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Linear combinations and the span as the smallest subspace reaching a set of vectors, and the pivot count that decides independence.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write a vector as a linear combination of others by solving the system whose unknowns are the coefficients, describe the span of a set and say why it is a subspace, decide independence by putting the vectors in as columns and counting pivots, read a dependence relation off the reduced form rather than merely reporting that one exists, argue without computation that more vectors than coordinates are always dependent, and run the same test in spaces of polynomials and functions where no coordinates are given.
You can reduce a matrix and read off its rank, its pivot columns and its free variables. Everything in this lesson is a question about vectors that turns into exactly that reduction. The new work is not computational; it is knowing which matrix to build and which count to read.
A linear combination of $v_1, \dots, v_k$ is a vector $c_1v_1 + \dots + c_kv_k$ with real coefficients. Their span is the set of all such combinations. The vectors are linearly dependent when some combination with coefficients not all zero is the zero vector — such a combination is a dependence relation — and linearly independent when the only one is the all-zeros combination.
Given vectors $v_1, \dots, v_k$ in $\mathbb{R}^m$, build the matrix $A$ whose columns are those vectors. Then $Ac$ is exactly the combination $c_1v_1 + \dots + c_kv_k$, and the two questions of this lesson become two questions about $A$:
The second question is therefore answered by the pivots alone. Reduce $A$ and count:
$$\text{independent} \iff \text{every column has a pivot} \iff \operatorname{rank} A = k \iff \text{no free variables}.$$
And because a pivot needs a row to itself, $\operatorname{rank} A \le m$. So more than $m$ vectors in $\mathbb{R}^m$ are always dependent — $k > m$ forces a free variable, a free variable gives a non-zero solution, and a non-zero solution is a dependence relation. No arithmetic is involved in that argument, only counting, and it is the most-used fact in the unit.
Span is a subspace. The span of any set passes all three conditions of the last lesson: $0$ is the combination with every coefficient zero, a sum of combinations is a combination, and a multiple of one is one. In fact the span is the smallest subspace containing the vectors — any subspace containing them must contain every combination of them, by closure.
Another way: picture
One non-zero vector in $\mathbb{R}^3$ spans a line; two independent ones span a plane; three independent ones span everything. Adding a fourth vector cannot enlarge the span, because there is nothing left to reach — and it is guaranteed to be a combination of the other three. Dependence is what redundancy looks like: the new vector already lay in the span of the old ones, so the span did not grow when it arrived.
Another way: steps
This is the step most often got backwards, and the reason is worth stating once rather than memorising.
The unknowns of the problem are the coefficients. In $c_1v_1 + c_2v_2 + c_3v_3 = 0$ there are three unknowns, one per vector, and the equation has one line per coordinate of the ambient space. So the system has one unknown per vector and one equation per coordinate — which is a matrix with one column per vector and one row per coordinate. The vectors are the columns.
Putting them in as rows solves a different problem. It is not useless — the non-zero rows of the reduced form give a basis for the row space, which is the same span — but the pivot positions then refer to coordinates rather than to vectors, so they cannot tell you which of the original vectors was redundant. The column arrangement can:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 7 \end{pmatrix} \longrightarrow \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
Pivots in columns 1 and 3, none in column 2: so $v_1$ and $v_3$ are independent and $v_2$ is the redundant one. Reading the reduced form further, $v_2 = 2v_1$, which is the dependence relation itself.
A free column names a redundant vector, and the reduced form names the combination that makes it redundant. That is more than a verdict, and it is why this arrangement is the one to learn.
Nothing in the definition of a span needs columns of numbers, and the useful cases often have none.
In $P_3$, the span of $\{1, x, x^2\}$ is the polynomials of degree at most $2$ — a subspace of $P_3$, and a proper one, since $x^3$ is not a combination of those three. Are $\{1, x, x^2\}$ independent? A combination $c_0 + c_1x + c_2x^2$ is the zero polynomial only when every coefficient is zero, because a non-zero polynomial of degree at most $2$ has at most two roots and so cannot vanish everywhere. Independent.
In the space of functions, are $\sin$ and $\cos$ independent? Suppose $a\sin t + b\cos t = 0$ for every $t$. Put $t = 0$: $b = 0$. Put $t = \pi/2$: $a = 0$. Independent, in two lines and with no matrix.
In $M_{2 \times 2}$, the four matrices with a single $1$ and three zeros span the whole space and are independent, so they are the natural basis of it — which is the next lesson's word.
The technique is always the same: assume a combination is zero, and squeeze the coefficients. In $\mathbb{R}^m$ the squeezing is elimination. Elsewhere it is substitution, evaluation at a convenient point, or comparing degrees — whatever the space makes cheap.
Checking pairs instead of the whole set. Three vectors can be dependent with no two proportional; $(1,0,0)$, $(0,1,0)$, $(1,1,0)$ is the example to keep. Dependence is a statement about a combination of them all.
Putting the vectors in as rows. It answers a related question and loses the information about which vector is redundant. The unknowns are the coefficients, so the vectors are the columns.
Thinking dependence means every vector is redundant. A dependence relation $c_1v_1 + c_2v_2 = 0$ lets you solve for any $v_i$ whose coefficient is non-zero — but not for one whose coefficient is zero. In $\{(1,0), (2,0), (0,1)\}$ the third vector is in no dependence relation and is not redundant at all.
Confusing "spans" with "is independent". They are opposite pressures. A set spans when it is large enough to reach everything; it is independent when it is small enough to waste nothing. Adding vectors helps the first and hurts the second, and a set doing both at once is what the next lesson calls a basis.
Including the zero vector and expecting independence. Any set containing $0$ is dependent, because $1 \cdot 0 = 0$ is a dependence relation with a non-zero coefficient.
Are $(1, 2, 1)$, $(2, 4, 3)$ and $(1, 2, 0)$ independent? Columns: $A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 3 & 0 \end{pmatrix}$.
Vectors as columns.
Row-reducing: row 2 minus twice row 1 is all zeros in the first two columns and $0$ in the third, and the reduced form is $\begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$.
Two pivots, three columns.
Column 3 is free, so the set is dependent, and the third column of the reduced form reads the relation off directly: $v_3 = 3v_1 - v_2$.
The reduced form names the redundancy.
Are $(3, 1)$, $(-2, 5)$, $(7, 0)$ and $(1, 1)$ independent in $\mathbb{R}^2$?
Four vectors in a two-dimensional space.
The matrix is $2 \times 4$, so it has at most $2$ pivots and at least $2$ free columns.
A pivot needs a row of its own.
Dependent, with no arithmetic done at all. Whenever the number of vectors exceeds the number of coordinates, the answer is settled before the numbers are read.
Count first; compute only if the count is inconclusive.
The two spanning vectors are proportional, so their span is a single line: all multiples of $(1, 2)$.
Look at the spanning set first.
Is $(4, 5)$ a multiple of $(1, 2)$? The first coordinate forces the multiplier to be $4$, and $4 \times 2 = 8 \ne 5$.
One coordinate fixes the multiplier; the other decides.
Not in the span. Note that two vectors were given and the span was still only one-dimensional — the number of vectors is an upper bound on the dimension of the span and nothing more.
Let $u = (1, 4)$ and $v = (0, 1)$. Write $w = (5, 8)$ as $pu + qv$ and enter the column whose entries are $p$ and $q$, in that order.
This task has no paper form; do it on a device.
You are asked whether $(7, 7, 1)$, $(1, 7, 7)$ and $(7, 1, 7)$ are independent. Put the steps in order.
Number the steps in order (write the number in the box):
$5$ vectors are given in $\mathbb{R}^{3}$. At most how many of them can be linearly independent?
Answer:
Three vectors in $\mathbb{R}^{3}$ are given, and no one of them is a scalar multiple of another. What follows?
Match each set of vectors in $\mathbb{R}^{2}$ to its verdict.
| Independent | Dependent: one is a multiple of the other | Dependent, although no two of them are multiples of each other | |
|---|---|---|---|
| $(6, 0)$ and $(0, 2)$ | |||
| $(6, 2)$ and $(12, 4)$ | |||
| $(6, 0)$, $(0, 2)$ and $(6, 2)$ |
For which real values of $k$ are $u = (3, 6)$ and $v = (9, k)$ linearly independent? Give the set of values of $k$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $u = (1, 4)$ and $v = (0, 1)$. Write $w = (6, 5)$ as $pu + qv$ and enter the column whose entries are $p$ and $q$, in that order.
This task has no paper form; do it on a device.
You can test a set of vectors for independence, name the redundant one and the relation that makes it redundant, and say whether a given vector lies in a span. Say in your own words why the vectors go in as the columns and not the rows. Next: the sets that span and are independent at once.
10. Your turn: is $(4, 5)$ in the span of $(1, 2)$ and $(2, 4)$?, step 3
Redundancy shrinks what a count promised.