A set of vectors is dependent when one can be written from the others, and independent when none can. The formal test: the only combination giving the zero vector uses every coefficient zero. For two vectors, dependent means one is a multiple of the other, which the determinant ad − bc detects. More vectors than dimensions are always dependent.
Step 1: Picture It
Start with something you already know.
A third route in a flat town
- In a flat town you can walk east or west along the streets.
- You can also walk north or south along the avenues.
- With those two, you can reach every corner in town.
- A friend shows you a diagonal path: one block east for every block north.
- Every place the path takes you, a walk east and then north already reached.
The three routes are dependent: the diagonal adds no place you could not reach before.
East and north alone are independent: drop either one and whole parts of town are out of reach.
Dependence belongs to the set, not to one route. Here you could drop any one of the three and still reach everywhere.
Vectors may also take part-blocks, but the reasoning is the same.
Write each route as a vector of blocks, (east, north):
| Route | As a vector |
|---|---|
| Along a street | ⟨1, 0⟩ |
| Along an avenue | ⟨0, 1⟩ |
| The diagonal path | ⟨1, 1⟩ = ⟨1, 0⟩ + ⟨0, 1⟩ |
Walk one street block, one avenue block, then the diagonal backward: ⟨1, 0⟩ + ⟨0, 1⟩ − ⟨1, 1⟩ = ⟨0, 0⟩. Steps that are not all zero bring you home, and that is the formal test failing.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
The vector ⟨2, 7⟩ can be built as a⟨1, 2⟩ + b⟨0, 1⟩. What is b?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Two vectors
⟨2, 4⟩ and ⟨3, 6⟩: ratios 3/2 and 6/4 are both 1.5, so ⟨3, 6⟩ = 1.5⟨2, 4⟩. Dependent. ⟨2, 4⟩ and ⟨3, 5⟩: ratios 1.5 and 1.25 differ. Independent.
The formal test
a⟨2, 4⟩ + b⟨3, 6⟩ = ⟨0, 0⟩ with a = 3, b = −2: ⟨6 − 6, 12 − 12⟩ = ⟨0, 0⟩. Non-zero coefficients reached zero, so the set is dependent. For an independent set, only a = b = 0 works.
The determinant
For columns ⟨a, c⟩ and ⟨b, d⟩, ad − bc = 0 exactly when one is a multiple of the other. ⟨2, 4⟩, ⟨3, 6⟩: 12 − 12 = 0, dependent. ⟨2, 4⟩, ⟨3, 5⟩: 10 − 12 = −2, independent.
Count first
Three vectors in the plane are always dependent, four in space, and so on. You cannot have more independent directions than dimensions, so counting settles the question before any arithmetic.
The zero vector spoils it
Any set containing ⟨0, 0⟩ is dependent: 1 × ⟨0, 0⟩ + 0 × everything else = ⟨0, 0⟩ with a non-zero coefficient. The zero vector adds no direction and fails the test on its own.
Why it matters
Independent vectors each earn their place: removing one shrinks the span. Dependent sets carry passengers. Chapter 6 builds bases from exactly the vectors that earn their place.
Step 3: Try It Yourself
Tap and try it out.
Test ⟨2, 4⟩ and ⟨3, 6⟩ for independence, then ⟨2, 4⟩ and ⟨3, 5⟩, by ratios, by a combination, and by the determinant
- 13/2 = 1.5 and 6/4 = 1.5: ⟨3, 6⟩ = 1.5⟨2, 4⟩a multiple: dependent
Step 0 of 4
- i-hat lands on(3, 1)
- j-hat lands on(1, 2)
- Determinant5
The shaded parallelogram is the image of the unit square, and its area is 5. That is exactly what the determinant measures.
Step 4: In Real Life
Redundant sensors
Three sensors, one of which always reads the average of the other two, give only two pieces of information. Independence says which measurements are actually earning their place.
Step 5: Watch an Example
One step at a time.
Watch Leo Test a Pair Three Ways
Leo asks whether ⟨1, 3⟩ and ⟨2, 1⟩ are independent. He checks it three ways, to see that they agree.
- Step 1
First the ratios. Dividing the second vector’s entries by the first’s gives 2/1 = 2 and 1/3, about 0.33. The two ratios differ, so no single number turns ⟨1, 3⟩ into ⟨2, 1⟩. Neither is a multiple of the other.
Watch Ravi Test a Pair, Then Finish It Yourself
Ravi asks whether ⟨−2, 6⟩ and ⟨1, −3⟩ are independent. He starts; the rest is yours.
- Step 1
Ravi tries the ratios first. Dividing the second vector’s entries by the first’s gives 1/(−2) = −0.5 and −3/6 = −0.5. They agree, so ⟨1, −3⟩ = −0.5⟨−2, 6⟩: the same line, pointing the other way.
Watch Noor Test Three Vectors
Noor asks whether ⟨1, 1⟩, ⟨1, −1⟩ and ⟨3, 1⟩ are independent.
- Step 1
Noor counts first. Three vectors sit in a plane, which holds at most two independent directions, so the set is dependent before any arithmetic.
Step 6: Your Turn
Practice makes it stick.
The Stationery Packs
Problem 1 of 17
A shop sells pack A, with 2 pens and 1 notebook, for $5, and pack B, with 1 pen and 3 notebooks, for $10. Pack C has 4 pens and 7 notebooks: exactly what some A packs and some B packs hold together. What do those A and B packs cost together, in dollars?
dollars
The Two-Move Robot
Problem 2 of 17
A floor robot has two moves, ⟨2, 1⟩ and ⟨4, 2⟩ meters (east, north). It can use any amount of each, forward or backward. It must stop at a point 6 meters east of its start. How many meters north of its start is that point?
meters
Test the Set
Ratios, the determinant, or weights that reach zero. Count first when there are many vectors.
Problem 3 of 17
With ⟨3, 1⟩ as the first column and ⟨2, 4⟩ as the second, what is ad − bc?
Problem 4 of 17
⟨−6, 3⟩ is a number times ⟨4, −2⟩. What is the number? Give it as a decimal.
Problem 5 of 17
For what number t are ⟨3, −2⟩ and ⟨9, t⟩ dependent?
Problem 6 of 17
a⟨2, 1⟩ + b⟨6, 3⟩ = ⟨0, 0⟩, and a = 3. What is b?
Problem 7 of 17
In space, ⟨2, −1, 5⟩ is built from ⟨1, 0, 0⟩, ⟨0, 1, 0⟩ and ⟨0, 0, 1⟩. What weight does ⟨0, 0, 1⟩ carry?
Whose Step?
One line on each page is a slip. Find it, then put it right.
Problem 8 of 17
Priya tested ⟨2, 6⟩ and ⟨1, 3⟩ with the determinant. One line is a slip. Which?
Priya’s test of ⟨2, 6⟩ and ⟨1, 3⟩
- 1columns ⟨2, 6⟩ and ⟨1, 3⟩
- 2ad − bc = 2 × 1 − 6 × 3
- 3= 2 − 18 = −16
- Answernot zero, so independent
Problem 9 of 17
Worked correctly, what is ad − bc for the columns ⟨2, 6⟩ and ⟨1, 3⟩?
Problem 10 of 17
Dev tested ⟨2, −1, 4⟩ and ⟨−6, 3, −12⟩ in space. One line is a slip. Which?
Dev’s test of ⟨2, −1, 4⟩ and ⟨−6, 3, −12⟩
- 1two vectors in three-dimensional space
- 22 is fewer than 3
- Answerso the two vectors are independent
Problem 11 of 17
Worked correctly: ⟨−6, 3, −12⟩ is a number times ⟨2, −1, 4⟩. What is the number?
Problem 12 of 17
Maya used the formal test on ⟨1, 4⟩ and ⟨2, 1⟩. One line is a slip. Which?
Maya’s formal test of ⟨1, 4⟩ and ⟨2, 1⟩
- 1a⟨1, 4⟩ + b⟨2, 1⟩ = ⟨0, 0⟩
- 2a = 0 and b = 0 satisfy it
- 30⟨1, 4⟩ + 0⟨2, 1⟩ = ⟨0, 0⟩
- Answera combination reaches zero, so dependent
Problem 13 of 17
Worked correctly: with ⟨1, 4⟩ as the first column and ⟨2, 1⟩ as the second, what is ad − bc?
Pick the Move
Choose the move that settles it. You do not need to work either question all the way through.
Problem 14 of 17
You want to know whether ⟨5, 1⟩, ⟨2, 7⟩ and ⟨−3, 4⟩ are independent. Which move settles it fastest?
Problem 15 of 17
You want to know whether ⟨1, 0, 2⟩, ⟨0, 1, 1⟩ and ⟨1, 1, 3⟩ in space are independent. Which move settles it?
Same Verdict?
Judge each one without long working. Select every entry that fits.
Problem 16 of 17
Select every set that is dependent.
Problem 17 of 17
A pair starts with ⟨1, 3⟩. Select every second vector that keeps the pair independent.
Step 7: Quick Check
Show what you know.
Question 1 of 2
⟨2, 3⟩ and ⟨4, 6⟩. Independent?
Question 2 of 2
What does linear dependence mean?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
For which positive number k are ⟨1, k⟩ and ⟨k, 4⟩ dependent?
Stretch 2 of 2
u = ⟨1, 2, 0⟩, v = ⟨0, 1, 1⟩ and w = ⟨2, 5, t⟩. For which number t is the set dependent?
What You Learned
- Vectors are dependent when one can be built from the others, and independent when none can.
- In the plane, two vectors are dependent exactly when one is a multiple of the other, and ad − bc = 0 detects it.
- The formal test asks whether a non-zero combination can reach the zero vector.
- More vectors than dimensions are always dependent, so count first.
For the grown-up
The misconception this lesson targets is that independence can be judged by looking: by counting the vectors, or by checking them two at a time. Both habits are half right. Counting proves dependence when there are more vectors than dimensions, but a set no bigger than the dimension can still be dependent, as two vectors in space that lie on one line are. Checking pairs for multiples is the whole test for two vectors, but three vectors can be dependent with no pair lined up, because one is built from the other two together. A third slip is reading the formal test backwards: all-zero weights always reach the zero vector, so they prove nothing; dependence needs weights that are not all zero. The habit that settles all three is to ask whether one vector can be built from the others, and to find the weights when it can.