1.1 Vectors:

Let v = ( \begin{matrix}1 \\ 2 \end{matrix} ).

If v = velocity in m/sec of an object, then the object is moving east at a rate of 1 m/sec and north at a rate of 2m/sec

Speed of the object =

A vector can be described by its Euclidean coordinates OR by its length and direction.

Let w = ( \begin{matrix}3 \\ −1 \end{matrix} ). Then v + w = ( \begin{matrix}1 \\ 2 \end{matrix} ) + ( \begin{matrix}3 \\ −1 \end{matrix} ) =

v − w = ( \begin{matrix}1 \\ 2 \end{matrix} ) − ( \begin{matrix}3 \\ −1 \end{matrix} ) =

2.1 Let f : X → Y where X ⊂ R^{n}, Y ⊂ R^{m} Graph of f = {(x, f(x)) | x ∈ X} ⊂ R^{n} × R^{m} Domain of f = X, Codomain of f = Y Range of f = Image of f = f(X)

= {y ∈ Y | there exists x ∈ X such that f(x) = y}.

f is a function if for all x in domain of f, f(x) has a unique value.

I.e, for all x, y ∈ X, if x = y, then f(x) = f(y) and for all x ∈ X, f(x) is defined.

f is 1:1 if f(x) = f(y) implies x = y.

f gives a one-to-one correspondence between X and f(X). Given b ∈ Y, f(x) = b has at most one solution

Side-note: f(x) = b has exactly one solution if b ∈ f(X). . Side-note: f(x) = b has no solution if b ∉ f(X).

f is onto if f(X) = Y (i.e., image of f = codomain of f). Given b ∈ Y, f(x) = b has at least one solution.

Ex 1: f : R^{n} → R, f(x) = ||x|| = \sqrt{x_{1}^{2} + x_{2}^{2} + ... + x_{n}^{2}} Domain = Codomain = Image = Is f 1:1?

Proof:

Is f onto?

Proof:

Alternate Proof:

Ex 2: g(x, y) = (x^{2}y, x^{4} − y, x^{6})

Domain = Codomain = Is g 1:1?

Proof:

Is g onto?

Proof:

Ex 3: h(x) = ( \begin{matrix}1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} ) ( \begin{matrix}x \\ y \\ z \end{matrix} )

I.e, h(x) = (x + 2y + 3z, 4x + 5y + 6z).

Domain = Codomain = Image = Is h onto? Is h 1:1?

How many solutions does h(x) = b have?

I.e., how many solutions does ( \begin{matrix}1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} ) ( \begin{matrix}x \\ y \\ z \end{matrix} ) = ( \begin{matrix}b_{1} \\ b_{2} \end{matrix} ) have?

I.e, how many solutions does the following system of equations have:

x + 2y + 3z = b_{1},
4x + 5y + 6z = b_{2}.

Does ( \begin{matrix}1 \\ 4 \end{matrix} ) x + ( \begin{matrix}2 \\ 5 \end{matrix} ) y + ( \begin{matrix}3 \\ 6 \end{matrix} ) z span all of R^{2}?

Is \{( \begin{matrix}1 \\ 4 \end{matrix} ) , ( \begin{matrix}2 \\ 5 \end{matrix} ) , ( \begin{matrix}3 \\ 6 \end{matrix} )\} linearly independent?

Definitions:

If the codomain of f is R (i.e., f : X → R), we say that f is real-valued or scalar valued.

Suppose f : X ⊂ R^{2} → R and c is a constant scalar.

The level curve at height c of f is the curve in R^{2} defined by f(x, y) = c. That is,

the level curve at height c of f = {(x, y) ∈ R^{2} | f(x, y) = c}.

The contour curve at height c of f is the curve in R^{3} defined by the two equations, z = f(x, y), z = c. That is,

the contour curve at height c of f =

= {(x, y, z) ∈ R^{2} | z = f(x, y) = c} = {(x, y, f(x, y)) ∈ R^{3} | f(x, y) = c}.

Recall the graph of f = {(x, y, z) | z = f(x, y)} = {(x, y, f(x, y)) | (x, y) ∈ X}⊂ R^{2} × R

The section of the graph of f by the plane x = c is the set of points in R^{3} defined by the two equations, z = f(x, y), x = c. That is,

the section by x = c is {(x, y, z) ∈ R^{2} | z = f(x, y), x = c} = {(c, y, f(c, y)) ∈ R^{3} | (c, y) ∈ X}.

The section by y = c is {(x, y, z) ∈ R^{2} | z = f(x, y), y = c} = {(x, c, f(x, c)) ∈ R^{3} | (x, c) ∈ X}.