An engineer with a pile of building blocks asks two questions. The Span page answered the first: what can I build with these? This page answers the second: do I need all of them, or is something in the pile redundant? A redundant vector costs you work (an extra unknown in every system, an extra column in every matrix) and buys you nothing. Here's the picture to keep in mind, starring the superhero from the Span page.
Remember the superhero from the Span page? In space, the hero had a hoverboard $h = (1, 2, 0)$, a magic carpet $m = (0, 1, 1)$, and a scooter $s = (2, 5, 1)$, and the scooter turned out to be useless: $s = 2h + m$, so anywhere the scooter goes, the other two could already go. This page is about spotting a useless gadget like that without guessing, and about the sets that have none.
The scooter was dead weight, and the giveaway was a trip that comes back home: $2h + m - s = 0$. With the jetpack instead, no trip comes home except standing still. That difference, a nontrivial way back to $0$ or none, is the whole page.
The Span page ended with this: adding $v$ to a set leaves the span unchanged exactly when $v$ was already in the span. So a vector is dead weight exactly when it's a linear combination of the others. The question is how to find one.
The obvious way is to check the vectors one at a time. Here's why that's a bad plan.
Let $v_1 = (3, 4, 5)$, $v_2 = (2, 9, 2)$, and $v_3 = (4, 18, 4)$. Is one of them a linear combination of the others?
Start with $v_1$. Is $v_1 = av_2 + bv_3$? Matching coordinates:
$$\begin{aligned} 2a + 4b &= 3 \\ 9a + 18b &= 4 \\ 2a + 4b &= 5. \end{aligned}$$The first and third equations say $2a + 4b$ is both $3$ and $5$. No solution, so $v_1$ is not a combination of the other two.
That doesn't settle it, though. Try $v_2$: is $v_2 = av_1 + bv_3$? Look closer and $v_3$ is exactly twice $v_2$, so $a = 0$, $b = \tfrac{1}{2}$ works:
$$v_2 = 0v_1 + \tfrac{1}{2}v_3.$$So yes, $v_2$ is dead weight (and so is $v_3$). But it took two systems to find out, and with ten vectors it could take ten.
Here it is in space. $v_2$ and $v_3$ point along one line, and every combination of them stays on that line, which is why $v_1$ failed.
Here's the trick that does it in one system. Take $v_2 = 0v_1 + \tfrac{1}{2}v_3$, double it, and move everything to one side:
$$0v_1 + 2v_2 - v_3 = 0.$$That's the zero vector written as a combination of all three, with coefficients that aren't all zero. And it runs backward too: from an equation like that, any vector with a nonzero coefficient can be solved for. So instead of asking "is some vector a combination of the others?" (one system per vector), ask "can $0$ be written as a combination of all of them, with coefficients that aren't all zero?" (one system). That question gets the names.
You can always get $0$ the lazy way, by using all zeros. That one gets a name first, so it can be ruled out.
For any vectors $u_1, u_2, \ldots, u_k$, the combination $0u_1 + 0u_2 + \cdots + 0u_k = 0$ is the trivial representation of $0$. Any other way of writing $0$ as a combination of them, with at least one coefficient not zero, is nontrivial.
Zero times everything equals zero. Groundbreaking. Somebody give this a name. …Oh, they did.
Rule out the trivial one, and what's left is the definition.
Vectors $u_1, u_2, \ldots, u_k$ in $\R^n$ are linearly dependent if $c_1u_1 + c_2u_2 + \cdots + c_ku_k = 0$ for some scalars $c_1, c_2, \ldots, c_k$ that are not all zero. That equation is called a linear relation among them, and a set of vectors is linearly dependent when its vectors are.
Arrows $u_1, u_2, \ldots, u_k$ in $\R^2$ or $\R^3$ are linearly dependent if some chain $c_1u_1 + c_2u_2 + \cdots + c_ku_k$ comes back home to $0$ without every coefficient being $0$. That equation is called a linear relation among them.
Two words in that definition do a lot of work.
Read "not all zero" carefully. It does not mean "all nonzero." In $0v_1 + 2v_2 - v_3 = 0$ the coefficient of $v_1$ is $0$, and that's fine: one nonzero coefficient is all it takes. The flip side: you can only solve for a vector whose coefficient is nonzero. Try to solve that equation for $v_1$ and you'd be dividing by $0$.
So deciding dependence comes down to one system: write the coefficients as unknowns, match coordinates, and set everything equal to zero. It's the Span page's routine with a right side of all zeros.
Show that $u_1 = (1, 3, -1)$, $u_2 = (2, -1, 2)$, $u_3 = (5, 1, 3)$ are linearly dependent, and write one of them as a combination of the others.
We need $c_1, c_2, c_3$, not all zero, with $c_1u_1 + c_2u_2 + c_3u_3 = 0$. Matching coordinates:
$$\begin{aligned} c_1 + 2c_2 + 5c_3 &= 0 \\ 3c_1 - c_2 + c_3 &= 0 \\ -c_1 + 2c_2 + 3c_3 &= 0. \end{aligned}$$$R_2 \to R_2 - 3R_1$ and $R_3 \to R_3 + R_1$ give $-7c_2 - 14c_3 = 0$ and $4c_2 + 8c_3 = 0$. Both say $c_2 = -2c_3$, so $c_3$ is free. Take $c_3 = -1$: then $c_2 = 2$, and the first equation gives $c_1 = -2c_2 - 5c_3 = -4 + 5 = 1$. Check it:
$$(1, 3, -1) + 2(2, -1, 2) - (5, 1, 3) = (0, 0, 0).$$So $u_1 + 2u_2 - u_3 = 0$ and the vectors are dependent. Every coefficient is nonzero, so any of them can be solved for. Solving for $u_3$: $u_3 = u_1 + 2u_2$.
In space, all three arrows lie in one plane, and the chain $u_1 + 2u_2$ lands exactly on $u_3$'s tip.
Nothing changes in $\R^4$, even when the vectors give more equations than unknowns.
Are $u_1 = (1, 0, 2, 1)$, $u_2 = (0, 1, 1, -1)$, $u_3 = (1, 1, 3, 0)$ in $\R^4$ linearly dependent?
Matching coordinates in $c_1u_1 + c_2u_2 + c_3u_3 = 0$ gives four equations:
$$\begin{aligned} c_1 + c_3 &= 0 \\ c_2 + c_3 &= 0 \\ 2c_1 + c_2 + 3c_3 &= 0 \\ c_1 - c_2 &= 0. \end{aligned}$$The first two say $c_1 = c_2 = -c_3$. Then the third is $-2c_3 - c_3 + 3c_3 = 0$ and the fourth is $0 = 0$, both true for every $c_3$. So $c_3$ is free: $c_3 = -1$ gives $u_1 + u_2 - u_3 = 0$. Dependent, with $u_3 = u_1 + u_2$. Four equations for three unknowns, and still a free unknown: the count of equations doesn't decide it, the elimination does.
The opposite, and the property you'll be hoping for most of the time, gets a name of its own.
Vectors $u_1, u_2, \ldots, u_k$ in $\R^n$ are linearly independent if they are not linearly dependent: the only way to get $c_1u_1 + c_2u_2 + \cdots + c_ku_k = 0$ is $c_1 = c_2 = \cdots = c_k = 0$.
Arrows $u_1, u_2, \ldots, u_k$ in $\R^2$ or $\R^3$ are linearly independent if no chain of them comes home except standing still: $c_1u_1 + c_2u_2 + \cdots + c_ku_k = 0$ only when every $c_i = 0$.
So dependence is a nontrivial representation of $0$, and independence says the trivial one is all there is. One bookkeeping note: a list with the same vector twice is dependent ($1u - 1u = 0$), and a set of vectors is dependent when some finite list of its distinct vectors is. In $\R^n$ an independent set is always finite (more than $n$ vectors are always dependent; see Quick Tests below), so finite lists are all this page needs.
In pictures: two arrows are dependent when they lie on one line, and three arrows are dependent when they lie in one plane, because then a chain of them can loop back home. Arrows that point in "genuinely different directions" are independent.
The step that wrote $u_3 = u_1 + 2u_2$ is worth stating properly, because it's how you'll use dependence: a relation tells you exactly which vectors can be solved for.
Two or more vectors $u_1, \ldots, u_k$ are linearly dependent exactly when one of them is a linear combination of the others. More precisely, if $c_1u_1 + \cdots + c_ku_k = 0$, then every $u_i$ with $c_i \neq 0$ is a linear combination of the others.
Suppose $c_1u_1 + \cdots + c_ku_k = 0$ and $c_i \neq 0$. Move every other term to the right side and divide by $c_i$:
$$u_i = -\frac{c_1}{c_i}u_1 - \cdots - \frac{c_{i-1}}{c_i}u_{i-1} - \frac{c_{i+1}}{c_i}u_{i+1} - \cdots - \frac{c_k}{c_i}u_k,$$a linear combination of the others. In particular, if the vectors are dependent, some $c_i$ is nonzero and that $u_i$ is a combination of the others.
Conversely, suppose $u_i = b_1u_1 + \cdots + b_{i-1}u_{i-1} + b_{i+1}u_{i+1} + \cdots + b_ku_k$. Subtracting $u_i$ from both sides gives a relation $b_1u_1 + \cdots - u_i + \cdots + b_ku_k = 0$ whose coefficient on $u_i$ is $-1 \neq 0$. So the vectors are dependent. $\blacksquare$
This is a visual argument, not a formal proof. A relation is a loop: a chain of stretched arrows that leaves home and comes back. Walk the loop and stop one leg early, and you're standing on the tip of the last arrow, so that arrow is a combination of the others. Since a loop has no real starting point, you can do this for any arrow that's actually in the loop.
An arrow with coefficient $0$ isn't in the loop at all, so the loop says nothing about it. And it works the other way: if $u_3 = u_1 + 2u_2$, then walking $u_1$, $2u_2$, and $u_3$ backward is a loop. The full proof is in the n version.
Now try both sides of it yourself, in the plane. Each mission gives you some vectors; slide the coefficients and the scaled vectors chain tip to tail from the origin. Get the chain back home to $0$ without cheating.
In the last mission $u_1$ and $u_2$ point different ways, so $c_1u_1$ and $c_2u_2$ can never cancel unless both are $0$. That's independence: no nontrivial way home. On paper, showing independence means showing the system forces every unknown to be $0$.
Show that $u_1 = (1, 1, 0)$, $u_2 = (0, 1, 1)$, $u_3 = (1, 0, 3)$ are linearly independent.
Suppose $c_1u_1 + c_2u_2 + c_3u_3 = 0$. Matching coordinates:
$$\begin{aligned} c_1 + c_3 &= 0 \\ c_1 + c_2 &= 0 \\ c_2 + 3c_3 &= 0. \end{aligned}$$The first gives $c_1 = -c_3$, and then the second gives $c_2 = -c_1 = c_3$. Put that in the third: $c_3 + 3c_3 = 4c_3 = 0$, so $c_3 = 0$, and then $c_1 = c_2 = 0$ too. The only solution is the trivial one, so the vectors are linearly independent.
In space, the three arrows span a real box (its volume is the triple product, $4$), and no chain of them comes home.
Independence depends on the scalars. Over $\mathbb{F}_2$, where $1 + 1 = 0$, the same three vectors are dependent: $(1, 1, 0) + (0, 1, 1) + (1, 0, 3) = (2, 2, 4)$, which is $(0, 0, 0)$ when every entry is read mod $2$. The step that failed is the last one: $4c_3 = 0$ no longer forces $c_3 = 0$, because $4 = 0$ in $\mathbb{F}_2$.
Here's the whole routine in one place.
The test, every time: set $c_1u_1 + \cdots + c_ku_k = 0$, match coordinates, and solve. A free unknown left over means a nonzero solution, so the vectors are dependent. Everything forced to $0$ means independent. Don't decide by eyeballing: "none of these is a multiple of another" is not the definition, and it isn't enough ($(1, 3, -1)$, $(2, -1, 2)$, $(5, 1, 3)$ above are dependent with no two multiples). And the all-zeros solution is always there, so finding it proves nothing.
Your turn. In free response, type the coefficients of a relation in order, separated by commas, like 1, -2, 1 (any nonzero multiple counts), or all zeros, like 0, 0, 0, if the vectors are independent.
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Solving the system always works. But some sets give themselves away before you write a single equation, and it pays to spot them.
(a) Any list of vectors that contains the zero vector is linearly dependent.
(b) A single vector $u$ is linearly independent exactly when $u \neq 0$.
(c) Two vectors are linearly dependent exactly when one of them is a multiple of the other.
(a) Give the zero vector coefficient $1$ and every other vector coefficient $0$. The combination is $1 \cdot 0 = 0$, and the coefficient $1$ isn't zero, so it's a nontrivial representation of $0$.
(b) If $u = 0$, it's dependent by (a). If $u$ is dependent, then $cu = 0$ for some $c \neq 0$, and multiplying by $c^{-1}$ gives $u = c^{-1}(cu) = c^{-1}0 = 0$.
(c) By the theorem above, two vectors are dependent exactly when one is a linear combination of the other, and a combination of one vector is a multiple of it. $\blacksquare$
This is a visual argument, not a formal proof. The zero vector is already home, so "one step of it" is a loop. A single arrow $u \neq 0$ slides along its line as $c$ changes, and it's home only at $c = 0$. Two arrows on one line can cancel; turn one of them off the line and they can't.
The picture is in the plane, but the same thing happens for two arrows in space: either they share a line or they don't. The full proof is in the n version.
One zero vector and the whole set is dependent. Talk about a bad influence.
Which of these are linearly dependent? Don't solve anything.
(a) $(1, 2, 3)$, $(0, 0, 0)$, $(4, 1, 1)$ (b) $(2, -4, 6)$, $(-1, 2, -3)$ (c) $(3, 1)$, $(1, 3)$
(a) Dependent: it holds the zero vector, and $0(1, 2, 3) + 1(0, 0, 0) + 0(4, 1, 1) = 0$ is nontrivial.
(b) Dependent: $(2, -4, 6) = -2(-1, 2, -3)$, so $(2, -4, 6) + 2(-1, 2, -3) = 0$. A negative multiple is still a multiple.
(c) Independent: a multiple of $(3, 1)$ has first entry three times its second, and $(1, 3)$ doesn't. Two vectors, neither a multiple of the other.
For three or more vectors, no shortcut about pairs is enough: $(1, 3, -1)$, $(2, -1, 2)$, $(5, 1, 3)$ are dependent, but no one of them is a multiple of another. What does work for any number of vectors is to put them in a matrix.
Let $A$ be the matrix whose columns are the vectors $u_1, \ldots, u_k$. These are all the same statement:
(a) $u_1, \ldots, u_k$ are linearly independent;
(b) $Ax = 0$ has only the solution $x = 0$;
(c) $\operatorname{rank}(A) = k$: every column of $A$'s row echelon form has a pivot.
By the column picture, $Ax = x_1u_1 + \cdots + x_ku_k$. So the solutions $x$ of $Ax = 0$ are exactly the coefficient lists of relations among the columns, and (a) and (b) say the same thing: the trivial relation is the only one.
For (b) and (c): a homogeneous system has a solution other than $0$ exactly when it has a free unknown, and the free unknowns are the columns of the row echelon form without a pivot (Systems). So $x = 0$ is the only solution exactly when all $k$ columns have pivots, that is, when $\operatorname{rank}(A) = k$ (Rank & Inverses). $\blacksquare$
This is a visual argument, not a formal proof. $Ax$ is a chain of $A$'s columns: $x_1$ of the first, $x_2$ of the second, $x_3$ of the third. So "$Ax = 0$ for some $x \neq 0$" says a chain of the columns comes home, which is a relation. The entries of $Ax$ are the left sides of the system, row by row, which is why solving the system finds the relation.
The pivot count is the same story told by elimination: a column without a pivot is a free unknown, and a free unknown is a nonzero way home. The full proof is in the n version.
For $n$ vectors in $\R^n$, $A$ is square, and this joins the list in Rank & Inverses: the columns of a square matrix are independent exactly when the matrix is invertible. That theorem also says something about too many vectors.
For three arrows in $\R^3$, $A$ is $3 \times 3$, and this joins the list in Rank & Inverses: the columns are independent exactly when $A$ flattens nothing. And it settles what happens with too many arrows.
Any list of more than $n$ vectors in $\R^n$ is linearly dependent.Three or more arrows in $\R^2$ are always linearly dependent, and so are four or more arrows in $\R^3$.
Let $u_1, \ldots, u_k$ be vectors in $\R^n$ with $k > n$. The equation $c_1u_1 + \cdots + c_ku_k = 0$ is a homogeneous system of $n$ equations (one per coordinate) in $k$ unknowns. It has fewer equations than unknowns, so it has a nontrivial solution (Systems), and that solution is a relation. $\blacksquare$
This is a visual argument, not a formal proof. If two of three arrows in the plane share a line, they're dependent already. If they don't, their grid of combinations covers the whole plane, so the third arrow lands on some grid point, and walking there and back is a loop.
In space it's the same with one more arrow: three arrows that don't lie in one plane fill all of $\R^3$, so a fourth lands among them, and if they do lie in one plane, they're dependent already. The picture leans on "two arrows that don't share a line fill the plane," which the Span page showed. The full proof is in the n version.
That theorem doesn't say which vectors are extra. For that, solve the system and read the relations off.
The five vectors $s_1 = (1, 0, 0)$, $s_2 = (0, 1, 0)$, $s_3 = (1, 2, 0)$, $s_4 = (0, -1, 1)$, $s_5 = (3, 3, 0)$ in $\R^3$ must be dependent (five is more than three). Find relations among them, and a set of three of them with the same span.
Matching coordinates in $c_1s_1 + \cdots + c_5s_5 = 0$:
$$\begin{aligned} c_1 + c_3 + 3c_5 &= 0 \\ c_2 + 2c_3 - c_4 + 3c_5 &= 0 \\ c_4 &= 0. \end{aligned}$$So $c_4 = 0$, $c_2 = -2c_3 - 3c_5$, and $c_1 = -c_3 - 3c_5$, with $c_3$ and $c_5$ free. Two free unknowns, two relations:
$c_3 = 1$, $c_5 = 0$ gives $-s_1 - 2s_2 + s_3 = 0$, so $s_3 = s_1 + 2s_2$.
$c_3 = 0$, $c_5 = 1$ gives $-3s_1 - 3s_2 + s_5 = 0$, so $s_5 = 3s_1 + 3s_2$.
Both $s_3$ and $s_5$ are combinations of the others, so by the Span page's theorem dropping them doesn't change the span. What's left is $s_1, s_2, s_4$, and it's independent: in $c_1s_1 + c_2s_2 + c_4s_4 = 0$ the third coordinate gives $c_4 = 0$, then the first two give $c_1 = c_2 = 0$. Notice $s_4$ has coefficient $0$ in every relation: it's the one vector that was never up for removal.
In space: four of the five lie flat on the floor $z = 0$, two of those are dead weight, and the three that are left span a box.
Here's that idea with a fourth arrow you control. Three arrows that fill space are fixed; the sliders set $w = au_1 + bu_2 + cu_3$, so $au_1 + bu_2 + cu_3 - w = 0$ is the relation. Each arrow is tagged with whether the span could do without it.
Every tag follows the theorem: an arrow can go exactly when its coefficient in the relation isn't $0$. And the last mission is the too-many theorem: four arrows in $\R^3$ always have a relation.
Two shortcuts from earlier pages, for when the vectors fill a square matrix. Two vectors $(a, c)$ and $(b, d)$ in $\R^2$ are independent exactly when $ad - bc \neq 0$ (the number from the $2 \times 2$ inverse). Three vectors in $\R^3$ are independent exactly when the triple product $u \cdot (v \times w)$ isn't $0$: the box they span isn't flat.
Your turn: pick out the one independent set. The answers are sets, so this one stays multiple choice.
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What happens to dependence when you add vectors to a set, or take some away? Two of the four cases are sure things.
Let $S$ and $T$ be finite sets of vectors with $S \subseteq T$. If $S$ is linearly dependent, then so is $T$. Equivalently, if $T$ is linearly independent, then so is $S$.
A nontrivial relation among the vectors of $S$ is also a relation among the vectors of $T$: give every vector of $T$ that isn't in $S$ the coefficient $0$. Its coefficients are still not all zero, so $T$ is dependent.
The second sentence is the first one read backward (its contrapositive): if $S$ were dependent, $T$ would be too. $\blacksquare$
This is a visual argument, not a formal proof. A loop stays a loop when more arrows show up: it just doesn't use them. And arrows that couldn't make a loop can't start making one when some of them leave. The other two cases depend on which arrows come or go.
The full proof is in the n version.
| remove vectors | add vectors | |
|---|---|---|
| independent | stays independent | could go either way |
| dependent | could go either way | stays dependent |
The five-vector example showed both "either way" cells: the dependent set $s_1, \ldots, s_5$ has the dependent subset $s_1, s_2, s_3$ and the independent subset $s_1, s_2, s_4$. When adding a vector to an independent set, there's an exact rule for which way it goes.
Let $u_1, \ldots, u_k$ be linearly independent, and let $v$ be another vector. Then $u_1, \ldots, u_k, v$ are linearly independent exactly when $v \notin \operatorname{span}\{u_1, \ldots, u_k\}$.
If $v$ is in the span, then $v = b_1u_1 + \cdots + b_ku_k$, and $b_1u_1 + \cdots + b_ku_k - v = 0$ is a relation whose coefficient on $v$ is $-1 \neq 0$. So the longer list is dependent.
Conversely, suppose the longer list is dependent: $c_1u_1 + \cdots + c_ku_k + cv = 0$ with coefficients not all zero. If $c$ were $0$, this would be a nontrivial relation among $u_1, \ldots, u_k$ alone, which can't exist because they're independent. So $c \neq 0$, and solving for $v$ gives $v = -\tfrac{c_1}{c}u_1 - \cdots - \tfrac{c_k}{c}u_k$, which is in the span. $\blacksquare$
This is a visual argument, not a formal proof. Two independent arrows $u$, $v$ span a plane. If a new arrow $w$ lies on that plane, it's some $au + bv$, and walking $au + bv - w$ is a loop. If $w$ sticks out of the plane, any chain that uses $w$ ends above or below the plane, so it can't get home; the only chains that get home skip $w$, and those are chains of $u$ and $v$, which can't loop.
Adding a third arrow to a single arrow works the same way, with a line in place of the plane. The full proof is in the n version.
The key step: a relation has to use the new vector, or it would be a relation among the old ones, and those can't have any. Drag $w$ through the plane of $u$ and $v$ and watch the switch happen.
The same idea from the other direction: an independent set has no dead weight at all, and a dependent set always has some.
Two or more vectors $u_1, \ldots, u_k$ are linearly independent exactly when removing any one of them makes the span smaller.
By the Span page's theorem, removing $u_i$ leaves the span unchanged exactly when $u_i$ is in the span of the others, that is, when $u_i$ is a linear combination of the others. By the theorem above, some $u_i$ is a combination of the others exactly when the vectors are dependent. So the vectors are dependent exactly when some removal leaves the span unchanged, which is the statement read backward. $\blacksquare$
This is a visual argument, not a formal proof. Three independent arrows fill space, and taking any one away leaves only the plane of the other two. Three dependent arrows lie in one plane, and taking away the one that's a combination of the others leaves that same plane.
The full proof is in the n version.
See it in the plane. Each set below lives in $\R^2$, and the purple is the span of whatever's left. Pull vectors out and watch which ones the span actually needed.
Set 2 is the coefficient-$0$ story again: a vector whose coefficient is $0$ in the relation can't be solved for, so it isn't a spare, and taking it out shrinks the span. In Set 3 there's no relation at all, so nothing is spare.
Once you know a vector is dead weight, you can rewrite any combination without it: solve the relation for it and substitute. Here's one with no coordinates at all, just the relation. In free response, type the coefficients of what's left, in order, like -3, -2.
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Put the two halves of this page together: a set that's linearly independent and spans the space. Keep throwing out dead weight and any finite spanning set shrinks to one (the five-vector example did exactly that). It's called a basis, and it has a superpower: every vector in the space is a combination of it in one and only one way. That's the next page.
Engineers meet dependence most often as a measurement that turned out to say nothing new. Each measurement is an equation, and an equation that's a combination of the others is dead weight: elimination turns it into $0 = 0$.
A factory needs the masses of its bolts, nuts, and washers: $a$, $b$, and $c$ grams. The parts only come in sealed sample boxes, so a technician weighs three boxes (subtracting the empty box's mass). One bolt and one nut weigh $7$ grams, so $a + b = 7$. One nut and one washer weigh $9$ grams: $b + c = 9$. One bolt, two nuts, and one washer weigh $16$ grams: $a + 2b + c = 16$. Can she find the masses? If not, what box should she have weighed instead?
The rows of the system are $(1, 1, 0)$, $(0, 1, 1)$, and $(1, 2, 1)$, and the third is the sum of the first two: $(1, 1, 0) + (0, 1, 1) - (1, 2, 1) = 0$. The rows are dependent. Elimination shows what that costs:
$$\begin{aligned} a + b &= 7 \\ b + c &= 9 \\ a + 2b + c &= 16 \end{aligned} \quad \xrightarrow{R_3 \to R_3 - R_1 - R_2} \quad \begin{aligned} a + b &= 7 \\ b + c &= 9 \\ 0 &= 0. \end{aligned}$$The third weighing added nothing (its right side, $16 = 7 + 9$, only confirms the others), and $c$ is free: $(a, b, c) = (c - 2, 9 - c, c)$ for any $c$. The masses aren't pinned down.
The fix is a box whose row isn't a combination of $(1, 1, 0)$ and $(0, 1, 1)$. One bolt and one washer works: $(1, 0, 1)$ is not in their span (any $s(1, 1, 0) + t(0, 1, 1)$ with first entry $1$ and third entry $1$ has middle entry $2$, not $0$). Say it reads $a + c = 8$. Then $(a, b, c) = (c - 2, 9 - c, c)$ gives $c - 2 + c = 8$, so $c = 5$, and the masses are $a = 3$, $b = 4$, $c = 5$ grams.
The same thing happens in surveying, circuit analysis, and sensor networks: ten measurements of a system with ten unknowns pin it down only when the ten rows are independent. A good experiment design checks that before anyone picks up a scale.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Type coefficients on one line, with commas between them, like 1, -2, 1. The which-set questions stay multiple choice.
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Made it this far? Time for something with a little heat. Four vectors, one relation, and exactly one vector the span can't do without. Find the relation first, and the answer falls out of it. In free response, type the vector's number, like 2 for $u_2$.
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