Linear independence and basis vectors — AI-ML notes

1. Linear combination — the atom of everything

WHAT it means: stretching (scaling) and adding vectors. HOW it builds space: with 2 non-parallel vectors in $\mathbb{R}^2$ you can reach every point — their span is the whole plane.

2. Linear independence — first-principles derivation

WHY this exact test? Let's derive why "only trivial solution" is the right criterion for "no redundancy."

Suppose one vector is redundant — say $\mathbf{v}_k$ can be written using the others: $\mathbf{v}_k = a_1\mathbf{v}_1 + \dots + a_{k-1}\mathbf{v}_{k-1}.$ Rearrange everything to one side: $a_1\mathbf{v}_1 + \dots + a_{k-1}\mathbf{v}_{k-1} - 1\cdot\mathbf{v}_k = \mathbf{0}.$ Look — we found coefficients $(a_1,\dots,a_{k-1},-1)$ that are not all zero (the last one is $-1$ ) yet give $\mathbf{0}$ . So redundancy ⟺ a non-trivial combination equals zero. That's exactly why the definition uses the zero-vector equation.

WHY the determinant? $\det(A)$ is the (signed) volume of the parallelepiped spanned by the columns. If the vectors are dependent, they lie in a lower-dimensional flat (zero volume) ⇒ $\det = 0$ .

Figure — Linear independence and basis vectors

3. Basis — the minimal complete kit

Derivation of uniqueness: Suppose two representations exist: $\mathbf{v}=\sum c_i\mathbf{b}_i = \sum d_i\mathbf{b}_i.$ Subtract: $\sum (c_i-d_i)\mathbf{b}_i = \mathbf{0}$ . By independence, every coefficient is zero: $c_i - d_i = 0 \Rightarrow c_i = d_i$ . So coordinates are unique. This is WHY bases give well-defined coordinate systems.

4. Worked examples

5. Steel-manned mistakes

6. Active recall

#flashcards/ai-ml

Define linear independence in one equation-based sentence.

The only solution to

c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k=\mathbf{0}

is all

c_i=0

What two properties must a basis satisfy?

Linear independence AND spanning the whole space.

Why does redundancy imply a non-trivial zero combination?

\mathbf{v}_k=\sum a_i\mathbf{v}_i

, then

\sum a_i\mathbf{v}_i-\mathbf{v}_k=\mathbf{0}

has coefficient

-1

, hence non-trivial.

For a square matrix of column vectors, the test for independence is?

\det(A)\neq 0

(equivalently full rank).

What does dimension equal?

The number of vectors in any basis of the space.

Why are coordinates in a basis unique?

Two representations subtract to

\sum(c_i-d_i)\mathbf{b}_i=\mathbf{0}

; independence forces

c_i=d_i

Can 3 vectors be independent in

\mathbb{R}^2

No — you can't have more independent vectors than the dimension (2).

Why does

\det=0

mean dependence geometrically?

The parallelepiped they span has zero volume, so they lie in a lower-dimensional flat.

ML consequence of dependent feature columns?

Multicollinearity / redundant info / non-invertible

X^\top X

Recall Feynman: explain to a 12-year-old

Think of LEGO direction-arrows. Each arrow lets you walk a certain way. If one arrow only sends you where two other arrows already could, it's a copycat — throw it out. The smallest team of arrows that can still walk you to every spot in the playground is the "basis." With that team, there's exactly one recipe of steps to reach any spot — no confusion. Extra copycat arrows just waste space and confuse the map.

Connections

Vector spaces and subspaces — where bases live.
Rank of a matrix — counts independent columns/rows.
Determinant and volume — the $\det=0$ dependence test.
Span and column space — what a set of vectors reaches.
Eigenvectors and eigenbasis — a basis chosen to diagonalize.
Multicollinearity in linear regression — dependence hurting ML.
Change of basis — translating coordinates between bases.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho aap ek room me kisi bhi point ko describe karna chahte ho kuch "direction arrows" se. Agar koi arrow aisi direction me point karti hai jahan aap already baaki arrows ko combine karke pahunch sakte ho, to wo arrow bekaar hai — usse koi nayi freedom nahi milti. Isko hum linear dependence kehte hain. Aur agar har arrow ek genuinely nayi direction deti hai, to wo linearly independent hain.

Formal test simple hai: agar $c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k=\mathbf{0}$ ka sirf solution hai sab $c_i=0$ (trivial), tab vectors independent hain. Agar koi non-zero coefficients ka combination bhi zero de de, matlab redundancy hai — dependent. Square matrix ke liye shortcut: $\det(A)\neq 0$ means independent, kyunki determinant volume hota hai aur dependent vectors ka volume zero ho jaata hai (wo ek line ya plane me squeeze ho jaate hain).

Basis matlab wo minimal team of vectors jo do cheezein saath kare: independent ho (koi waste nahi) aur poore space ko span kare (har point tak pahunche). Basis ka fayda — har vector ke coordinates unique hote hain, ek hi recipe. Number of basis vectors = dimension.

ML me ye kyun important hai? Aapke features agar linearly dependent hain, to wo redundant information de rahe hain — regression me multicollinearity aata hai, aur $X^\top X$ invert nahi hoti. Basis ka concept batata hai ki minimal, non-redundant feature set kaunsa hai. So "zyada features = zyada info" ye galat soch hai; independence maayne rakhti hai.

Test yourself — Linear Algebra Essentials

Connections