ is $\bmat{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \emat$.

Now that we know the matrix, we can compute rotations of arbitrary vectors. For example, to rotate the point $(2, -1)$ by $60^\circ$: \kern-7ex \begin{aligned} R_{60} \coll 2 {-1} &= \bmat{rr} \cos 60^\circ & -\sin 60^\circ \\ \sin 60^\circ & \cos 60^\circ \emat \coll 2 {-1} \\ &= \bmat{rr} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \emat \coll 2 {-1} = \coll {(2+\sqrt{3})/2} {(2 \sqrt{3}-1)/2} \end{aligned}

Rotations will be one of our main examples.

### New linear transformations from old

If $T : \R^m \to \R^\red{n}$ and $S : \R^\red{n} \to \R^p$, then $S(T(\vx))$ makes sense for $\vx$ in $\R^m$. The composition of $S$ and $T$ is the transformation $S \circ T : \R^m \to \R^p$ defined by $$(S \circ T)(\vx) = S(T(\vx)) .$$ If $S$ and $T$ are linear, it is easy to check that this new transformation $S \circ T$ is automatically linear. For example, \kern-8ex \begin{aligned} (S \circ T)(\vu + \vv) &= S(T(\vu + \vv)) = S(T(\vu) + T(\vv)) \\ &= S(T(\vu)) + S(T(\vv)) = (S \circ T)(\vu) + (S \circ T)(\vv) . \end{aligned} It's because of this that matrix multiplication is defined how it is! Notice also that the condition on the sizes of matrices in a product matches the requirement that $S$ and $T$ be composable.