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Geometry Reference for Linear Algebra

This page covers the essential geometric ideas you need for your linear algebra course. No prior geometry background required — just read top to bottom.

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1. Vectors as Arrows in Space

A vector is just an ordered list of numbers. But geometrically, think of it as an arrow starting at the origin.

  • A vector in $\mathbb{R}^2$ has 2 components → lives in 2D (a flat plane)
    • Example: $(3, 2)$ means “go 3 units right, 2 units up”
  • A vector in $\mathbb{R}^3$ has 3 components → lives in 3D (space around you)
    • Example: $(1, 2, 3)$ means “go 1 right, 2 forward, 3 up”
  • A vector in $\mathbb{R}^n$ has $n$ components → lives in $n$-dimensional space

The number of components = the dimension of the space it belongs to.

How to tell which $\mathbb{R}^n$ a vector belongs to: Just count its entries.

\[\begin{pmatrix}1\\2\end{pmatrix} \in \mathbb{R}^2, \quad \begin{pmatrix}1\\1\\1\end{pmatrix} \in \mathbb{R}^3, \quad \begin{pmatrix}1\\0\\3\\7\end{pmatrix} \in \mathbb{R}^4\]

2. The Origin

The origin is the “zero point” of the coordinate system:

  • In $\mathbb{R}^2$: the origin is $(0, 0)$
  • In $\mathbb{R}^3$: the origin is $(0, 0, 0)$
  • In $\mathbb{R}^n$: the origin is $(0, 0, \ldots, 0)$

Every vector “starts” at the origin (as an arrow). The zero vector $\vec{0}$ is the origin itself — an arrow of length zero.

Why “through the origin” matters in linear algebra: Every subspace must contain the zero vector. That’s why spans, null spaces, column spaces, etc. always pass through the origin. A line or plane that doesn’t pass through the origin is not a subspace.


3. Lines

3a. Lines Through the Origin

A line through the origin is all scalar multiples of a single nonzero vector:

\[L = \{c\vec{v} : c \in \mathbb{R}\}\]
  • When $c > 0$: you go in the direction of $\vec{v}$
  • When $c = 0$: you’re at the origin
  • When $c < 0$: you go in the opposite direction

This is like stretching an arrow forward and backward indefinitely.

Example: $\vec{v} = (1, 2)$ in $\mathbb{R}^2$.

${c(1,2) : c \in \mathbb{R}} = {(c, 2c)}$ — this is the line $y = 2x$, passing through the origin.

Some points on this line: $(0,0)$, $(1,2)$, $(-1,-2)$, $(0.5, 1)$, $(3, 6)$, …

3b. General Lines (not necessarily through origin)

A general line has the form:

\[L = \{\vec{p} + t\vec{d} : t \in \mathbb{R}\}\]

where $\vec{p}$ is a point on the line and $\vec{d}$ is the direction vector.

If $\vec{p} = \vec{0}$, this reduces to a line through the origin.


4. Planes

4a. Planes Through the Origin

A plane through the origin in $\mathbb{R}^3$ is all linear combinations of two independent (non-parallel) vectors:

\[P = \{c_1\vec{v}_1 + c_2\vec{v}_2 : c_1, c_2 \in \mathbb{R}\}\]

Think of it this way: You can slide freely along $\vec{v}_1$ and independently slide along $\vec{v}_2$. Since they point in different directions, you sweep out a flat 2D surface passing through the origin.

4b. Plane Equation

Every plane in $\mathbb{R}^3$ can be written as:

\[ax + by + cz = d\]
  • The vector $(a, b, c)$ is called the normal vector — it’s perpendicular to the plane.
  • If $d = 0$, the plane passes through the origin: $ax + by + cz = 0$.

Example: The plane $x - 2y + z = 0$.

  • Normal vector: $(1, -2, 1)$
  • Passes through origin (since $d = 0$)
  • A point like $(1, 1, 1)$: check $1 - 2(1) + 1 = 0$ ✓ — it’s on the plane
  • A point like $(1, 0, 0)$: check $1 - 0 + 0 = 1 \neq 0$ ✗ — not on the plane

5. Dot Product (Inner Product)

The dot product of two vectors:

\[\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + \cdots + u_nv_n\]

Geometric meaning:

\[\vec{u} \cdot \vec{v} = \|\vec{u}\| \cdot \|\vec{v}\| \cdot \cos\theta\]

where $\theta$ is the angle between the vectors.

Key consequences:

Condition Meaning
$\vec{u} \cdot \vec{v} = 0$ $\vec{u}$ and $\vec{v}$ are perpendicular (orthogonal)
$\vec{u} \cdot \vec{v} > 0$ Angle between them is acute ($< 90°$)
$\vec{u} \cdot \vec{v} < 0$ Angle between them is obtuse ($> 90°$)

Example:

$(1, 0) \cdot (0, 1) = 0 + 0 = 0$ → perpendicular ✓ (the $x$-axis and $y$-axis are perpendicular)

$(1, 1) \cdot (1, -1) = 1 - 1 = 0$ → these are perpendicular too! (the diagonals of a square)

$(1, 2, 3) \cdot (4, 5, 6) = 4 + 10 + 18 = 32 > 0$ → acute angle between them


6. Cross Product (Only in $\mathbb{R}^3$)

The cross product $\vec{u} \times \vec{v}$ gives a vector that is perpendicular to both $\vec{u}$ and $\vec{v}$.

Formula:

\[(u_1, u_2, u_3) \times (v_1, v_2, v_3) = (u_2v_3 - u_3v_2, \;\; u_3v_1 - u_1v_3, \;\; u_1v_2 - u_2v_1)\]

Memory trick (determinant form):

\[\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}\]

Example: Finding the normal to a plane spanned by $(1,1,1)$ and $(1,2,3)$.

\[(1,1,1) \times (1,2,3) = (1\cdot3 - 1\cdot2, \;\; 1\cdot1 - 1\cdot3, \;\; 1\cdot2 - 1\cdot1) = (1, -2, 1)\]

So the normal is $(1, -2, 1)$, and the plane equation is $x - 2y + z = 0$.

Why this works: The cross product gives a vector perpendicular to both input vectors. Since both input vectors lie in the plane, the cross product points out of the plane — that’s exactly what a normal vector does.

Cross product only works in $\mathbb{R}^3$! There’s no cross product in $\mathbb{R}^2$ or $\mathbb{R}^4$.


7. Lengths and Distances

The length (or norm) of a vector:

\[\|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + \cdots + v_n^2}\]

This is just the Pythagorean theorem extended to $n$ dimensions.

Vector Length
$(3, 4)$ $\sqrt{9 + 16} = 5$
$(1, 1, 1)$ $\sqrt{1 + 1 + 1} = \sqrt{3}$
$(1, 0, 0)$ $1$ (a unit vector)

A unit vector has length 1. To make any vector into a unit vector: $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$.


8. Projection

The projection of $\vec{u}$ onto $\vec{v}$ is the “shadow” of $\vec{u}$ cast along the direction of $\vec{v}$:

\[\text{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}} \vec{v}\]

This is used extensively in Gram-Schmidt orthogonalization.

Example: Project $(3, 4)$ onto $(1, 0)$ (the $x$-axis).

\[\text{proj}_{(1,0)} (3,4) = \frac{(3)(1) + (4)(0)}{(1)(1) + (0)(0)} (1, 0) = \frac{3}{1}(1,0) = (3, 0)\]

Makes sense — the “shadow” of $(3,4)$ onto the $x$-axis is $(3, 0)$.


9. The Big Picture — Span and Geometry

This is the key table that connects linear algebra to geometry:

# linearly independent vectors in $\mathbb{R}^3$ Span is Why?
0 The origin (a point) No vectors → only $\vec{0}$ is reachable
1 A line through the origin Scaling one vector forward/backward traces a line
2 A plane through the origin Two non-parallel directions → sweep out a flat surface
3 All of $\mathbb{R}^3$ Three non-coplanar directions → reach everywhere

Generalization to $\mathbb{R}^n$: The span of $k$ linearly independent vectors in $\mathbb{R}^n$ is a $k$-dimensional subspace (a “$k$-dimensional flat surface” through the origin).


10. Perpendicularity (Orthogonality)

Two vectors are orthogonal (perpendicular) when their dot product is zero:

\[\vec{u} \perp \vec{v} \iff \vec{u} \cdot \vec{v} = 0\]

Where this shows up in linear algebra:

  • Orthogonal basis: Basis vectors are all perpendicular to each other → simplifies computations
  • Gram-Schmidt: Converts any basis into an orthogonal one
  • Normal vectors: Perpendicular to a plane/subspace
  • Null space & row space: $\text{Null}(A) \perp \text{Row}(A)$ — every vector in the null space is perpendicular to every row of $A$

11. Parallel vs Independent — Visual Test

Relationship Geometric Meaning Algebraic Test
Parallel (dependent) Same or opposite direction One is a scalar multiple of the other: $\vec{v}_2 = k\vec{v}_1$
Not parallel (independent, for 2 vectors) Point in genuinely different directions No single $k$ works for all components
Coplanar (dependent, for 3 vectors) All three lie in the same plane One vector is a combination of the other two
Not coplanar (independent, for 3 vectors in $\mathbb{R}^3$) They “spread out” into full 3D The $3 \times 3$ determinant $\neq 0$

Example: Are $(1,1,1)$ and $(1,2,3)$ parallel?

If $(1,2,3) = k(1,1,1)$, then from the 1st component: $k = 1$. But from the 2nd: $k = 2$. Contradiction!

So they are not parallellinearly independent → their span is a plane (not just a line).


Quick Reference Card

Concept Formula / Test Geometric Meaning
Dot product $\vec{u} \cdot \vec{v} = \sum u_iv_i$ Measures angle between vectors
Orthogonal $\vec{u} \cdot \vec{v} = 0$ Perpendicular
Cross product $\vec{u} \times \vec{v}$ Vector perpendicular to both
Length $|\vec{v}| = \sqrt{\sum v_i^2}$ Arrow length
Unit vector $\hat{v} = \vec{v}/|\vec{v}|$ Direction without magnitude
Projection $\frac{\vec{u}\cdot\vec{v}}{\vec{v}\cdot\vec{v}}\vec{v}$ Shadow of $\vec{u}$ along $\vec{v}$
Plane normal $(a,b,c)$ in $ax+by+cz=d$ Perpendicular to the plane
Span dimension $k$ independent vectors → $k$-dim subspace Line / Plane / …

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