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Linear Algebra

Linear independence

Given a list of vectors, can any of them be built from the others, and how can you tell without testing each one in turn?

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Why this exists

Rokeya, Nusrat and Imtiaz sell spice blends. A blend is a vector: (3, 1, 0, 0) means 3 grams of cumin, 1 of coriander and none of turmeric or chili per scoop, and mixing scoops forms a linear combination. A blend is redundant when it can be mixed from the others. Keeping it adds nothing new to what the shelf can make, and it makes recipes ambiguous: the same mixture can then be reached in more than one way.

Checking each blend against the rest means one system per blend. This lesson replaces them with a single system whose right side is \mathbf{0}, and shows that its answer also decides whether recipes are unique.

Definition

Let \mathbf{v}_1, \dots, \mathbf{v}_k be vectors in \mathbb{R}^n. Weights all zero always give 0\mathbf{v}_1 + \dots + 0\mathbf{v}_k = \mathbf{0}, the trivial combination. The list is linearly independent when that is the only way to reach \mathbf{0}:

c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k = \mathbf{0} \ \text{ only when } \ c_1 = \dots = c_k = 0.

The list is linearly dependent when some weights, not all zero, give \mathbf{0}; such an equation is a dependence relation. Independence belongs to the whole list, not to one vector in it. A list with a vector repeated is dependent, since \mathbf{v} - \mathbf{v} = \mathbf{0}, which is why this lesson says “list”; “set” is the common word when the vectors are distinct.

With \mathbf{u}_1 = (1, 2, 0, 3), \mathbf{u}_2 = (2, -1, 4, 1) and \mathbf{u}_3 = (0, 5, -4, 5) in \mathbb{R}^4,

\begin{aligned} 2\mathbf{u}_1 - \mathbf{u}_2 - \mathbf{u}_3 &= (2, 4, 0, 6) - (2, -1, 4, 1) \\ &\quad - (0, 5, -4, 5) = (0, 0, 0, 0), \end{aligned}

so the list is dependent: \mathbf{u}_3 = 2\mathbf{u}_1 - \mathbf{u}_2 was built from the other two.

Intuition

Read c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k as a walk: c_1 steps along \mathbf{v}_1, then c_2 along \mathbf{v}_2, and so on. A dependence relation is a walk that ends where it began without standing still. Independent means the only walk that comes home is the one that never leaves.

\mathbf{u}
\mathbf{v}
\mathbf{w}
\mathbf{v}
Figure 1Walk along \mathbf{u}, then along \mathbf{v}, then backward along \mathbf{w}, and you are home. That closed walk is the dependence relation \mathbf{u} + \mathbf{v} - \mathbf{w} = \mathbf{0}.

In the plane, \mathbf{u} = (4, 1) and \mathbf{v} = (1, 3) point along different lines, and no walk on them alone comes home: c_1\mathbf{u} + c_2\mathbf{v} = \mathbf{0} with c_1 \neq 0 would put \displaystyle \mathbf{u} = -\frac{c_2}{c_1}\mathbf{v} on the line of \mathbf{v}. Add \mathbf{w} = (5, 4) and a closed walk appears, \mathbf{u} + \mathbf{v} - \mathbf{w} = \mathbf{0}. Any third vector in the plane would do the same (Result 4).

Two vectors on one line through \mathbf{0} are dependent: out along one, back along the other. Below, \mathbf{n} = -2\mathbf{m}, so 2\mathbf{m} + \mathbf{n} = \mathbf{0}, and the list \mathbf{m}, \mathbf{n}, \mathbf{y} is dependent too. Yet \mathbf{y} is off the line, so it is not a combination of \mathbf{m} and \mathbf{n}. Dependent means some vector is built from the others, not every one.

\mathbf{m}
\mathbf{n}
\mathbf{y}
Figure 2\mathbf{n} = -2\mathbf{m}, so \mathbf{m}, \mathbf{n}, \mathbf{y} is dependent, yet \mathbf{y} is not a combination of \mathbf{m} and \mathbf{n}.

One size up, three vectors in \mathbb{R}^3 are dependent exactly when all three lie in one plane through \mathbf{0}. If they are dependent, one is a combination of the other two (Result 1), so it lies in any plane through \mathbf{0} that holds those two. If all three lie in the plane \operatorname{span}\{\mathbf{a}, \mathbf{b}\}, write each as x_i\mathbf{a} + y_i\mathbf{b}: three pairs (x_i, y_i) in \mathbb{R}^2 have a relation (Result 4), and the same weights give a relation among the three vectors. Past three entries there is no picture, and the algebra below takes over.

Results

1. Dependence is redundancy. A list of two or more vectors is linearly dependent exactly when at least one of its vectors is a linear combination of the others.

2. Small lists. A list containing \mathbf{0} is dependent. One vector \mathbf{v} alone is independent exactly when \mathbf{v} \neq \mathbf{0}. Two vectors are dependent exactly when one is a multiple of the other.

3. The test is one homogeneous system. Put \mathbf{v}_1, \dots, \mathbf{v}_k as the columns of a matrix and row reduce it to an echelon form. The list is independent exactly when every column has a pivot. A column with no pivot gives a free weight, and setting it to 1 gives a dependence relation.

4. Too many vectors. More than n vectors in \mathbb{R}^n are always dependent.

5. Some vector depends on earlier ones. If \mathbf{v}_1, \dots, \mathbf{v}_k is dependent, then \mathbf{v}_1 = \mathbf{0} or some \mathbf{v}_j with j \geq 2 is a combination of \mathbf{v}_1, \dots, \mathbf{v}_{j-1}.

6. Independence is uniqueness of weights. A list is independent exactly when every combination of it has only one list of weights. For a dependent list, every combination has infinitely many.

Results 1 and 2 are the first day’s work; 3 to 6 are the second day’s.

Why it is true

Result 1. Suppose c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k = \mathbf{0} with some c_j \neq 0. Move the other terms to the right side and divide by c_j:

\mathbf{v}_j = \sum_{i \neq j} \left(-\frac{c_i}{c_j}\right)\mathbf{v}_i.

Conversely, if \mathbf{v}_j = \sum_{i \neq j} d_i\mathbf{v}_i, then \sum_{i \neq j} d_i\mathbf{v}_i - \mathbf{v}_j = \mathbf{0} is a relation whose weight on \mathbf{v}_j is -1.

Result 2. If \mathbf{v}_j = \mathbf{0}, weight 1 on \mathbf{v}_j and 0 elsewhere gives \mathbf{0}. For one vector, c\mathbf{v} = \mathbf{0} with c \neq 0 gives \displaystyle \mathbf{v} = \frac{1}{c}(c\mathbf{v}) = \mathbf{0}. For two, apply Result 1: the combinations of a single vector are its multiples.

Result 3. Matching entries turns c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k = \mathbf{0} into a linear system with one equation per entry, whose coefficient matrix has columns \mathbf{v}_1, \dots, \mathbf{v}_k. Row operations do not change its solutions (Solving a system by row reduction: one, none or infinitely many solutions) and keep a zero right side zero, so that column can be left off. In echelon form, a column with no pivot belongs to a free weight: set it to 1 and any other free weights to 0, and back substitution gives a solution with a nonzero weight. If every column has a pivot, the k-th row reads p\,c_k = 0 with p \neq 0, so c_k = 0, and each row above then forces the next weight to 0.

Result 4. An echelon form has at most one pivot per row, so at most n pivots. With k > n columns some column has none, and Result 3 gives a relation.

Result 5. In a dependence relation, let j be the largest index with c_j \neq 0, so the relation reads c_1\mathbf{v}_1 + \dots + c_j\mathbf{v}_j = \mathbf{0}. If j = 1, then c_1\mathbf{v}_1 = \mathbf{0} and \mathbf{v}_1 = \mathbf{0} as in Result 2. If j \geq 2, solve for \mathbf{v}_j as in Result 1; only earlier vectors appear.

Result 6. If \mathbf{b} = \sum c_i\mathbf{v}_i = \sum d_i\mathbf{v}_i, subtracting gives \sum (c_i - d_i)\mathbf{v}_i = \mathbf{0}, so for an independent list every c_i = d_i. For a dependent list with relation \sum a_i\mathbf{v}_i = \mathbf{0}, a_i not all zero, every number t gives

\mathbf{b} = \sum_{i=1}^{k} (c_i + t a_i)\mathbf{v}_i,

a different list for each t. Since \mathbf{0} is a combination of every list, a dependent list always has such a \mathbf{b}. This settles what Linear combination saw by example: weights are unique for every vector the list reaches or for none, and there are never exactly two lists.

Worked examples

Example 1

Day 1: every relation among three vectors in \mathbb{R}^4

Find every dependence relation among \mathbf{u}_1 = (1, 2, 0, 3), \mathbf{u}_2 = (2, -1, 4, 1) and \mathbf{u}_3 = (0, 5, -4, 5). Which of them are combinations of the other two? Is \mathbf{u}_1, \mathbf{u}_2 independent?

Solution

Matching entries in c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + c_3\mathbf{u}_3 = \mathbf{0} gives

\begin{aligned} c_1 + 2c_2 &= 0 \\ 2c_1 - c_2 + 5c_3 &= 0 \\ 4c_2 - 4c_3 &= 0 \\ 3c_1 + c_2 + 5c_3 &= 0. \end{aligned}

The third equation gives c_2 = c_3, and then the first gives c_1 = -2c_3. Substituting, the second reads -4c_3 - c_3 + 5c_3 = 0 and the fourth -6c_3 + c_3 + 5c_3 = 0, true for every c_3. So c_3 = t is free, and the relations are

-2t\,\mathbf{u}_1 + t\,\mathbf{u}_2 + t\,\mathbf{u}_3 = \mathbf{0},

all multiples of the one in the Definition. Every weight is nonzero when t \neq 0, so by Result 1 each vector can be solved for:

\mathbf{u}_1 = \tfrac{1}{2}\mathbf{u}_2 + \tfrac{1}{2}\mathbf{u}_3, \quad \mathbf{u}_2 = 2\mathbf{u}_1 - \mathbf{u}_3.

Check the first: \mathbf{u}_2 + \mathbf{u}_3 = (2, 4, 0, 6) = 2\mathbf{u}_1. A relation among \mathbf{u}_1, \mathbf{u}_2 alone is one of these with c_3 = 0, so t = 0 and every weight is 0: the pair is independent.

Example 2

Day 1: special cases at a glance

Decide, without solving a system, whether each list is independent: (a) (3, 0, -1), (0, 0, 0), (5, 2, 2); (b) (6, -9, 3), (-4, 6, -2); (c) (2, 5, -3), (4, 10, 1).

Solution

(a) Dependent, because the list contains \mathbf{0}: 0(3, 0, -1) + 1(0, 0, 0) + 0(5, 2, 2) = \mathbf{0} has a nonzero weight. The other two vectors play no part.

(b) For two vectors the question is whether one is a multiple of the other. The first entries allow only the factor \displaystyle -\frac{2}{3}, and \displaystyle -\frac{2}{3}(6, -9, 3) = (-4, 6, -2) matches in every entry. Dependent, with relation 2(6, -9, 3) + 3(-4, 6, -2) = \mathbf{0}.

(c) If (4, 10, 1) = k(2, 5, -3), the first entries force k = 2, but 2(-3) = -6 \neq 1. The reverse factor would be \displaystyle \frac{1}{2}, and \displaystyle \frac{1}{2}(1) \neq -3. Neither is a multiple of the other, so the pair is independent.

Example 3

Day 1: Nusrat’s shelf

Nusrat’s blends, in grams per scoop of (cumin, coriander, turmeric, chili), are \mathbf{a} = (1, 2, 1, 0), \mathbf{b} = (2, 4, 2, 0) and \mathbf{c} = (0, 1, 3, 2). Is the list independent, and which blends are redundant?

Solution

\mathbf{b} = 2\mathbf{a}, a double-strength \mathbf{a}, so 2\mathbf{a} - \mathbf{b} + 0\mathbf{c} = \mathbf{0} and the list is dependent. Both \mathbf{b} (two scoops of \mathbf{a}) and \mathbf{a} (half a scoop of \mathbf{b}) are redundant.

\mathbf{c} is not. The chili entry of c_1\mathbf{a} + c_2\mathbf{b} + c_3\mathbf{c} is 2c_3, so every relation has c_3 = 0, and by the proof of Result 1, \mathbf{c} cannot be solved for. Directly: every combination of \mathbf{a} and \mathbf{b} has no chili, and \mathbf{c} has 2 grams. The shelf has the shape of the second figure: Result 1 promises one redundant vector, not that all are.

Example 4

Day 2: testing by elimination

Decide whether \mathbf{x}_1 = (1, 0, -1, 2), \mathbf{x}_2 = (2, 1, 0, 3) and \mathbf{x}_3 = (0, 2, 3, 1) are independent.

Solution

By Result 3, row reduce the matrix with these columns. R_3 + R_1 and R_4 - 2R_1 clear the first column:

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ -1 & 0 & 3 \\ 2 & 3 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 3 \\ 0 & -1 & 1 \end{bmatrix}.

Then R_3 - 2R_2 and R_4 + R_2 leave rows (0, 0, -1) and (0, 0, 3), and R_4 + 3R_3 clears the last:

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix}.

Every column has a pivot, so the list is independent. Read from the bottom: -c_3 = 0, then c_2 + 2c_3 = 0 gives c_2 = 0, then c_1 + 2c_2 = 0 gives c_1 = 0. The zero row only says 0 = 0; pivots per column decide, not rows.

Example 5

Day 2: a value that makes a list dependent

For which h are \mathbf{v}_1 = (1, -2, 1), \mathbf{v}_2 = (2, -3, 4) and \mathbf{v}_3 = (0, 1, h) dependent? For that h, write \mathbf{v}_3 as a combination of \mathbf{v}_1 and \mathbf{v}_2.

Solution

R_2 + 2R_1 and R_3 - R_1, then R_3 - 2R_2:

\begin{bmatrix} 1 & 2 & 0 \\ -2 & -3 & 1 \\ 1 & 4 & h \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 2 & h \end{bmatrix}
\to \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & h - 2 \end{bmatrix}.

The first two columns have pivots for every h; the third has one exactly when h - 2 \neq 0. So the list is dependent exactly when h = 2. Then c_3 = t is free, the second row gives c_2 = -t, and the first gives c_1 = -2c_2 = 2t. With t = 1, 2\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 = \mathbf{0}, so

\mathbf{v}_3 = \mathbf{v}_2 - 2\mathbf{v}_1 = (2, -3, 4) - (2, -4, 2) = (0, 1, 2).
Example 6

Day 2: four vectors in \mathbb{R}^3

Let \mathbf{w}_1 = (1, 3, 1), \mathbf{w}_2 = (2, 7, 1), \mathbf{w}_3 = (0, -1, 1) and \mathbf{w}_4 = (1, 0, 3). (a) Why must the list be dependent? (b) Find every dependence relation. (c) Find the first vector that is a combination of earlier ones, and show the list without it is independent.

Solution

(a) Four vectors in \mathbb{R}^3: Result 4, with no computing.

(b) R_2 - 3R_1 and R_3 - R_1, then R_3 + R_2:

\begin{bmatrix} 1 & 2 & 0 & 1 \\ 3 & 7 & -1 & 0 \\ 1 & 1 & 1 & 3 \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & -1 & -3 \\ 0 & 0 & 0 & -1 \end{bmatrix}.

Pivots sit in columns 1, 2 and 4, so c_3 = t is free. The bottom row gives c_4 = 0, the second c_2 = c_3 + 3c_4 = t, the first c_1 = -2c_2 - c_4 = -2t:

-2t\,\mathbf{w}_1 + t\,\mathbf{w}_2 + t\,\mathbf{w}_3 + 0\,\mathbf{w}_4 = \mathbf{0}.

(c) \mathbf{w}_1 \neq \mathbf{0}, and \mathbf{w}_2 is not a multiple of \mathbf{w}_1: the first entries would force 2\mathbf{w}_1 = (2, 6, 2). Next, t = 1 solves for \mathbf{w}_3:

\mathbf{w}_3 = 2\mathbf{w}_1 - \mathbf{w}_2 = (2, 6, 2) - (2, 7, 1) = (0, -1, 1).

This is the vector Result 5 promised: the largest index with a nonzero weight is 3. The same row operations reduce the matrix with columns \mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_4 to the echelon form above with column 3 deleted, which has a pivot in every column, so \mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_4 is independent. Since \mathbf{w}_4 has weight 0 in every relation, it is not a combination of the others.

Example 7

Day 2: one recipe or many

Rokeya keeps \mathbf{p} = (3, 1, 0, 0), \mathbf{q} = (0, 2, 1, 0) and \mathbf{r} = (0, 0, 2, 1), in grams of (cumin, coriander, turmeric, chili) per scoop. (a) Show the list is independent and find every recipe for the mixture (6, 4, 7, 3). (b) Imtiaz adds the blend \mathbf{s} = (3, 1, 2, 1). Find every recipe for (6, 4, 7, 3) now, and those with no negative amounts.

Solution

(a) In order (cumin, coriander, turmeric, chili),

c_1\mathbf{p} + c_2\mathbf{q} + c_3\mathbf{r} = (3c_1, \ c_1 + 2c_2, \ c_2 + 2c_3, \ c_3).

If this is \mathbf{0}, cumin forces c_1 = 0, then coriander c_2 = 0, and chili c_3 = 0: independent. For (6, 4, 7, 3), cumin gives c_1 = 2, coriander 2 + 2c_2 = 4 so c_2 = 1, chili c_3 = 3, and turmeric checks: 1 + 6 = 7. So the mixture is 2\mathbf{p} + \mathbf{q} + 3\mathbf{r}, and by Result 6 no other recipe exists; there is nothing more to search for.

(b) \mathbf{s} = (3, 1, 0, 0) + (0, 0, 2, 1) = \mathbf{p} + \mathbf{r}, so \mathbf{p} + \mathbf{r} - \mathbf{s} = \mathbf{0} and the new list is dependent. Adding t times this relation to the recipe in (a) gives, for every t,

(2 - t)\mathbf{p} + \mathbf{q} + (3 - t)\mathbf{r} + t\,\mathbf{s} = (6, 4, 7, 3).

These are all: in any recipe c_1\mathbf{p} + c_2\mathbf{q} + c_3\mathbf{r} + t\mathbf{s}, replacing t\mathbf{s} by t\mathbf{p} + t\mathbf{r} leaves a recipe from \mathbf{p}, \mathbf{q}, \mathbf{r}, which by (a) has c_1 + t = 2, c_2 = 1, c_3 + t = 3. No amount is negative exactly when 0 \leq t \leq 2. The new jar makes nothing new (Span) and costs the recipe its uniqueness.

Boundaries

  • A property of the list. Removing vectors from an independent list keeps it independent; adding vectors to a dependent list keeps it dependent (Exercises 9 and 10). Changing the list can change the answer.
  • Dependent does not mean every vector is redundant, only at least one (Examples 3 and 6).
  • Few vectors prove nothing. At most n vectors in \mathbb{R}^n may be independent or dependent; only more than n settles it without work.
  • Weights are any real numbers. A blend that needs a negative scoop is still a combination. Asking for nonnegative weights is a different, stricter question (Example 7).
  • Finite lists. Everything here concerns finitely many vectors. By convention, with an empty sum taken to be \mathbf{0}, the empty list counts as independent: it has no weight that could be nonzero.

Where this goes next

An independent list that also reaches every vector of a space is a basis, and its size is the dimension (Basis and dimension). Result 5 is how a list that reaches everything is trimmed to a basis: drop each vector that is a combination of earlier ones, which leaves the span unchanged. Unique weights are what make coordinates possible (Coordinates in a basis and change of basis). In matrix language, the columns of A are independent exactly when A\mathbf{x} = \mathbf{0} only for \mathbf{x} = \mathbf{0} (Multiplying a matrix by a vector), and Reduced row echelon form reads off every relation at once. Eigenvectors for different eigenvalues turn out to be independent (Eigenvalues and eigenvectors). The whole area is mapped in Vectors in n dimensions.

Exercises

Day 1: deciding from the definition

  1. Are (4, -6, 2, 10) and (-6, 9, -3, -15) independent? If not, give a dependence relation.
  2. Are (1, 4, -3) and (3, 12, -8) independent?
  3. Give a dependence relation among (5, 1), (0, 0), (2, 7).
  4. Find a dependence relation among \mathbf{a} = (2, 1, 1), \mathbf{b} = (1, -1, 0), \mathbf{c} = (4, -1, 1), and write each as a combination of the other two.
  5. Show that (1, 0, 0, 2), (0, 1, 0, -1), (0, 0, 1, 3) are independent without row reducing.
  6. Find k so that (2, k, 6) and (-1, 3, -3) are dependent.

Day 1: spice blends

  1. Nusrat’s second shelf holds (2, 1, 1, 0), (0, 1, 1, 2) and (1, 1, 1, 1). Give a dependence relation, and say which blend can be mixed from the others with no negative amounts.
  2. Imtiaz’s shelf holds (3, 0, 1, 0), (0, 2, 0, 1) and (1, 1, 0, 0). Show that no blend is redundant.

Day 1: reasoning

  1. Prove that removing vectors from an independent list leaves an independent list.
  2. Prove that adding vectors to a dependent list leaves a dependent list.
  3. Give three dependent vectors in \mathbb{R}^3, no one of them a multiple of another.

Day 2: testing by elimination

  1. Decide whether (1, -1, 0, 2), (2, 0, 1, 3), (0, 2, 1, -1) are independent; if not, give a relation.
  2. Decide whether (2, 1, 0), (1, 4, 1), (0, 1, 3) are independent.
  3. For which h are (1, 4, -1), (2, 9, 0), (1, 6, h) dependent? For that h, write the third as a combination of the first two.
  4. Explain without computing why (1, 2), (3, -1), (0, 4) are dependent, then find a relation with integer weights.
  5. In the list (3, 0, 1), (6, 0, 2), (0, 1, -1), (3, 1, 0), find the first vector that is a combination of earlier ones. Drop it, and repeat until the list is independent.

Day 2: unique weights

  1. Show that (1, 1, 0), (0, 2, 1), (2, 0, 1) are independent, and write (4, 0, 0) as a combination of them. Why are there no other weights?
  2. With Imtiaz’s shelf from Exercise 8, find every recipe for (7, 3, 2, 1), and show (3, 3, 3, 3) has none.
  3. Imtiaz adds the blend (4, 1, 1, 0). Find every recipe for (7, 3, 2, 1) now, and those with no negative amounts.

Day 2: reasoning

  1. Prove that if \mathbf{u}, \mathbf{v}, \mathbf{w} are independent, so are \mathbf{u} + \mathbf{v}, \mathbf{v} + \mathbf{w}, \mathbf{u} + \mathbf{w}.
  2. Prove that nonzero vectors that are pairwise orthogonal (Dot product, length and orthogonality in n dimensions) are independent.
  3. Prove that if one vector \mathbf{b} has exactly one list of weights as a combination of \mathbf{v}_1, \dots, \mathbf{v}_k, then the list is independent.
  4. Give one independent and one dependent list of three vectors in \mathbb{R}^4, and explain why five vectors in \mathbb{R}^4 are never independent.
Answers
  1. Dependent. The second is \displaystyle -\frac{3}{2} times the first, so 3(4, -6, 2, 10) + 2(-6, 9, -3, -15) = \mathbf{0}.
  2. Independent. The first entries force the factor 3, but 3(-3) = -9 \neq -8; the reverse factor \displaystyle \frac{1}{3} fails the same entry.
  3. 0(5, 1) + 1(0, 0) + 0(2, 7) = \mathbf{0}; any nonzero weight on (0, 0) works.
  4. \mathbf{a} + 2\mathbf{b} - \mathbf{c} = \mathbf{0}, since (2, 1, 1) + (2, -2, 0) = (4, -1, 1). So \mathbf{c} = \mathbf{a} + 2\mathbf{b}, \mathbf{a} = \mathbf{c} - 2\mathbf{b} and \displaystyle \mathbf{b} = \frac{1}{2}\mathbf{c} - \frac{1}{2}\mathbf{a}.
  5. The first three entries of c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 are c_1, c_2, c_3, so it is \mathbf{0} only when all three are 0.
  6. k = -6. Neither vector is \mathbf{0}, so one is a multiple of the other; the first entries force (2, k, 6) = -2(-1, 3, -3) = (2, -6, 6), and the third entries agree.
  7. (2, 1, 1, 0) + (0, 1, 1, 2) - 2(1, 1, 1, 1) = \mathbf{0}. All weights are nonzero, so each blend is a combination of the other two, but only \displaystyle (1, 1, 1, 1) = \frac{1}{2}(2, 1, 1, 0) + \frac{1}{2}(0, 1, 1, 2) uses no negative amount.
  8. With weights c_1, c_2, c_3 the mixture is (3c_1 + c_3, 2c_2 + c_3, c_1, c_2). If it is \mathbf{0}, turmeric gives c_1 = 0, chili c_2 = 0, then cumin c_3 = 0. Independent, so by Result 1 no blend is a combination of the others.
  9. A relation among the remaining vectors, with weight 0 put on each removed vector, is a relation among the original list, so all its weights are 0.
  10. A dependence relation of the original list, with weight 0 on each added vector, is a relation of the longer list and keeps its nonzero weight.
  11. For example (1, 0, 0), (0, 1, 0), (1, 1, 0): the third is the sum of the first two, and comparing entries shows no one is a multiple of another.
  12. Dependent: 2(1, -1, 0, 2) - (2, 0, 1, 3) + (0, 2, 1, -1) = \mathbf{0}. After R_2 + R_1 and R_4 - 2R_1, rows two to four are multiples of (0, 1, 1), so column 3 has no pivot; c_3 = t, c_2 = -t, c_1 = 2t.
  13. Independent. Swap R_1 and R_2, then R_2 - 2R_1 gives (0, -7, -2); swap it with (0, 1, 3), and R_3 + 7R_2 gives (0, 0, 19). Every column has a pivot.
  14. h = 3, and then (1, 6, 3) = 2(2, 9, 0) - 3(1, 4, -1). R_2 - 4R_1, R_3 + R_1, R_3 - 2R_2 leave rows (1, 2, 1), (0, 1, 2), (0, 0, h - 3).
  15. Three vectors in \mathbb{R}^2 (Result 4). From c_1 + 3c_2 = 0 and 2c_1 - c_2 + 4c_3 = 0 with c_2 = 4: -12(1, 2) + 4(3, -1) + 7(0, 4) = \mathbf{0}.
  16. First (6, 0, 2) = 2(3, 0, 1). Without it, (3, 1, 0) = (3, 0, 1) + (0, 1, -1) is next. What remains, (3, 0, 1) and (0, 1, -1), is independent: the first two entries show neither is a multiple of the other.
  17. (4, 0, 0) = 2(1, 1, 0) - (0, 2, 1) + (2, 0, 1). R_2 - R_1 gives (0, 2, -2) and then \displaystyle R_3 - \frac{1}{2}R_2 gives (0, 0, 2), a pivot in every column. The list is independent, so by Result 6 these are the only weights.
  18. (7, 3, 2, 1) = 2(3, 0, 1, 0) + (0, 2, 0, 1) +{} (1, 1, 0, 0), the only recipe. The mixture (3c_1 + c_3, 2c_2 + c_3, c_1, c_2) from Exercise 8 needs turmeric c_1 = 2, chili c_2 = 1, cumin 6 + c_3 = 7 so c_3 = 1, and coriander checks, 2 + 1 = 3. For (3, 3, 3, 3) the same entries force c_1 = 3, c_2 = 3, c_3 = -6, but then coriander is 0 \neq 3.
  19. (4, 1, 1, 0) = (3, 0, 1, 0) + (1, 1, 0, 0), so the list is now dependent, and as in Example 7(b) the recipes are (2 - t)(3, 0, 1, 0) + (0, 2, 0, 1) + (1 - t)(1, 1, 0, 0) + t(4, 1, 1, 0) for every t. No amount is negative exactly when 0 \leq t \leq 1.
  20. If a(\mathbf{u} + \mathbf{v}) + b(\mathbf{v} + \mathbf{w}) + c(\mathbf{u} + \mathbf{w}) = \mathbf{0}, regrouping gives (a + c)\mathbf{u} + (a + b)\mathbf{v} + (b + c)\mathbf{w} = \mathbf{0}, so a + c = a + b = b + c = 0. The first two give b = c, the third b = c = 0, and then a = 0.
  21. Dot c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k = \mathbf{0} with \mathbf{v}_j. Each \mathbf{v}_i \cdot \mathbf{v}_j with i \neq j is 0, leaving c_j\lVert \mathbf{v}_j \rVert^2 = 0. Since \mathbf{v}_j \neq \mathbf{0}, c_j = 0, for every j.
  22. If the list had a relation \sum a_i\mathbf{v}_i = \mathbf{0} with a_i not all zero, and \mathbf{b} = \sum c_i\mathbf{v}_i, then \mathbf{b} = \sum (c_i + a_i)\mathbf{v}_i would be a second list of weights.
  23. (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) are independent, by Exercise 5’s argument; (1, 0, 0, 0), (2, 0, 0, 0), (0, 1, 0, 0) are dependent, the second being twice the first. Five vectors in \mathbb{R}^4: a 4-row echelon form has at most 4 pivots (Result 4).
kg1128 · updated 2026-10-22 · CC-BY-SA-4.0