🧮 Linear Algebra Kingdom · Linear Algebra
Linear Independence
Whether a set carries any repetition: test two or three vectors for independence, find the dependence relation when there is one, spot the value that makes a set dependent, read independence off a pivot count, and know why more than n vectors in n dimensions can never be independent.
In short
- A set is independent when the only weights giving c1*u + c2*v + ... = 0 are all zero, and dependent as soon as any other choice works.
- For exactly two vectors, dependent means one is a multiple of the other; with three or more, checking them in pairs is not enough.
- Put the vectors in as columns and reduce: a pivot in every column is independence, and a column without one is a free weight and a relation.
- Any set with more vectors than the space has dimensions is dependent, and so is any set holding the zero vector.
A third pole that names nothing new
Vela's two poles at Basis Camp already name every field on the isle. Drive a third pole into the ground and something odd happens: the new pole can itself be named in terms of the first two, so it tells you nothing you could not already say. It is not wrong. It is simply already implied.
A set of vectors is linearly dependent when at least one of them is like that third pole — reachable from the others. It is linearly independent when none of them is, so every vector in the set opens a direction the rest cannot reach.
The official test is written as one equation, and it is worth reading slowly:
c1*u + c2*v + c3*w = 0
Every set can satisfy this with c1 = c2 = c3 = 0, which is called the trivial solution and proves nothing. The question is whether there is any other way.
- If zero weights are the only way, the set is independent.
- If some choice of weights that are not all zero also gives 0, the set is dependent, and that choice is called a dependence relation.
The two ideas are the same thing seen from two sides. If 2u - 3v + w = 0 then w = -2u + 3v, so w was reachable from u and v all along.
Two vectors, and the shapes you can see at a glance
For exactly two vectors there is only one way to be dependent: one of them is a multiple of the other. Divide entry by entry and see whether every slot gives the same number.
u = [3, -2, 5], v = [-9, 6, -15]: every slot gives -3, so v = -3u -> dependent u = [3, -2, 5], v = [-9, 7, -15]: the middle slot gives -3.5 -> independent
Careful: that shortcut is only valid for two vectors. With three or more, a set can be dependent with no two of them multiples of each other at all — the third might be a combination of the other two, which no pairwise check will ever see.
Three shapes settle a set on sight, without any arithmetic:
- A zero vector anywhere in the set makes it dependent. Give the zero vector the weight 1 and every other vector the weight 0. The sum is 0 and the weights are not all zero, which is a dependence relation.
- Two vectors on one line make it dependent. Weight that pair to cancel each other and give everything else 0.
- More vectors than dimensions makes it dependent. A set of 5 vectors in R4 is dependent whatever the numbers are, and so is any set of n + 1 vectors in Rn. The next section says why.
Two warnings about reasons that sound right and are not. Perpendicular is not the same as independent — vectors can meet at any angle and still be independent, so a dot product settles nothing here. And being written differently is not enough — three distinct vectors, no two alike, can still be dependent.
Reading the answer off the pivots
The general test is mechanical. Put the vectors in as the columns of a matrix A, so that c1*u + c2*v + c3*w = 0 is exactly Ac = 0, and reduce.
a pivot in every column -> the only solution is c = 0 -> independent a column with no pivot -> that weight is free -> dependent
A column without a pivot is a weight nobody pins down. Set it to 1, solve for the rest, and you have built a dependence relation, so the set cannot be independent.
Two consequences fall out of this at once.
Independence is about columns; spanning is about rows. Spanning Rm needs a pivot in every row; independence needs a pivot in every column. They are different questions and they can disagree, and swapping them is the classic slip of this whole topic.
More vectors than dimensions is always dependent. A matrix with n rows has at most n pivots, because every pivot needs a row of its own. Hand it more than n columns and at least one of them is left without a pivot, whatever the entries are.
For a square case there is a shortcut. Two vectors of R2, or three of R3, are dependent exactly when the determinant of the matrix holding them is 0. That is what makes "find the k that makes this set dependent" a one-line problem: write the determinant, which comes out as a number times k plus a number, set it to 0, and solve.
Finding the relation, and the largest independent set
When a set is dependent it is often not enough to say so — the question wants the relation itself, the actual weights.
For {u, v, w} where w is the one that was implied, solve w = c1*u + c2*v. Each slot is one ordinary equation in c1 and c2, so take two slots that pin both weights down, solve that pair, and then check the weights in the slots you did not use. Finding c1 = 2 and c2 = -3 means
w = 2u - 3v, and equivalently 2u - 3v - w = 0
is the dependence relation. Both forms say the same thing; read the question to see which one it wants, because the signs differ between them.
The last idea is a counting one. How many vectors can an independent set in Rn hold? At most n, because n pivots is all there is room for — and n is reached, by the standard basis. So R3 tops out at 3.
Given a particular collection, the largest independent subset you can choose out of it has exactly rank many vectors — the pivot count again. Six vectors in R4 that reduce to 3 pivots means you can keep 3 of them and must leave 3 out, and the ones to keep are the ones whose columns carry pivots.
Worked examples
Example 1
Is the set {u, v} independent, where u = [4, -6, 10] and v = [-6, 9, -15]?
- With exactly two vectors, the only way to be dependent is for one to be a multiple of the other.
- Divide slot by slot: -6/4 = -1.5, 9/-6 = -1.5, -15/10 = -1.5.
- Every slot gives the same number, so v = -1.5u — a fractional multiple is still a multiple.
- The set is dependent, and 3u + 2v = 0 is a dependence relation with weights that are not both zero.
Example 2
u = [1, 2], v = [3, 1] and w = [9, 8]. Find the weights c1 and c2 with w = c1*u + c2*v.
- Write one equation per slot: c1 + 3c2 = 9 from the first, and 2c1 + c2 = 8 from the second.
- From the first equation c1 = 9 - 3c2; substituting into the second gives 18 - 6c2 + c2 = 8.
- That is -5c2 = -10, so c2 = 2, and then c1 = 9 - 6 = 3.
- Check: 3*[1, 2] + 2*[3, 1] = [9, 8] = w, so 3u + 2v - w = 0 is the dependence relation.
Example 3
The columns of A are three vectors of R4, and reduction of A gives R = [[1, 0, 4], [0, 1, -2], [0, 0, 0], [0, 0, 0]]. Is the set independent?
- Independence is about the columns, so look along the columns of the reduced form for pivots.
- Column 1 and column 2 each carry a pivot; column 3 carries none.
- That third weight is free, so setting it to 1 gives the relation 4*(first) - 2*(second) + 1*(third) = 0.
- The weights are not all zero, so the set is dependent — the rows of zeros are beside the point.
Practice problems, with solutions
Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.
Problem 1
Difficulty 1 of 5u = [2, 2] v = [8, 8] Is the set {u, v} linearly independent?
- Dependent — v = 4u, so one of them is a multiple of the other.
- Independent — u and v are not perpendicular to each other.
- Independent — neither vector is a multiple of the other.
- Dependent — the only weights that give c1*u + c2*v = 0 are c1 = 0 and c2 = 0.
Answer: A. Dependent — v = 4u, so one of them is a multiple of the other.
- Slot by slot, v over u gives 8/2, 8/2.
- Every slot gives 4, so v = 4u.
- Then 4u - v = 0 is a relation with weights that are not both zero, so the set is dependent.
Problem 2
Difficulty 3 of 5u = [3, -5] v = [5, 0] w = [1, 3] p = [-1, 1] Is the set {u, v, w, p} linearly independent?
- Dependent — two of them are multiples of each other.
- Independent — no one of them is a multiple of any other.
- Dependent — a set of 4 vectors in R2 holds more vectors than R2 has dimensions, so no such set can be independent.
- It depends on the numbers: some sets of 4 vectors in R2 are independent.
Answer: C. Dependent — a set of 4 vectors in R2 holds more vectors than R2 has dimensions, so no such set can be independent.
- As columns the set makes a matrix with 2 rows and 4 columns.
- A pivot needs a row of its own, so there are at most 2 pivots however the numbers fall.
- That leaves at least one column with no pivot, so its weight is free and a non-zero relation exists: the set is dependent.
Problem 3
Difficulty 4 of 5u = [3, 6, 0] v = [-6, -5, 3] w = [-3, 1, 3] The set {u, v, w} is dependent, because w is a combination of the other two. Find the weights c1 and c2 with w = c1*u + c2*v. Type them as a pair, for example (2, -3).
Answer: (1, 1)
- The 1st slot says 3c1 - 6c2 = -3, and the 2nd slot says 6c1 - 5c2 = 1.
- Solving those two equations together gives c1 = 1 and c2 = 1.
- Check: 1*u + 1*v = [-3, 1, 3] = w, so 1u + 1v - w = 0 is the dependence relation.
Common mistakes
- Checking the vectors two at a time and calling a set independent because no two are multiples — a third vector can still be a combination of the other two.
- Testing whether the vectors are perpendicular, which answers a different question entirely: independence has nothing to do with the angle between them.
- Counting pivots along the rows instead of the columns, which is the test for spanning rather than for independence.
- Writing the dependence relation with the signs of the wrong arrangement: w = 2u - 3v and 2u - 3v - w = 0 say the same thing, but 2u - 3v + w = 0 does not.
What you should be able to do
- Decide whether a set of vectors is linearly independent, and justify the answer.
- Find the weights of a dependence relation when a set is dependent.
- Find the value of a parameter that makes a set dependent.
- Read independence off the pivot positions, and give the largest independent set n dimensions allow.
Where this fits in the curriculum
Common Core
- HSN-VM.C.10
High school — Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers; the determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse.
HSN-VM.C.10 is a (+) standard — beyond the college- and career-ready threshold, i.e. precalculus rather than Algebra II. HSN-VM.C.10 names the determinant and the existence of an inverse, which is what independence looks like for n vectors in n dimensions. Independence as a property of any set of vectors, and the dependence relation that comes with it, is first-year university content; the Common Core stops before it.