# Linear Algebra - Linear Dependency

### Table of Contents

## 1 - About

Vectors v1, . . . , vn are linearly dependent if the zero vector can be written as a nontrivial linear combination of the vectors: <MATH>0 = \alpha_1 v_1 + \dots + \alpha_n v_n</MATH>

In this case, we refer to the linear combination as a linear dependency in v1, . . . , vn. On the other hand, if the only linear combination that equals the zero vector is the trivial linear combination, we say v1, . . . , vn are linearly independent.

## 2 - Articles Related

## 3 - Example

The vectors [1, 0, 0], [0, 2, 0], and [2, 4, 0] are linearly dependent, as shown by the following equation: <MATH>2 [1, 0, 0] + 2 [0, 2, 0] - 1 [2, 4, 0] = [0, 0, 0]</MATH> Therefore: <math>2 [1, 0, 0] + 2 [0, 2, 0] - 1 [2, 4, 0]</math> is a linear dependency in <math>[1, 0, 0], [0, 2, 0], [2, 4, 0]</math>.

## 4 - When

When a list of vector <math>S = [v_0, \dots , v_n]</math> are:

- not lineary independant:

<math> \begin{array}{lll} \href{dimension}{\text{dim }} \href{span}{\text{Span }} \{v_0, \dots , v_n\} < n + 1 \\ \href{rank}{rank} (v_0, \dots , v_n) < len(v_0, \dots , v_n) \end{array} </math>

- are lineary independant:

<math> \begin{array}{ll} \href{dimension}{\text{dim }} \href{span}{\text{Span }} \{v_0, \dots , v_n\} = n + 1 \\ \href{rank}{rank}(v_0, \dots , v_n) = len(v_0, \dots , v_n) \end{array} </math>

## 5 - Computation

def is_independent(L): ''' input: a list L of vectors (using vec class) output: True if the vectors form a linearly independent list. ''' for i in range(len(L)): if isABasis(L, i): return True return False

## 6 - How to know if vectors are linearly independent

### 6.1 - Easy

The vectors [1, 0, 0], [0, 2, 0], and [0, 0, 4] are linearly independent.

Since each vector has a nonzero entry where the others have zeroes.

Consider any linear combination <MATH>\alpha_1 [1, 0, 0] + \alpha_1 [0, 2, 0] + \alpha_1 [0, 0, 4]</MATH> This equals to <MATH> [\alpha_1, 2\alpha_2, 4\alpha_3]</MATH>

If this is the zero vector, it must be that <math>\alpha_1 = \alpha_2 = \alpha_3 = 0</math>

That is, the linear combination is trivial.

Thus, the only linear combination that equals the zero vector is the trivial linear combination.

### 6.2 - Difficult

By linear-combinations definition, <math>v_1, \dots , v_n</math> are linearly dependent iff there is a nonzero vector <math> \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_n \end{bmatrix} </math> such that <math> \begin{bmatrix} \begin{array}{r|r|} & & \\ v_1 & \dots & v_n \\ & & \end{array} \end{bmatrix} \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_n \end{bmatrix} = 0 </math>

Therefore, v1, . . . , vn are linearly dependent iff the null space of the matrix is nontrivial.

This shows that the question:

- How can we tell if vectors v1, . . . , vn are linearly dependent?

is the same as a question we asked earlier:

- How can we tell if the null space of a matrix is trivial?

is the same as :

- How can we tell if the solution set of a homogeneous linear system is trivial?

is the same as:

- How can we tell if a solution u1 to a linear system is the only solution?

Answering these questions requires an algorithm. Computational Problem: Testing linear dependence

- input: a list [v1, . . . , vn] of vectors
- output: DEPENDENDENT if the vectors are linearly dependent, and INDEPENDENT otherwise.

### 6.3 - if a subset of S form a cycle

We can get the zero vector by adding together vectors corresponding to edges that form a cycle: in such a sum, for each entry x, there are exactly two vectors having 1’s in position x.

Example: the vectors corresponding to {Main, Wriston},{Main, Keeney} {Keeney, Wriston } sum to the zero vector.

Therefore if a subset of S form a cycle then S is linearly dependent.

Example: The vectors corresponding to {Main, Keeney}, {Main, Wriston}, {Keeney, Wriston }, {Wriston, Gregorian} are linearly dependent because these edges include a cycle.

The zero vector is equal to the nontrivial linear combination:

Pembroke | Athletic | Bio-Med | Main | Keeney | Wriston | Gregorian | |
---|---|---|---|---|---|---|---|

1 * | 1 | 1 | |||||

+ 1 * | 1 | 1 | |||||

+ 1 * | 1 | 1 | |||||

+ 0 * | 1 | 1 |

### 6.4 - a set of edges contains no cycle

On the other hand, if a set of edges contains no cycle (i.e. is a forest) then the corresponding set of vectors is linearly independent.

## 7 - Algorithm

If they successfully finish, the Grow algorithm and the Shrink algorithm each find a set of vectors spanning the vector space V. In each case, the set of vectors found is linearly independent.

### 7.1 - Grow-Algorithm Corollary

Analyzing the Grow algorithm

def Grow(V) S = 0; repeat while possible: find a vector v in V that is not in Span S, and put it in S.

Grow-Algorithm Corollary: The vectors obtained by the Grow algorithm are linearly independent.

In graphs, this means that the solution obtained by the Grow algorithm has no cycles (is a forest).

The Grow-Algorithm Corollary implies that, **if the Grow algorithm terminates**, the
set of vectors it has selected is a basis for the vector space V.

### 7.2 - Shrink-Algorithm Corollary

def Shrink(V) S = some finite set of vectors that spans V repeat while possible: find a vector v in S such that Span (S − {v}) = V, and remove v from S.

This algorithm use the superfluous-vector_lemma.

Shrink-Algorithm Corollary: The vectors obtained by the Shrink algorithm are linearly independent.

In graphs, this means that the Shrink algorithm outputs a solution that is a forest.

The Shrink-Algorithm Corollary implies that, if we can run the Shrink algorithm starting with a finite set of vectors that spans V, upon termination it will have selected a basis for V.