# Linear Independence

Mathematics I · Linear Algebra · https://tryals.app/learn/mathematics-i/linear-independence

A set of vectors is **linearly independent** if none of them can be written as a linear combination of the others, meaning no vector is redundant.

Formally, $v_1, \dots, v_k$ are **linearly independent** when the **only** way to satisfy $c_1 v_1 + \dots + c_k v_k = \mathbf{0}$ is with **all coefficients zero** ($c_1 = \dots = c_k = 0$).

If any **nontrivial combination** (not all zero) gives $\mathbf{0}$, they are **linearly dependent**, and one vector can be expressed in terms of the others.

| Property | Meaning | Test |
| --- | --- | --- |
| **Pairwise** | Two vectors | Dependent iff one is a **scalar multiple** of the other |
| **Zero Vector** | Any set containing $\mathbf{0}$ | **Always dependent** (coefficient 1 for $\mathbf{0}$, 0 elsewhere) |
| **Dimension Limit** | More than $n$ vectors in $\mathbb{R}^n$ | **Always dependent** (too many directions) |

## Testing Independence

Checking independence is a **homogeneous linear system**: $c_1 v_1 + \dots + c_k v_k = \mathbf{0}$. The vectors are independent when this system has **only the trivial solution**, meaning the matrix formed by the vectors as columns has **full column rank** (a pivot in every column).

> **Common pitfall:** Thinking independence just means no vector is a multiple of another. That only works for **two** vectors! With three or more, a set can be dependent even if no vector is a multiple of any other—for instance, if the third vector is the **sum** of the first two.

Independence means **no wasted vectors**. Combined with spanning, it forms the foundation for a **basis** in the next lesson.

## Practice questions

6 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Sort each set of vectors by whether it is linearly independent.

**Answer:**

- Independent: {(1, 0), (0, 1)}, {(1, 0, 0), (0, 1, 0)}
- Dependent: {(1, 2), (2, 4)}, {(1, 0), (0, 1), (1, 1)}, {(0, 0), (1, 3)}, {(1, 1, 0), (0, 1, 1), (1, 2, 1)}

**Why:** The last set is the important one: no vector there is a multiple of another, yet (1,1,0) + (0,1,1) = (1,2,1), a nontrivial combination giving zero when moved to one side. Dependence means SOME combination vanishes, not that two vectors are parallel.

Page: https://tryals.app/practice/mathematics-i/linear-independence/sort-each-set-of-vectors-by-whether-it-is-linearly-independent

### 2. The set {u, v, w} is linearly DEPENDENT. Select every statement that must then be true.

A. At least one of the three is a linear combination of the other two
B. One of the three is a scalar multiple of another
C. The homogeneous system with u, v, w as columns has a nonzero solution
D. Some combination of u, v, w with not all coefficients zero equals the zero vector
E. All three vectors are equal

**Answer:** A. At least one of the three is a linear combination of the other two; C. The homogeneous system with u, v, w as columns has a nonzero solution; D. Some combination of u, v, w with not all coefficients zero equals the zero vector

**Why:** Dependence means a nontrivial vanishing combination exists, which is the same as the homogeneous system having a nonzero solution, and solving that relation for the vector with a nonzero coefficient exhibits it as a combination of the others. But no vector need be a MULTIPLE of another, the sum-of-two-others example shows why.

Page: https://tryals.app/practice/mathematics-i/linear-independence/the-set-u-v-w-is-linearly-dependent-select-every-statement-that

### 3. Order the steps of testing whether three vectors in space are linearly independent.

**Answer:**

1. Set up the equation c1 v1 + c2 v2 + c3 v3 = 0
2. Write the three vectors as the columns of a matrix
3. Row reduce the matrix to echelon form
4. If every column contains a pivot, only the trivial solution exists and the vectors are independent

**Why:** A pivot in every column means no free variables, so the only solution is the trivial one and the vectors are independent. A free variable produces a nontrivial relation and therefore exhibits the redundancy explicitly.

Page: https://tryals.app/practice/mathematics-i/linear-independence/order-the-steps-of-testing-whether-three-vectors-in-space-are

### 4. In a machine learning dataset, two features are recorded: temperature in Celsius and the same temperature in Fahrenheit. What does linear algebra say about the model that follows?

A. The model will be more accurate, since the same information appears twice
B. Nothing, the extra column is ignored automatically
C. The columns are linearly dependent, so the fit is unidentifiable
D. The model will fail to run, since dependent columns cannot be processed

**Answer:** C. The columns are linearly dependent, so the fit is unidentifiable

**Why:** One column is an affine function of the other, so it contributes no new direction and the weight can be split infinitely many ways. Predictions stay fine and the coefficients become uninterpretable, which is multicollinearity, and why 'the model works but the coefficients flip sign every refit' is a dependency problem rather than a mystery.

Page: https://tryals.app/practice/mathematics-i/linear-independence/in-a-machine-learning-dataset-two-features-are-recorded-temperature

### 5. Three vectors in three-dimensional space, no one of them a scalar multiple of any other, can still be linearly dependent. Why is pairwise non-proportionality not enough?

A. Because the third can lie in the plane spanned by the first two.
B. Because independence requires the vectors to be mutually orthogonal, which is stronger than being non-proportional.
C. Because non-proportionality only rules out dependence when the vectors have integer entries.
D. Because three vectors in three-dimensional space are always dependent, whatever their values.

**Answer:** A. Because the third can lie in the plane spanned by the first two.

**Why:** The counterexample settles it: among (1,0,0), (0,1,0) and (1,1,0) no vector is a multiple of another, and the third is the sum of the first two. Pairwise checking asks about one other vector; dependence asks whether any non-trivial combination of all of them vanishes. Independence does not require orthogonality either, (1,0) and (1,1) are independent and not perpendicular.

Page: https://tryals.app/practice/mathematics-i/linear-independence/three-vectors-in-three-dimensional-space-no-one-of-them-a-scalar

### 6. A set of three vectors in three-dimensional space in which no vector is a scalar multiple of any other must be linearly independent.

**Answer:** False

**Why:** False. (1,0,0), (0,1,0) and (1,1,0) contain no parallel pair yet are dependent, because the third is the sum of the first two, all three lie in one plane. Dependence is about the whole set, not about pairs.

Page: https://tryals.app/practice/mathematics-i/linear-independence/a-set-of-three-vectors-in-three-dimensional-space-in-which-no-vector