# What is a line?

y = mx + c
y is a function of x
mx - y + c = 0
y = mx + c
x and y satisfy this equation
Geometrically: a line is a collection of points (x,y) which satisfy the given equation

### (switch to geogebra)

Algebraically: Degree 1 polynomial
Geometrically: A flat surface
Notes: through origin, not through origin, parallel lines, intersecting lines

# What is a plane?

z = ax + by + c
z is a function of x, y
ax + by -z + c = 0
x, y and z satisfy this equation

### (switch to geogebra)

Algebraically: Degree 1 polynomial
Geometrically: A flat surface
Notes: through origin, not through origin, parallel planes, intersecting planes, equation with x co-efficient as 0
Geometrically: a plane is a collection of points (x,y,z) which satisfy the given equation

# The (absolute) basics

### A point is a 0 dimensional object in this 2 dimensional space

3x-2y = 0
(3,2)

# The (absolute) basics

### A point is a 0 dimensional object in this 3 dimensional space

3x-2y-z = 0

# The (absolute) basics

### Consider an n dimensional space

3x-2y-z = 0

# What is a hyperplane?

x_n = a_1x_1 + a_2x_2 + \cdots + a_{n-1}x_{n-1} + c
is a function of
x_1, x_2, \dots, x_n
satisfy this equation
Algebraically: Degree 1 polynomial
Geometrically: A flat surface
Geometrically: a hyperplane is a collection of n-dimensional points which satisfy the given equation
x_n
x_1, x_2, \dots, x_{n-1}
a_1x_1 + a_2x_2 + \cdots + a_{n-1}x_{n-1} - x_n + c = 0

# Some more basics

### Intersection of two planes* (2d) is a line (1d)

* assuming they are independent & not parallel

### (demo in geogebra)

Two lines(1D) intersect at a point(0D)

Two planes(2D) intersect at a line(1D)

3 Planes(2D) intersect at a point(0D)

# Puzzle

### This point will satisfy the equation of all the n planes

* assuming they are independent & not parallel

# Back to system of linear equations

a_{11}x_1 + a_{12}x_2 + a_{13}x_3 + \dots + a_{1n}x_n= b_1
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 + \dots + a_{2n}x_n= b_2
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 + \dots + a_{3n}x_n= b_3
a_{m1}x_1 + a_{m2}x_2 + a_{m3}x_3 + \dots + a_{mn}x_n= b_m
\dots
equations of m (n-1) dimensional planes

### Our Quest: Find n-dimensional point(s)                                  which satisfy all the given equations (i.e., lie on all the given planes)

(x_1, x_2, x_3, \dots , x_n)

# Back to system of linear equations

equations of m (n-1) dimensional planes

### Our Quest: Find n-dimensional point(s)                                  which satisfy all the given equations (i.e., lie on all the given planes)

(x_1, x_2, x_3, \dots , x_n)
\begin{bmatrix} a_{11}&a_{12}&a_{13}&\cdots&a_{1n}\\ a_{21}&a_{22}&a_{23}&\cdots&a_{2n}\\ a_{31}&a_{32}&a_{33}&\cdots&a_{3n}\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ a_{m1}&a_{m2}&a_{m3}&\cdots&a_{mn}\\ \end{bmatrix}
\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \cdots\\ x_{n}\\ \end{bmatrix}
=\begin{bmatrix} b_{1}\\ b_{2}\\ b_{3}\\ \cdots\\ b_{m}\\ \end{bmatrix}
A
\mathbf{x}
\mathbf{b}

# How many solutions are possible?

\begin{bmatrix} 1&1\\ 1&1\\ \end{bmatrix}
\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}
= \begin{bmatrix} 1\\ 2 \end{bmatrix}
\begin{bmatrix} 1&1\\ 1&1\\ \end{bmatrix}
\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}
= \begin{bmatrix} 1\\ 1 \end{bmatrix}
\begin{bmatrix} 1&1\\ 1&-1\\ \end{bmatrix}
\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}
= \begin{bmatrix} 3\\ 1 \end{bmatrix}
* this slide is just for building intuition, there is much more to the answer than what is being revealed on the slide
0~solutions
1~solution
\infty~solutions
x_1 + x_2 = 1
x_1 + x_2 = 2
x_1 + x_2 = 3
x_1 - x_2 = 1
x_1 + x_2 = 1
x_1 + x_2 = 1

### (switch to geogebra: 2d case (above examples), 3d case, a magic trick)

No solution in 3D

Magic trick(many solution)

# How many solutions are possible?

* this slide is just for building intuition, there is much more to the answer than what is being revealed on the slide
Is~~m = n?

### Depends on

Is~~m > n?
Is~~m < n?
How~many~unique~equations?
(the long answer)
(the short answer)
the~rank~of~the~matrix

### But (the beauty of it)

there~will~always~be~0, 1~or~\infty~solutions!

(0, 1~or~\infty)

(1~or~\infty)

# The row picture

x_1 + 2x_2 = 3
x_1 - 2x_2 = -1
x_1 + 2x_2 - x_3= 4
2x_1 - x_2 + x_3= 1
-x_1 + x_2 + 2x_3= 3

### We are looking for a point which satisfies each equation (row)

x_1 + 2x_2 - x_3 + x_4 = 5
2x_1 - x_2 + x_3 - 2x_4 = -1
-x_1 + x_2 + 2x_3 - x_4= 2

### Easy to solve geometrically by plotting the corresponding lines ...

-x_1 - x_2 + 2x_3 +2x_4= 1

Row picture(2D)

Row Picture(3D)

# The column picture

x_1 + 2x_2 = 3
x_1 - 2x_2 = -1
x_1 + 2x_2 - x_3= 4
2x_1 - x_2 + x_3= 1
-x_1 + x_2 + 2x_3= 3

### We are looking for the right linear combination of the columns of A

x_1 + 2x_2 - x_3 + x_4 = 5
2x_1 - x_2 + x_3 - 2x_4 = -1
-x_1 + x_2 + 2x_3 - x_4= 2
-x_1 - x_2 + 2x_3 +2x_4= 1

### .... easy to imagine even in higher dimensions

\begin{bmatrix} 1\\ 1 \end{bmatrix}
+
x_1
\begin{bmatrix} 2\\ -2 \end{bmatrix}
x_2
=\begin{bmatrix} 3\\ -1\\ \end{bmatrix}
\begin{bmatrix} 1&2\\ 1&-2 \end{bmatrix}
=\begin{bmatrix} 3\\ -1\\ \end{bmatrix}
\begin{bmatrix} x_1\\ x_2 \end{bmatrix}
\begin{bmatrix} 1\\ 2\\ -1 \end{bmatrix}
+
x_1
\begin{bmatrix} 2\\ -1\\ 1 \end{bmatrix}
x_2
=\begin{bmatrix} 4\\ 1\\ 3\\ \end{bmatrix}
+
\begin{bmatrix} -1\\ 1\\ 2 \end{bmatrix}
x_3

Linear combination of vectors in 2D

Linear Combination of vectors in 3D

# The column picture

### We are looking for the right linear combination of the columns of A

x_1 + 2x_2 - x_3 + x_4 = 5
2x_1 - x_2 + x_3 - 2x_4 = -1
-x_1 + x_2 + 2x_3 - x_4= 2
-x_1 - x_2 + 2x_3 +2x_4= 1

### .... easy to imagine even in higher dimensions

\begin{bmatrix} 1\\ 2\\ -1\\ -1 \end{bmatrix}
+
x_1
\begin{bmatrix} 2\\ -1\\ 1\\ -1 \end{bmatrix}
x_2
=\begin{bmatrix} 5\\ -1\\ 2\\ 1 \end{bmatrix}
+
\begin{bmatrix} -1\\ 1\\ 2\\ 2 \end{bmatrix}
x_3
+
\begin{bmatrix} 1\\ -2\\ -1\\ 2 \end{bmatrix}
x_4

# The column picture

### Can you think of a b such that this system of equations will not have a solution?

x_1 + x_2 = b_1
x_1 + x_2 = b_2
\begin{bmatrix} 1\\ 1\\ \end{bmatrix}
+
x_1
\begin{bmatrix} 1\\ 1\\ \end{bmatrix}
x_2
=\begin{bmatrix} b_1\\ b_2\\ \end{bmatrix}
any~\mathbf{b}~where~b_1 \neq b_2
e.g., \begin{bmatrix} 1\\ 2 \end{bmatrix}
, \begin{bmatrix} 2\\ -1 \end{bmatrix}
, \begin{bmatrix} 3\\ 2 \end{bmatrix}
, \dots

# The column picture

### Can you read off (at least) one solution just by staring at the matrices?

x_1 + 2x_2 -x_3 = 1
-x_1 + 2x_2 -3x_2 = -1
\begin{bmatrix} 1\\ -1\\ 2 \end{bmatrix}
+
x_1
\begin{bmatrix} -1\\ -3\\ 1 \end{bmatrix}
x_3
=\begin{bmatrix} 1\\ -1\\ 2 \end{bmatrix}
\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}
, \begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix}
, \begin{bmatrix} 1\\ -2\\ 2 \end{bmatrix}
, \dots
2x_1 + x_2 +x_3 = 2
\begin{bmatrix} 2\\ 2\\ 1 \end{bmatrix}
x_1
+
, \begin{bmatrix} 1\\ -c\\ c \end{bmatrix}

# Solving a system of linear equations

### Basic principle:

x_1 + 2x_2 - x_3 + x_4 = 5
2x_1 - x_2 + x_3 - 2x_4 = -1
-x_1 + x_2 + 2x_3 - x_4= 2
-x_1 - x_2 + 2x_3 +2x_4= 1
(high school style)

### Choose p and q such that some variables disappear in the new equation

a_{11}x_1 + a_{12}x_2 + a_{13}x_3 + a_{14}x_4 = b_1
a_{22}x_2 + a_{23}x_3 + a_{24}x_4 = b_2
a_{33}x_3 + a_{34}x_4 = b_3
a_{44}x_4 = b_4
a= b
c= d
pa + qc= pb + qd

# Solving a system of linear equations

(high school style)
x_1 + 2x_2 = 3
x_1 - 2x_2 = -1
(equation 2 - equation 1)
x_1 + 2x_2 = 3
- 4x_2 = -4
x_2 = \frac{-4}{-4} = 1
back substitute
x_1 + 2(1) = 3
\therefore x_1 = 1
(row 2 - row 1)
\begin{bmatrix} 1&2\\ 0&-4 \end{bmatrix}
=\begin{bmatrix} 3\\ -4\\ \end{bmatrix}
\begin{bmatrix} x_1\\ x_2 \end{bmatrix}
\begin{bmatrix} 1&2\\ 1&-2 \end{bmatrix}
=\begin{bmatrix} 3\\ -1\\ \end{bmatrix}
\begin{bmatrix} x_1\\ x_2 \end{bmatrix}

# Solving a system of linear equations

(high school style)
\begin{bmatrix} 1&2&-1\\ -1&2&-3\\ 2&1&2 \end{bmatrix}
=\begin{bmatrix} 1\\ -5\\ 6 \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}
x_1 + 2x_2 -x_3 = 1
-x_1 + 2x_2 -3x_3 = -5
2x_1 + x_2 + 2x_3 = 6

# Solving a system of linear equations

(high school style)
\begin{bmatrix} 1&2&-1\\ -1&2&-3\\ 2&1&2 \end{bmatrix}
=\begin{bmatrix} 1\\ -5\\ 6 \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}
(equation 2 + equation 1)
(equation 3 - 2*equation 1)
\begin{bmatrix} 1&2&-1\\ 0&4&-4\\ 0&-3&4 \end{bmatrix}
=\begin{bmatrix} 1\\ -4\\ 4 \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}
\begin{bmatrix} 1&2&-1\\ 0&4&-4\\ 0&-3&4 \end{bmatrix}
=\begin{bmatrix} 1\\ -4\\ 4 \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}
(equation 3 + 3/4*equation 2)
\begin{bmatrix} 1&2&-1\\ 0&4&-4\\ 0&0&1 \end{bmatrix}
=\begin{bmatrix} 1\\ -4\\ 1 \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}

# Solving a system of linear equations

(high school style)
\begin{bmatrix} 1&2&-1\\ 0&4&-4\\ 0&0&1 \end{bmatrix}
=\begin{bmatrix} 1\\ -4\\ 1 \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}
x_1 + 2x_2 -x_3 = 1
4x_2 -4x_3 = -4
x_3 = 1
back substitute
x_3 = 1
4x_2 -4(1) = -4
x_2 = 0
x_1 + 2(0) - (1) = 1
x_1 = 2

# A puzzle for you

\begin{bmatrix} 1&2&-1\\ -1&2&-3\\ 2&1&2 \end{bmatrix}
=\begin{bmatrix} 1\\ -5\\ 6 \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}
\begin{bmatrix} 1&2&-1\\ 0&4&-4\\ 0&0&1 \end{bmatrix}
=\begin{bmatrix} 1\\ -4\\ 1 \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}

### How will you represent the above as matrix operations?

(row 2 + row 1)
(row 3 - 2*row 1)
(row 3 + 3/4*row 2)